To which degree does the C preprocessor regard integer literal suffixes?Standards for the C and/or C++...
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To which degree does the C preprocessor regard integer literal suffixes?
Standards for the C and/or C++ preprocessors?Purpose of Unions in C and C++Which is better option to use for dividing an integer number by 2?Why does the C preprocessor interpret the word “linux” as the constant “1”?Why does GCC generate 15-20% faster code if I optimize for size instead of speed?How should a 64-bit integer literal be represented in C++?Promotion when evaluating constant integer expressions in preprocessor directives - GCCWhat does the integer suffix J mean?Are the L and LL integer suffixes ever needed?Defending “U” suffix after Hex literalsIs there still a reason to use `int` in C++ code?
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}
Today, I stumbled over something like this:
#define FOO 2u
#if (FOO == 2)
unsigned int foo = FOO;
#endif
Regardless of why the code is as it is (let's not question the why
), I was wondering to which degree the preprocessor can even handle integer literal suffixes. I was actually surprised that it works at all.
After doing some experiments with GCC and C99 with this code ...
#include <stdio.h>
int main()
{
#if (1u == 1)
printf("1u == 1n");
#endif
#if (1u + 1l == 2ll)
printf("1u + 1l == 2lln");
#endif
#if (1ull - 2u == -1)
printf("1ull - 2u == -1n");
#endif
#if (1u - 2u == 0xFFFFFFFFFFFFFFFF)
printf("1u - 2u == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1l == 0xFFFFFFFFFFFFFFFF)
printf("-1l == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1ll == 0xFFFFFFFFFFFFFFFF)
printf("-1ll == 0xFFFFFFFFFFFFFFFFn");
#endif
}
... which just prints all the statements:
1u == 1
1u + 1l == 2ll
1ull - 2u == -1
1u - 2u == 0xFFFFFFFFFFFFFFFF
-1 == 0xFFFFFFFFFFFFFFFF
-1l == 0xFFFFFFFFFFFFFFFF
-1ll == 0xFFFFFFFFFFFFFFFF
... I guess the preprocessor simply ignores integer literal suffixes altogether and probably always does arithmetics and comparisons in the native integer size, in this case 64 bit?
So, here is the stuff I'd like to know:
- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
- Where is all that specified/documented?
I wanted to find out by myself and checked out
Wikipedia and the C standard (working paper). I found information about integer suffixes and information about the preprocessor, but none about the combination of these. Obviously, I have also googled it but didn't get any useful results.
I have seen this Stack Overflow question that clarifies where it should be specified, but yet, I couldn't find an answer for my questions.
c++ c integer c-preprocessor suffix
add a comment |
Today, I stumbled over something like this:
#define FOO 2u
#if (FOO == 2)
unsigned int foo = FOO;
#endif
Regardless of why the code is as it is (let's not question the why
), I was wondering to which degree the preprocessor can even handle integer literal suffixes. I was actually surprised that it works at all.
After doing some experiments with GCC and C99 with this code ...
#include <stdio.h>
int main()
{
#if (1u == 1)
printf("1u == 1n");
#endif
#if (1u + 1l == 2ll)
printf("1u + 1l == 2lln");
#endif
#if (1ull - 2u == -1)
printf("1ull - 2u == -1n");
#endif
#if (1u - 2u == 0xFFFFFFFFFFFFFFFF)
printf("1u - 2u == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1l == 0xFFFFFFFFFFFFFFFF)
printf("-1l == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1ll == 0xFFFFFFFFFFFFFFFF)
printf("-1ll == 0xFFFFFFFFFFFFFFFFn");
#endif
}
... which just prints all the statements:
1u == 1
1u + 1l == 2ll
1ull - 2u == -1
1u - 2u == 0xFFFFFFFFFFFFFFFF
-1 == 0xFFFFFFFFFFFFFFFF
-1l == 0xFFFFFFFFFFFFFFFF
-1ll == 0xFFFFFFFFFFFFFFFF
... I guess the preprocessor simply ignores integer literal suffixes altogether and probably always does arithmetics and comparisons in the native integer size, in this case 64 bit?
So, here is the stuff I'd like to know:
- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
- Where is all that specified/documented?
I wanted to find out by myself and checked out
Wikipedia and the C standard (working paper). I found information about integer suffixes and information about the preprocessor, but none about the combination of these. Obviously, I have also googled it but didn't get any useful results.
I have seen this Stack Overflow question that clarifies where it should be specified, but yet, I couldn't find an answer for my questions.
c++ c integer c-preprocessor suffix
8
C11 §6.10.1 Conditional inclusion ¶4: … For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the typesintmax_t
anduintmax_t
defined in the header<stdint.h>
. …
– Jonathan Leffler
9 hours ago
add a comment |
Today, I stumbled over something like this:
#define FOO 2u
#if (FOO == 2)
unsigned int foo = FOO;
#endif
Regardless of why the code is as it is (let's not question the why
), I was wondering to which degree the preprocessor can even handle integer literal suffixes. I was actually surprised that it works at all.
After doing some experiments with GCC and C99 with this code ...
#include <stdio.h>
int main()
{
#if (1u == 1)
printf("1u == 1n");
#endif
#if (1u + 1l == 2ll)
printf("1u + 1l == 2lln");
#endif
#if (1ull - 2u == -1)
printf("1ull - 2u == -1n");
#endif
#if (1u - 2u == 0xFFFFFFFFFFFFFFFF)
printf("1u - 2u == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1l == 0xFFFFFFFFFFFFFFFF)
printf("-1l == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1ll == 0xFFFFFFFFFFFFFFFF)
printf("-1ll == 0xFFFFFFFFFFFFFFFFn");
#endif
}
... which just prints all the statements:
1u == 1
1u + 1l == 2ll
1ull - 2u == -1
1u - 2u == 0xFFFFFFFFFFFFFFFF
-1 == 0xFFFFFFFFFFFFFFFF
-1l == 0xFFFFFFFFFFFFFFFF
-1ll == 0xFFFFFFFFFFFFFFFF
... I guess the preprocessor simply ignores integer literal suffixes altogether and probably always does arithmetics and comparisons in the native integer size, in this case 64 bit?
So, here is the stuff I'd like to know:
- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
- Where is all that specified/documented?
I wanted to find out by myself and checked out
Wikipedia and the C standard (working paper). I found information about integer suffixes and information about the preprocessor, but none about the combination of these. Obviously, I have also googled it but didn't get any useful results.
I have seen this Stack Overflow question that clarifies where it should be specified, but yet, I couldn't find an answer for my questions.
c++ c integer c-preprocessor suffix
Today, I stumbled over something like this:
#define FOO 2u
#if (FOO == 2)
unsigned int foo = FOO;
#endif
Regardless of why the code is as it is (let's not question the why
), I was wondering to which degree the preprocessor can even handle integer literal suffixes. I was actually surprised that it works at all.
After doing some experiments with GCC and C99 with this code ...
#include <stdio.h>
int main()
{
#if (1u == 1)
printf("1u == 1n");
#endif
#if (1u + 1l == 2ll)
printf("1u + 1l == 2lln");
#endif
#if (1ull - 2u == -1)
printf("1ull - 2u == -1n");
#endif
#if (1u - 2u == 0xFFFFFFFFFFFFFFFF)
printf("1u - 2u == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1l == 0xFFFFFFFFFFFFFFFF)
printf("-1l == 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1ll == 0xFFFFFFFFFFFFFFFF)
printf("-1ll == 0xFFFFFFFFFFFFFFFFn");
#endif
}
... which just prints all the statements:
1u == 1
1u + 1l == 2ll
1ull - 2u == -1
1u - 2u == 0xFFFFFFFFFFFFFFFF
-1 == 0xFFFFFFFFFFFFFFFF
-1l == 0xFFFFFFFFFFFFFFFF
-1ll == 0xFFFFFFFFFFFFFFFF
... I guess the preprocessor simply ignores integer literal suffixes altogether and probably always does arithmetics and comparisons in the native integer size, in this case 64 bit?
So, here is the stuff I'd like to know:
- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
- Where is all that specified/documented?
I wanted to find out by myself and checked out
Wikipedia and the C standard (working paper). I found information about integer suffixes and information about the preprocessor, but none about the combination of these. Obviously, I have also googled it but didn't get any useful results.
I have seen this Stack Overflow question that clarifies where it should be specified, but yet, I couldn't find an answer for my questions.
c++ c integer c-preprocessor suffix
c++ c integer c-preprocessor suffix
edited 7 mins ago
Peter Mortensen
14.1k1988114
14.1k1988114
asked 9 hours ago
MichiMichi
414317
414317
8
C11 §6.10.1 Conditional inclusion ¶4: … For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the typesintmax_t
anduintmax_t
defined in the header<stdint.h>
. …
– Jonathan Leffler
9 hours ago
add a comment |
8
C11 §6.10.1 Conditional inclusion ¶4: … For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the typesintmax_t
anduintmax_t
defined in the header<stdint.h>
. …
– Jonathan Leffler
9 hours ago
8
8
C11 §6.10.1 Conditional inclusion ¶4: … For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the types
intmax_t
and uintmax_t
defined in the header <stdint.h>
. …– Jonathan Leffler
9 hours ago
C11 §6.10.1 Conditional inclusion ¶4: … For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the types
intmax_t
and uintmax_t
defined in the header <stdint.h>
. …– Jonathan Leffler
9 hours ago
add a comment |
4 Answers
4
active
oldest
votes
- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
The type suffixes of integer constants are not inherently meaningful to the preprocessor, but they are an inherent part of the corresponding preprocessing tokens, not separate. The standard has this to say about them:
A preprocessing number begins with a digit optionally preceded by a
period (.) and may be followed by valid identifier characters and the
character sequences e+, e-, E+, E-, p+, p-, P+, or P-.
Preprocessing number tokens lexically include all floating and
integer constant tokens.
(C11 6.4.8/2-3; emphasis added)
For the most part, the preprocessor doesn't treat preprocessing tokens of this type any differently than any other. The exception is in the controlling expressions of #if
directives, which are evaluated by performing macro expansion, replacing identifiers with 0, and then converting each preprocessing token into a token before evaluating the result according to C rules. Converting to tokens accounts for the type suffixes, yielding bona fide integer constants.
This does not necessarily produce results identical to those you would get from runtime evaluation of the same expressions, however, because
For the purposes of this token conversion and evaluation, all signed
integer types and all unsigned integer types act as if they have the
same representation as, respectively, the typesintmax_t
anduintmax_t
.
(C2011, 6.10.1/4)
You go on to ask
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
The only direct dependency is the implementation's definitions of intmax_t
and uintmax_t
. These are not directly tied to language choice or machine architecture, though there may be correlations with those.
- Where is all that specified/documented?
In the respective languages' language specifications, of course. I've cited the two of the more relevant sections of the C11 specification, and linked you to a late draft of that standard. (The current C is C18, but it hasn't changed in any of these regards.)
1
The suffixes of integer constants are meaningful to the preprocessor because they affect signedness.-1u < 0
evaluates to false, but-1l < 0
evaluates to true.
– Eric Postpischil
8 hours ago
Yes, @EricPostpischil, they affect evaluation of the control expressions of#if
directives. I'm perhaps splitting hairs too finely here, but although that's meaningful to C, I do not consider it meaningful to the preprocessor.
– John Bollinger
8 hours ago
they affect evaluation of the control expressions of #if and I do not consider it meaningful to the preprocessor seem incongruous.
– ryyker
8 hours ago
Clearly the meanings of-1u < 0
and-1l < 0
in a#if
statement are different, since they evaluate to different results, and the selection of which statements to include thus differs. And the only difference is the suffix. Therefore, the suffix is meaningful to preprocessing.
– Eric Postpischil
8 hours ago
1
As I said, I am splitting hairs rather finely. In particular, I am placing the evaluation of the expression outside the scope of the preprocessor proper, on account of (1) it operating on tokens, not preprocessing tokens, and (2) it being governed by the regular C rules for evaluation, as opposed to special rules pertaining to the preprocessor. This is appropos to the question, which expresses surprise about the C syntax for integer constants being recognized and handled. Yes it is, because no other is defined for the purpose.
– John Bollinger
7 hours ago
|
show 1 more comment
As I noted in a comment, this is defined in the C standard. Here's the complete text of §6.10.1 ¶4 (and the two footnotes):
C11 §6.10.1 Conditional inclusion
…
¶4 Prior to evaluation, macro invocations in the list of preprocessing tokens that will become the controlling constant expression are replaced (except for those macro names modified by the
defined
unary operator), just as in normal text. If the token defined is generated as a result of this replacement process or use of thedefined
unary operator does not match one of the two specified forms prior to macro replacement, the behavior is undefined. After all replacements due to macro expansion and thedefined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number 0, and then each preprocessing token is converted into a token. The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the typesintmax_t
anduintmax_t
defined in the header<stdint.h>
.167) This includes interpreting character constants, which may involve converting escape sequences into execution character set members. Whether the numeric value for these character constants matches the value obtained when an identical character constant occurs in an expression (other than within a#if
or#elif
directive) is implementation-defined.168) Also, whether a single-character character constant may have a negative value is implementation-defined.
167 167) Thus, on an implementation where
INT_MAX
is0x7FFF
andUINT_MAX
is0xFFFF
, the constant0x8000
is signed and positive within a#if
expression even though it would be unsigned in translation phase 7.
168 Thus, the constant expression in the following
#if
directive andif
statement is not guaranteed to evaluate to the same value in these two contexts.
#if 'z' - 'a' == 25
if ('z' - 'a' == 25)
Section 6.6 is §6.6 Constant expressions, which details the differences between the full expressions in section §6.5 Expressions and constant expressions.
In effect, the preprocessor largely ignores the suffixes. Hexadecimal constants are unsigned. The results you show are to be expected on a machine where intmax_t
and uintmax_t
are 64-bit quantities. If the limits on the intmax_t
and uintmax_t
were larger, some of the expressions might change.
1
The preprocessor cannot ignore the suffixes. It might ignore their width effects, but not the signedness.
– Eric Postpischil
8 hours ago
add a comment |
C 2018 6.10.1 deals with conditional inclusion (#if
and related statements and the defined
operator). Paragraph 1 says:
The expression that controls conditional inclusion shall be an integer constant expression except that: identifiers (including those lexically identical to keywords) are interpreted as described below; and it may contain unary operator expressions of the form
defined
identifier
or
defined
(
identifier)
…
Integer constant expression is defined in 6.6 6:
An integer constant expression shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants,
sizeof
expressions whose results are integer constants,_Alignof
expressions, and floating constants that are the immediate operands of casts. Cast operators in an integer constant expression shall only convert arithmetic types to integer types, except as part of an operand to thesizeof
or_Alignof
operator.
That paragraph is for C generally, not just the preprocessor. So the expressions that can appear in #if
statements are the same as the integer constant expressions that can appear generally in C. However, as stated in the quote above, sizeof
and _Alignof
are just identifiers; they are not recognized as C operators. In particular, 6.10.1 4 tells us:
… After all replacements due to macro expansion and the
defined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number0
,…
So, where sizeof
or _Alignof
appear in a #if
expression, it becomes 0
. Thus, a #if
expression can only have operands that are constants and defined
expressions.
Paragraph 4 goes on to say:
… The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the types
intmax_t
anduintmax_t
defined in the header<stdint.h>
.…
6.6 is the section for constant expressions.
So, the compiler will accept integer suffixes in #if
expressions, and that does not depend on the C implementation (for the suffixes required in the core C language; implementations could allow extensions). However, all the arithmetic will be performed using intmax_t
or uintmax_t
, and those do depend on the implementation. If your expressions do not depend on the width of integers above the minimum required1, they should be evaluated the same in any C implementation.
Additionally, paragraph 4 goes on to say there may be some variations with character constants and values, which I omit here as it is not relevant to this question.
Footnote
1intmax_t
designates a signed type capable of representing any value of any signed integer type (7.20.1.5 1), and long long int
is a signed type that must be at least 64 bits (5.2.4.2.1 1), so any conforming C implementation must provide 64-bit integer arithmetic in the preprocessor.
About So the expressions that can appear in #if statements are the same as the integer constant expressions that can appear generally in C.: I guess not quite. A footnote in chapter "Constant expressions" reads Further constraints that apply to the integer constant expressions used in conditional-inclusion preprocessing directives are discussed in 6.10.1. and then in a footnote of 6.10.1: Because the controlling constant expression is evaluated during translation phase 4, all identifiers either are or are not macro names — there simply are no keywords, enumeration constants, etc.
– Michi
6 hours ago
@Michi: That is discussed starting with the sentence immediately after where you stopped the quote, beginning “However” and concluding “Thus, a#if
expression can only have operands that are constants and defined expressions.”.
– Eric Postpischil
2 hours ago
add a comment |
TLDR dumbed down version:
l
and ll
are effectively (not literally!) ignored by the preprocessor conditionals (basically, everything is treated as if it had a ll
suffix), however u
is considered (normally, as for every C integer constant)!
After reading all the marvelous answers, I created some more examples that reveal some expected but yet interesting behavior:
#include <stdio.h>
int main()
{
#if (1 - 2u > 0) // If one operand is unsigned, the result is unsigned.
// Usual implicit type conversion.
printf("1 - 2u > 0n");
#endif
#if (0 < 0xFFFFFFFFFFFFFFFF)
printf("0 < 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 < 0)
printf("-1 < 0n");
#endif
#if (-1 < 0xFFFFFFFFFFFFFFFF)
printf("-1 < 0xFFFFFFFFFFFFFFFFn"); // nope
#elif (-1 > 0xFFFFFFFFFFFFFFFF)
printf("-1 > 0xFFFFFFFFFFFFFFFFn"); // nope, obviously
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFF (!!!)n");
#endif
}
With this output:
1 - 2u > 0
0 < 0xFFFFFFFFFFFFFFFF
-1 < 0
-1 == 0xFFFFFFFFFFFFFFFF (!!!)
add a comment |
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4 Answers
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4 Answers
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- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
The type suffixes of integer constants are not inherently meaningful to the preprocessor, but they are an inherent part of the corresponding preprocessing tokens, not separate. The standard has this to say about them:
A preprocessing number begins with a digit optionally preceded by a
period (.) and may be followed by valid identifier characters and the
character sequences e+, e-, E+, E-, p+, p-, P+, or P-.
Preprocessing number tokens lexically include all floating and
integer constant tokens.
(C11 6.4.8/2-3; emphasis added)
For the most part, the preprocessor doesn't treat preprocessing tokens of this type any differently than any other. The exception is in the controlling expressions of #if
directives, which are evaluated by performing macro expansion, replacing identifiers with 0, and then converting each preprocessing token into a token before evaluating the result according to C rules. Converting to tokens accounts for the type suffixes, yielding bona fide integer constants.
This does not necessarily produce results identical to those you would get from runtime evaluation of the same expressions, however, because
For the purposes of this token conversion and evaluation, all signed
integer types and all unsigned integer types act as if they have the
same representation as, respectively, the typesintmax_t
anduintmax_t
.
(C2011, 6.10.1/4)
You go on to ask
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
The only direct dependency is the implementation's definitions of intmax_t
and uintmax_t
. These are not directly tied to language choice or machine architecture, though there may be correlations with those.
- Where is all that specified/documented?
In the respective languages' language specifications, of course. I've cited the two of the more relevant sections of the C11 specification, and linked you to a late draft of that standard. (The current C is C18, but it hasn't changed in any of these regards.)
1
The suffixes of integer constants are meaningful to the preprocessor because they affect signedness.-1u < 0
evaluates to false, but-1l < 0
evaluates to true.
– Eric Postpischil
8 hours ago
Yes, @EricPostpischil, they affect evaluation of the control expressions of#if
directives. I'm perhaps splitting hairs too finely here, but although that's meaningful to C, I do not consider it meaningful to the preprocessor.
– John Bollinger
8 hours ago
they affect evaluation of the control expressions of #if and I do not consider it meaningful to the preprocessor seem incongruous.
– ryyker
8 hours ago
Clearly the meanings of-1u < 0
and-1l < 0
in a#if
statement are different, since they evaluate to different results, and the selection of which statements to include thus differs. And the only difference is the suffix. Therefore, the suffix is meaningful to preprocessing.
– Eric Postpischil
8 hours ago
1
As I said, I am splitting hairs rather finely. In particular, I am placing the evaluation of the expression outside the scope of the preprocessor proper, on account of (1) it operating on tokens, not preprocessing tokens, and (2) it being governed by the regular C rules for evaluation, as opposed to special rules pertaining to the preprocessor. This is appropos to the question, which expresses surprise about the C syntax for integer constants being recognized and handled. Yes it is, because no other is defined for the purpose.
– John Bollinger
7 hours ago
|
show 1 more comment
- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
The type suffixes of integer constants are not inherently meaningful to the preprocessor, but they are an inherent part of the corresponding preprocessing tokens, not separate. The standard has this to say about them:
A preprocessing number begins with a digit optionally preceded by a
period (.) and may be followed by valid identifier characters and the
character sequences e+, e-, E+, E-, p+, p-, P+, or P-.
Preprocessing number tokens lexically include all floating and
integer constant tokens.
(C11 6.4.8/2-3; emphasis added)
For the most part, the preprocessor doesn't treat preprocessing tokens of this type any differently than any other. The exception is in the controlling expressions of #if
directives, which are evaluated by performing macro expansion, replacing identifiers with 0, and then converting each preprocessing token into a token before evaluating the result according to C rules. Converting to tokens accounts for the type suffixes, yielding bona fide integer constants.
This does not necessarily produce results identical to those you would get from runtime evaluation of the same expressions, however, because
For the purposes of this token conversion and evaluation, all signed
integer types and all unsigned integer types act as if they have the
same representation as, respectively, the typesintmax_t
anduintmax_t
.
(C2011, 6.10.1/4)
You go on to ask
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
The only direct dependency is the implementation's definitions of intmax_t
and uintmax_t
. These are not directly tied to language choice or machine architecture, though there may be correlations with those.
- Where is all that specified/documented?
In the respective languages' language specifications, of course. I've cited the two of the more relevant sections of the C11 specification, and linked you to a late draft of that standard. (The current C is C18, but it hasn't changed in any of these regards.)
1
The suffixes of integer constants are meaningful to the preprocessor because they affect signedness.-1u < 0
evaluates to false, but-1l < 0
evaluates to true.
– Eric Postpischil
8 hours ago
Yes, @EricPostpischil, they affect evaluation of the control expressions of#if
directives. I'm perhaps splitting hairs too finely here, but although that's meaningful to C, I do not consider it meaningful to the preprocessor.
– John Bollinger
8 hours ago
they affect evaluation of the control expressions of #if and I do not consider it meaningful to the preprocessor seem incongruous.
– ryyker
8 hours ago
Clearly the meanings of-1u < 0
and-1l < 0
in a#if
statement are different, since they evaluate to different results, and the selection of which statements to include thus differs. And the only difference is the suffix. Therefore, the suffix is meaningful to preprocessing.
– Eric Postpischil
8 hours ago
1
As I said, I am splitting hairs rather finely. In particular, I am placing the evaluation of the expression outside the scope of the preprocessor proper, on account of (1) it operating on tokens, not preprocessing tokens, and (2) it being governed by the regular C rules for evaluation, as opposed to special rules pertaining to the preprocessor. This is appropos to the question, which expresses surprise about the C syntax for integer constants being recognized and handled. Yes it is, because no other is defined for the purpose.
– John Bollinger
7 hours ago
|
show 1 more comment
- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
The type suffixes of integer constants are not inherently meaningful to the preprocessor, but they are an inherent part of the corresponding preprocessing tokens, not separate. The standard has this to say about them:
A preprocessing number begins with a digit optionally preceded by a
period (.) and may be followed by valid identifier characters and the
character sequences e+, e-, E+, E-, p+, p-, P+, or P-.
Preprocessing number tokens lexically include all floating and
integer constant tokens.
(C11 6.4.8/2-3; emphasis added)
For the most part, the preprocessor doesn't treat preprocessing tokens of this type any differently than any other. The exception is in the controlling expressions of #if
directives, which are evaluated by performing macro expansion, replacing identifiers with 0, and then converting each preprocessing token into a token before evaluating the result according to C rules. Converting to tokens accounts for the type suffixes, yielding bona fide integer constants.
This does not necessarily produce results identical to those you would get from runtime evaluation of the same expressions, however, because
For the purposes of this token conversion and evaluation, all signed
integer types and all unsigned integer types act as if they have the
same representation as, respectively, the typesintmax_t
anduintmax_t
.
(C2011, 6.10.1/4)
You go on to ask
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
The only direct dependency is the implementation's definitions of intmax_t
and uintmax_t
. These are not directly tied to language choice or machine architecture, though there may be correlations with those.
- Where is all that specified/documented?
In the respective languages' language specifications, of course. I've cited the two of the more relevant sections of the C11 specification, and linked you to a late draft of that standard. (The current C is C18, but it hasn't changed in any of these regards.)
- To which degree does the preprocessor regard integer literal suffixes? Or does it just ignore them?
The type suffixes of integer constants are not inherently meaningful to the preprocessor, but they are an inherent part of the corresponding preprocessing tokens, not separate. The standard has this to say about them:
A preprocessing number begins with a digit optionally preceded by a
period (.) and may be followed by valid identifier characters and the
character sequences e+, e-, E+, E-, p+, p-, P+, or P-.
Preprocessing number tokens lexically include all floating and
integer constant tokens.
(C11 6.4.8/2-3; emphasis added)
For the most part, the preprocessor doesn't treat preprocessing tokens of this type any differently than any other. The exception is in the controlling expressions of #if
directives, which are evaluated by performing macro expansion, replacing identifiers with 0, and then converting each preprocessing token into a token before evaluating the result according to C rules. Converting to tokens accounts for the type suffixes, yielding bona fide integer constants.
This does not necessarily produce results identical to those you would get from runtime evaluation of the same expressions, however, because
For the purposes of this token conversion and evaluation, all signed
integer types and all unsigned integer types act as if they have the
same representation as, respectively, the typesintmax_t
anduintmax_t
.
(C2011, 6.10.1/4)
You go on to ask
- Are there any dependencies or different behaviors with different environments, e.g. different compilers, C vs. C++, 32 bit vs. 64 bit machine, etc.? I.e., what does the preprocessor's behavior depend on?
The only direct dependency is the implementation's definitions of intmax_t
and uintmax_t
. These are not directly tied to language choice or machine architecture, though there may be correlations with those.
- Where is all that specified/documented?
In the respective languages' language specifications, of course. I've cited the two of the more relevant sections of the C11 specification, and linked you to a late draft of that standard. (The current C is C18, but it hasn't changed in any of these regards.)
edited 2 mins ago
Peter Mortensen
14.1k1988114
14.1k1988114
answered 8 hours ago
John BollingerJohn Bollinger
89k84684
89k84684
1
The suffixes of integer constants are meaningful to the preprocessor because they affect signedness.-1u < 0
evaluates to false, but-1l < 0
evaluates to true.
– Eric Postpischil
8 hours ago
Yes, @EricPostpischil, they affect evaluation of the control expressions of#if
directives. I'm perhaps splitting hairs too finely here, but although that's meaningful to C, I do not consider it meaningful to the preprocessor.
– John Bollinger
8 hours ago
they affect evaluation of the control expressions of #if and I do not consider it meaningful to the preprocessor seem incongruous.
– ryyker
8 hours ago
Clearly the meanings of-1u < 0
and-1l < 0
in a#if
statement are different, since they evaluate to different results, and the selection of which statements to include thus differs. And the only difference is the suffix. Therefore, the suffix is meaningful to preprocessing.
– Eric Postpischil
8 hours ago
1
As I said, I am splitting hairs rather finely. In particular, I am placing the evaluation of the expression outside the scope of the preprocessor proper, on account of (1) it operating on tokens, not preprocessing tokens, and (2) it being governed by the regular C rules for evaluation, as opposed to special rules pertaining to the preprocessor. This is appropos to the question, which expresses surprise about the C syntax for integer constants being recognized and handled. Yes it is, because no other is defined for the purpose.
– John Bollinger
7 hours ago
|
show 1 more comment
1
The suffixes of integer constants are meaningful to the preprocessor because they affect signedness.-1u < 0
evaluates to false, but-1l < 0
evaluates to true.
– Eric Postpischil
8 hours ago
Yes, @EricPostpischil, they affect evaluation of the control expressions of#if
directives. I'm perhaps splitting hairs too finely here, but although that's meaningful to C, I do not consider it meaningful to the preprocessor.
– John Bollinger
8 hours ago
they affect evaluation of the control expressions of #if and I do not consider it meaningful to the preprocessor seem incongruous.
– ryyker
8 hours ago
Clearly the meanings of-1u < 0
and-1l < 0
in a#if
statement are different, since they evaluate to different results, and the selection of which statements to include thus differs. And the only difference is the suffix. Therefore, the suffix is meaningful to preprocessing.
– Eric Postpischil
8 hours ago
1
As I said, I am splitting hairs rather finely. In particular, I am placing the evaluation of the expression outside the scope of the preprocessor proper, on account of (1) it operating on tokens, not preprocessing tokens, and (2) it being governed by the regular C rules for evaluation, as opposed to special rules pertaining to the preprocessor. This is appropos to the question, which expresses surprise about the C syntax for integer constants being recognized and handled. Yes it is, because no other is defined for the purpose.
– John Bollinger
7 hours ago
1
1
The suffixes of integer constants are meaningful to the preprocessor because they affect signedness.
-1u < 0
evaluates to false, but -1l < 0
evaluates to true.– Eric Postpischil
8 hours ago
The suffixes of integer constants are meaningful to the preprocessor because they affect signedness.
-1u < 0
evaluates to false, but -1l < 0
evaluates to true.– Eric Postpischil
8 hours ago
Yes, @EricPostpischil, they affect evaluation of the control expressions of
#if
directives. I'm perhaps splitting hairs too finely here, but although that's meaningful to C, I do not consider it meaningful to the preprocessor.– John Bollinger
8 hours ago
Yes, @EricPostpischil, they affect evaluation of the control expressions of
#if
directives. I'm perhaps splitting hairs too finely here, but although that's meaningful to C, I do not consider it meaningful to the preprocessor.– John Bollinger
8 hours ago
they affect evaluation of the control expressions of #if and I do not consider it meaningful to the preprocessor seem incongruous.
– ryyker
8 hours ago
they affect evaluation of the control expressions of #if and I do not consider it meaningful to the preprocessor seem incongruous.
– ryyker
8 hours ago
Clearly the meanings of
-1u < 0
and -1l < 0
in a #if
statement are different, since they evaluate to different results, and the selection of which statements to include thus differs. And the only difference is the suffix. Therefore, the suffix is meaningful to preprocessing.– Eric Postpischil
8 hours ago
Clearly the meanings of
-1u < 0
and -1l < 0
in a #if
statement are different, since they evaluate to different results, and the selection of which statements to include thus differs. And the only difference is the suffix. Therefore, the suffix is meaningful to preprocessing.– Eric Postpischil
8 hours ago
1
1
As I said, I am splitting hairs rather finely. In particular, I am placing the evaluation of the expression outside the scope of the preprocessor proper, on account of (1) it operating on tokens, not preprocessing tokens, and (2) it being governed by the regular C rules for evaluation, as opposed to special rules pertaining to the preprocessor. This is appropos to the question, which expresses surprise about the C syntax for integer constants being recognized and handled. Yes it is, because no other is defined for the purpose.
– John Bollinger
7 hours ago
As I said, I am splitting hairs rather finely. In particular, I am placing the evaluation of the expression outside the scope of the preprocessor proper, on account of (1) it operating on tokens, not preprocessing tokens, and (2) it being governed by the regular C rules for evaluation, as opposed to special rules pertaining to the preprocessor. This is appropos to the question, which expresses surprise about the C syntax for integer constants being recognized and handled. Yes it is, because no other is defined for the purpose.
– John Bollinger
7 hours ago
|
show 1 more comment
As I noted in a comment, this is defined in the C standard. Here's the complete text of §6.10.1 ¶4 (and the two footnotes):
C11 §6.10.1 Conditional inclusion
…
¶4 Prior to evaluation, macro invocations in the list of preprocessing tokens that will become the controlling constant expression are replaced (except for those macro names modified by the
defined
unary operator), just as in normal text. If the token defined is generated as a result of this replacement process or use of thedefined
unary operator does not match one of the two specified forms prior to macro replacement, the behavior is undefined. After all replacements due to macro expansion and thedefined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number 0, and then each preprocessing token is converted into a token. The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the typesintmax_t
anduintmax_t
defined in the header<stdint.h>
.167) This includes interpreting character constants, which may involve converting escape sequences into execution character set members. Whether the numeric value for these character constants matches the value obtained when an identical character constant occurs in an expression (other than within a#if
or#elif
directive) is implementation-defined.168) Also, whether a single-character character constant may have a negative value is implementation-defined.
167 167) Thus, on an implementation where
INT_MAX
is0x7FFF
andUINT_MAX
is0xFFFF
, the constant0x8000
is signed and positive within a#if
expression even though it would be unsigned in translation phase 7.
168 Thus, the constant expression in the following
#if
directive andif
statement is not guaranteed to evaluate to the same value in these two contexts.
#if 'z' - 'a' == 25
if ('z' - 'a' == 25)
Section 6.6 is §6.6 Constant expressions, which details the differences between the full expressions in section §6.5 Expressions and constant expressions.
In effect, the preprocessor largely ignores the suffixes. Hexadecimal constants are unsigned. The results you show are to be expected on a machine where intmax_t
and uintmax_t
are 64-bit quantities. If the limits on the intmax_t
and uintmax_t
were larger, some of the expressions might change.
1
The preprocessor cannot ignore the suffixes. It might ignore their width effects, but not the signedness.
– Eric Postpischil
8 hours ago
add a comment |
As I noted in a comment, this is defined in the C standard. Here's the complete text of §6.10.1 ¶4 (and the two footnotes):
C11 §6.10.1 Conditional inclusion
…
¶4 Prior to evaluation, macro invocations in the list of preprocessing tokens that will become the controlling constant expression are replaced (except for those macro names modified by the
defined
unary operator), just as in normal text. If the token defined is generated as a result of this replacement process or use of thedefined
unary operator does not match one of the two specified forms prior to macro replacement, the behavior is undefined. After all replacements due to macro expansion and thedefined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number 0, and then each preprocessing token is converted into a token. The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the typesintmax_t
anduintmax_t
defined in the header<stdint.h>
.167) This includes interpreting character constants, which may involve converting escape sequences into execution character set members. Whether the numeric value for these character constants matches the value obtained when an identical character constant occurs in an expression (other than within a#if
or#elif
directive) is implementation-defined.168) Also, whether a single-character character constant may have a negative value is implementation-defined.
167 167) Thus, on an implementation where
INT_MAX
is0x7FFF
andUINT_MAX
is0xFFFF
, the constant0x8000
is signed and positive within a#if
expression even though it would be unsigned in translation phase 7.
168 Thus, the constant expression in the following
#if
directive andif
statement is not guaranteed to evaluate to the same value in these two contexts.
#if 'z' - 'a' == 25
if ('z' - 'a' == 25)
Section 6.6 is §6.6 Constant expressions, which details the differences between the full expressions in section §6.5 Expressions and constant expressions.
In effect, the preprocessor largely ignores the suffixes. Hexadecimal constants are unsigned. The results you show are to be expected on a machine where intmax_t
and uintmax_t
are 64-bit quantities. If the limits on the intmax_t
and uintmax_t
were larger, some of the expressions might change.
1
The preprocessor cannot ignore the suffixes. It might ignore their width effects, but not the signedness.
– Eric Postpischil
8 hours ago
add a comment |
As I noted in a comment, this is defined in the C standard. Here's the complete text of §6.10.1 ¶4 (and the two footnotes):
C11 §6.10.1 Conditional inclusion
…
¶4 Prior to evaluation, macro invocations in the list of preprocessing tokens that will become the controlling constant expression are replaced (except for those macro names modified by the
defined
unary operator), just as in normal text. If the token defined is generated as a result of this replacement process or use of thedefined
unary operator does not match one of the two specified forms prior to macro replacement, the behavior is undefined. After all replacements due to macro expansion and thedefined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number 0, and then each preprocessing token is converted into a token. The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the typesintmax_t
anduintmax_t
defined in the header<stdint.h>
.167) This includes interpreting character constants, which may involve converting escape sequences into execution character set members. Whether the numeric value for these character constants matches the value obtained when an identical character constant occurs in an expression (other than within a#if
or#elif
directive) is implementation-defined.168) Also, whether a single-character character constant may have a negative value is implementation-defined.
167 167) Thus, on an implementation where
INT_MAX
is0x7FFF
andUINT_MAX
is0xFFFF
, the constant0x8000
is signed and positive within a#if
expression even though it would be unsigned in translation phase 7.
168 Thus, the constant expression in the following
#if
directive andif
statement is not guaranteed to evaluate to the same value in these two contexts.
#if 'z' - 'a' == 25
if ('z' - 'a' == 25)
Section 6.6 is §6.6 Constant expressions, which details the differences between the full expressions in section §6.5 Expressions and constant expressions.
In effect, the preprocessor largely ignores the suffixes. Hexadecimal constants are unsigned. The results you show are to be expected on a machine where intmax_t
and uintmax_t
are 64-bit quantities. If the limits on the intmax_t
and uintmax_t
were larger, some of the expressions might change.
As I noted in a comment, this is defined in the C standard. Here's the complete text of §6.10.1 ¶4 (and the two footnotes):
C11 §6.10.1 Conditional inclusion
…
¶4 Prior to evaluation, macro invocations in the list of preprocessing tokens that will become the controlling constant expression are replaced (except for those macro names modified by the
defined
unary operator), just as in normal text. If the token defined is generated as a result of this replacement process or use of thedefined
unary operator does not match one of the two specified forms prior to macro replacement, the behavior is undefined. After all replacements due to macro expansion and thedefined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number 0, and then each preprocessing token is converted into a token. The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the typesintmax_t
anduintmax_t
defined in the header<stdint.h>
.167) This includes interpreting character constants, which may involve converting escape sequences into execution character set members. Whether the numeric value for these character constants matches the value obtained when an identical character constant occurs in an expression (other than within a#if
or#elif
directive) is implementation-defined.168) Also, whether a single-character character constant may have a negative value is implementation-defined.
167 167) Thus, on an implementation where
INT_MAX
is0x7FFF
andUINT_MAX
is0xFFFF
, the constant0x8000
is signed and positive within a#if
expression even though it would be unsigned in translation phase 7.
168 Thus, the constant expression in the following
#if
directive andif
statement is not guaranteed to evaluate to the same value in these two contexts.
#if 'z' - 'a' == 25
if ('z' - 'a' == 25)
Section 6.6 is §6.6 Constant expressions, which details the differences between the full expressions in section §6.5 Expressions and constant expressions.
In effect, the preprocessor largely ignores the suffixes. Hexadecimal constants are unsigned. The results you show are to be expected on a machine where intmax_t
and uintmax_t
are 64-bit quantities. If the limits on the intmax_t
and uintmax_t
were larger, some of the expressions might change.
edited 6 hours ago
answered 8 hours ago
Jonathan LefflerJonathan Leffler
580k966941049
580k966941049
1
The preprocessor cannot ignore the suffixes. It might ignore their width effects, but not the signedness.
– Eric Postpischil
8 hours ago
add a comment |
1
The preprocessor cannot ignore the suffixes. It might ignore their width effects, but not the signedness.
– Eric Postpischil
8 hours ago
1
1
The preprocessor cannot ignore the suffixes. It might ignore their width effects, but not the signedness.
– Eric Postpischil
8 hours ago
The preprocessor cannot ignore the suffixes. It might ignore their width effects, but not the signedness.
– Eric Postpischil
8 hours ago
add a comment |
C 2018 6.10.1 deals with conditional inclusion (#if
and related statements and the defined
operator). Paragraph 1 says:
The expression that controls conditional inclusion shall be an integer constant expression except that: identifiers (including those lexically identical to keywords) are interpreted as described below; and it may contain unary operator expressions of the form
defined
identifier
or
defined
(
identifier)
…
Integer constant expression is defined in 6.6 6:
An integer constant expression shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants,
sizeof
expressions whose results are integer constants,_Alignof
expressions, and floating constants that are the immediate operands of casts. Cast operators in an integer constant expression shall only convert arithmetic types to integer types, except as part of an operand to thesizeof
or_Alignof
operator.
That paragraph is for C generally, not just the preprocessor. So the expressions that can appear in #if
statements are the same as the integer constant expressions that can appear generally in C. However, as stated in the quote above, sizeof
and _Alignof
are just identifiers; they are not recognized as C operators. In particular, 6.10.1 4 tells us:
… After all replacements due to macro expansion and the
defined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number0
,…
So, where sizeof
or _Alignof
appear in a #if
expression, it becomes 0
. Thus, a #if
expression can only have operands that are constants and defined
expressions.
Paragraph 4 goes on to say:
… The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the types
intmax_t
anduintmax_t
defined in the header<stdint.h>
.…
6.6 is the section for constant expressions.
So, the compiler will accept integer suffixes in #if
expressions, and that does not depend on the C implementation (for the suffixes required in the core C language; implementations could allow extensions). However, all the arithmetic will be performed using intmax_t
or uintmax_t
, and those do depend on the implementation. If your expressions do not depend on the width of integers above the minimum required1, they should be evaluated the same in any C implementation.
Additionally, paragraph 4 goes on to say there may be some variations with character constants and values, which I omit here as it is not relevant to this question.
Footnote
1intmax_t
designates a signed type capable of representing any value of any signed integer type (7.20.1.5 1), and long long int
is a signed type that must be at least 64 bits (5.2.4.2.1 1), so any conforming C implementation must provide 64-bit integer arithmetic in the preprocessor.
About So the expressions that can appear in #if statements are the same as the integer constant expressions that can appear generally in C.: I guess not quite. A footnote in chapter "Constant expressions" reads Further constraints that apply to the integer constant expressions used in conditional-inclusion preprocessing directives are discussed in 6.10.1. and then in a footnote of 6.10.1: Because the controlling constant expression is evaluated during translation phase 4, all identifiers either are or are not macro names — there simply are no keywords, enumeration constants, etc.
– Michi
6 hours ago
@Michi: That is discussed starting with the sentence immediately after where you stopped the quote, beginning “However” and concluding “Thus, a#if
expression can only have operands that are constants and defined expressions.”.
– Eric Postpischil
2 hours ago
add a comment |
C 2018 6.10.1 deals with conditional inclusion (#if
and related statements and the defined
operator). Paragraph 1 says:
The expression that controls conditional inclusion shall be an integer constant expression except that: identifiers (including those lexically identical to keywords) are interpreted as described below; and it may contain unary operator expressions of the form
defined
identifier
or
defined
(
identifier)
…
Integer constant expression is defined in 6.6 6:
An integer constant expression shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants,
sizeof
expressions whose results are integer constants,_Alignof
expressions, and floating constants that are the immediate operands of casts. Cast operators in an integer constant expression shall only convert arithmetic types to integer types, except as part of an operand to thesizeof
or_Alignof
operator.
That paragraph is for C generally, not just the preprocessor. So the expressions that can appear in #if
statements are the same as the integer constant expressions that can appear generally in C. However, as stated in the quote above, sizeof
and _Alignof
are just identifiers; they are not recognized as C operators. In particular, 6.10.1 4 tells us:
… After all replacements due to macro expansion and the
defined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number0
,…
So, where sizeof
or _Alignof
appear in a #if
expression, it becomes 0
. Thus, a #if
expression can only have operands that are constants and defined
expressions.
Paragraph 4 goes on to say:
… The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the types
intmax_t
anduintmax_t
defined in the header<stdint.h>
.…
6.6 is the section for constant expressions.
So, the compiler will accept integer suffixes in #if
expressions, and that does not depend on the C implementation (for the suffixes required in the core C language; implementations could allow extensions). However, all the arithmetic will be performed using intmax_t
or uintmax_t
, and those do depend on the implementation. If your expressions do not depend on the width of integers above the minimum required1, they should be evaluated the same in any C implementation.
Additionally, paragraph 4 goes on to say there may be some variations with character constants and values, which I omit here as it is not relevant to this question.
Footnote
1intmax_t
designates a signed type capable of representing any value of any signed integer type (7.20.1.5 1), and long long int
is a signed type that must be at least 64 bits (5.2.4.2.1 1), so any conforming C implementation must provide 64-bit integer arithmetic in the preprocessor.
About So the expressions that can appear in #if statements are the same as the integer constant expressions that can appear generally in C.: I guess not quite. A footnote in chapter "Constant expressions" reads Further constraints that apply to the integer constant expressions used in conditional-inclusion preprocessing directives are discussed in 6.10.1. and then in a footnote of 6.10.1: Because the controlling constant expression is evaluated during translation phase 4, all identifiers either are or are not macro names — there simply are no keywords, enumeration constants, etc.
– Michi
6 hours ago
@Michi: That is discussed starting with the sentence immediately after where you stopped the quote, beginning “However” and concluding “Thus, a#if
expression can only have operands that are constants and defined expressions.”.
– Eric Postpischil
2 hours ago
add a comment |
C 2018 6.10.1 deals with conditional inclusion (#if
and related statements and the defined
operator). Paragraph 1 says:
The expression that controls conditional inclusion shall be an integer constant expression except that: identifiers (including those lexically identical to keywords) are interpreted as described below; and it may contain unary operator expressions of the form
defined
identifier
or
defined
(
identifier)
…
Integer constant expression is defined in 6.6 6:
An integer constant expression shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants,
sizeof
expressions whose results are integer constants,_Alignof
expressions, and floating constants that are the immediate operands of casts. Cast operators in an integer constant expression shall only convert arithmetic types to integer types, except as part of an operand to thesizeof
or_Alignof
operator.
That paragraph is for C generally, not just the preprocessor. So the expressions that can appear in #if
statements are the same as the integer constant expressions that can appear generally in C. However, as stated in the quote above, sizeof
and _Alignof
are just identifiers; they are not recognized as C operators. In particular, 6.10.1 4 tells us:
… After all replacements due to macro expansion and the
defined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number0
,…
So, where sizeof
or _Alignof
appear in a #if
expression, it becomes 0
. Thus, a #if
expression can only have operands that are constants and defined
expressions.
Paragraph 4 goes on to say:
… The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the types
intmax_t
anduintmax_t
defined in the header<stdint.h>
.…
6.6 is the section for constant expressions.
So, the compiler will accept integer suffixes in #if
expressions, and that does not depend on the C implementation (for the suffixes required in the core C language; implementations could allow extensions). However, all the arithmetic will be performed using intmax_t
or uintmax_t
, and those do depend on the implementation. If your expressions do not depend on the width of integers above the minimum required1, they should be evaluated the same in any C implementation.
Additionally, paragraph 4 goes on to say there may be some variations with character constants and values, which I omit here as it is not relevant to this question.
Footnote
1intmax_t
designates a signed type capable of representing any value of any signed integer type (7.20.1.5 1), and long long int
is a signed type that must be at least 64 bits (5.2.4.2.1 1), so any conforming C implementation must provide 64-bit integer arithmetic in the preprocessor.
C 2018 6.10.1 deals with conditional inclusion (#if
and related statements and the defined
operator). Paragraph 1 says:
The expression that controls conditional inclusion shall be an integer constant expression except that: identifiers (including those lexically identical to keywords) are interpreted as described below; and it may contain unary operator expressions of the form
defined
identifier
or
defined
(
identifier)
…
Integer constant expression is defined in 6.6 6:
An integer constant expression shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants,
sizeof
expressions whose results are integer constants,_Alignof
expressions, and floating constants that are the immediate operands of casts. Cast operators in an integer constant expression shall only convert arithmetic types to integer types, except as part of an operand to thesizeof
or_Alignof
operator.
That paragraph is for C generally, not just the preprocessor. So the expressions that can appear in #if
statements are the same as the integer constant expressions that can appear generally in C. However, as stated in the quote above, sizeof
and _Alignof
are just identifiers; they are not recognized as C operators. In particular, 6.10.1 4 tells us:
… After all replacements due to macro expansion and the
defined
unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number0
,…
So, where sizeof
or _Alignof
appear in a #if
expression, it becomes 0
. Thus, a #if
expression can only have operands that are constants and defined
expressions.
Paragraph 4 goes on to say:
… The resulting tokens compose the controlling constant expression which is evaluated according to the rules of 6.6. For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the types
intmax_t
anduintmax_t
defined in the header<stdint.h>
.…
6.6 is the section for constant expressions.
So, the compiler will accept integer suffixes in #if
expressions, and that does not depend on the C implementation (for the suffixes required in the core C language; implementations could allow extensions). However, all the arithmetic will be performed using intmax_t
or uintmax_t
, and those do depend on the implementation. If your expressions do not depend on the width of integers above the minimum required1, they should be evaluated the same in any C implementation.
Additionally, paragraph 4 goes on to say there may be some variations with character constants and values, which I omit here as it is not relevant to this question.
Footnote
1intmax_t
designates a signed type capable of representing any value of any signed integer type (7.20.1.5 1), and long long int
is a signed type that must be at least 64 bits (5.2.4.2.1 1), so any conforming C implementation must provide 64-bit integer arithmetic in the preprocessor.
edited 8 hours ago
answered 8 hours ago
Eric PostpischilEric Postpischil
83.8k891171
83.8k891171
About So the expressions that can appear in #if statements are the same as the integer constant expressions that can appear generally in C.: I guess not quite. A footnote in chapter "Constant expressions" reads Further constraints that apply to the integer constant expressions used in conditional-inclusion preprocessing directives are discussed in 6.10.1. and then in a footnote of 6.10.1: Because the controlling constant expression is evaluated during translation phase 4, all identifiers either are or are not macro names — there simply are no keywords, enumeration constants, etc.
– Michi
6 hours ago
@Michi: That is discussed starting with the sentence immediately after where you stopped the quote, beginning “However” and concluding “Thus, a#if
expression can only have operands that are constants and defined expressions.”.
– Eric Postpischil
2 hours ago
add a comment |
About So the expressions that can appear in #if statements are the same as the integer constant expressions that can appear generally in C.: I guess not quite. A footnote in chapter "Constant expressions" reads Further constraints that apply to the integer constant expressions used in conditional-inclusion preprocessing directives are discussed in 6.10.1. and then in a footnote of 6.10.1: Because the controlling constant expression is evaluated during translation phase 4, all identifiers either are or are not macro names — there simply are no keywords, enumeration constants, etc.
– Michi
6 hours ago
@Michi: That is discussed starting with the sentence immediately after where you stopped the quote, beginning “However” and concluding “Thus, a#if
expression can only have operands that are constants and defined expressions.”.
– Eric Postpischil
2 hours ago
About So the expressions that can appear in #if statements are the same as the integer constant expressions that can appear generally in C.: I guess not quite. A footnote in chapter "Constant expressions" reads Further constraints that apply to the integer constant expressions used in conditional-inclusion preprocessing directives are discussed in 6.10.1. and then in a footnote of 6.10.1: Because the controlling constant expression is evaluated during translation phase 4, all identifiers either are or are not macro names — there simply are no keywords, enumeration constants, etc.
– Michi
6 hours ago
About So the expressions that can appear in #if statements are the same as the integer constant expressions that can appear generally in C.: I guess not quite. A footnote in chapter "Constant expressions" reads Further constraints that apply to the integer constant expressions used in conditional-inclusion preprocessing directives are discussed in 6.10.1. and then in a footnote of 6.10.1: Because the controlling constant expression is evaluated during translation phase 4, all identifiers either are or are not macro names — there simply are no keywords, enumeration constants, etc.
– Michi
6 hours ago
@Michi: That is discussed starting with the sentence immediately after where you stopped the quote, beginning “However” and concluding “Thus, a
#if
expression can only have operands that are constants and defined expressions.”.– Eric Postpischil
2 hours ago
@Michi: That is discussed starting with the sentence immediately after where you stopped the quote, beginning “However” and concluding “Thus, a
#if
expression can only have operands that are constants and defined expressions.”.– Eric Postpischil
2 hours ago
add a comment |
TLDR dumbed down version:
l
and ll
are effectively (not literally!) ignored by the preprocessor conditionals (basically, everything is treated as if it had a ll
suffix), however u
is considered (normally, as for every C integer constant)!
After reading all the marvelous answers, I created some more examples that reveal some expected but yet interesting behavior:
#include <stdio.h>
int main()
{
#if (1 - 2u > 0) // If one operand is unsigned, the result is unsigned.
// Usual implicit type conversion.
printf("1 - 2u > 0n");
#endif
#if (0 < 0xFFFFFFFFFFFFFFFF)
printf("0 < 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 < 0)
printf("-1 < 0n");
#endif
#if (-1 < 0xFFFFFFFFFFFFFFFF)
printf("-1 < 0xFFFFFFFFFFFFFFFFn"); // nope
#elif (-1 > 0xFFFFFFFFFFFFFFFF)
printf("-1 > 0xFFFFFFFFFFFFFFFFn"); // nope, obviously
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFF (!!!)n");
#endif
}
With this output:
1 - 2u > 0
0 < 0xFFFFFFFFFFFFFFFF
-1 < 0
-1 == 0xFFFFFFFFFFFFFFFF (!!!)
add a comment |
TLDR dumbed down version:
l
and ll
are effectively (not literally!) ignored by the preprocessor conditionals (basically, everything is treated as if it had a ll
suffix), however u
is considered (normally, as for every C integer constant)!
After reading all the marvelous answers, I created some more examples that reveal some expected but yet interesting behavior:
#include <stdio.h>
int main()
{
#if (1 - 2u > 0) // If one operand is unsigned, the result is unsigned.
// Usual implicit type conversion.
printf("1 - 2u > 0n");
#endif
#if (0 < 0xFFFFFFFFFFFFFFFF)
printf("0 < 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 < 0)
printf("-1 < 0n");
#endif
#if (-1 < 0xFFFFFFFFFFFFFFFF)
printf("-1 < 0xFFFFFFFFFFFFFFFFn"); // nope
#elif (-1 > 0xFFFFFFFFFFFFFFFF)
printf("-1 > 0xFFFFFFFFFFFFFFFFn"); // nope, obviously
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFF (!!!)n");
#endif
}
With this output:
1 - 2u > 0
0 < 0xFFFFFFFFFFFFFFFF
-1 < 0
-1 == 0xFFFFFFFFFFFFFFFF (!!!)
add a comment |
TLDR dumbed down version:
l
and ll
are effectively (not literally!) ignored by the preprocessor conditionals (basically, everything is treated as if it had a ll
suffix), however u
is considered (normally, as for every C integer constant)!
After reading all the marvelous answers, I created some more examples that reveal some expected but yet interesting behavior:
#include <stdio.h>
int main()
{
#if (1 - 2u > 0) // If one operand is unsigned, the result is unsigned.
// Usual implicit type conversion.
printf("1 - 2u > 0n");
#endif
#if (0 < 0xFFFFFFFFFFFFFFFF)
printf("0 < 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 < 0)
printf("-1 < 0n");
#endif
#if (-1 < 0xFFFFFFFFFFFFFFFF)
printf("-1 < 0xFFFFFFFFFFFFFFFFn"); // nope
#elif (-1 > 0xFFFFFFFFFFFFFFFF)
printf("-1 > 0xFFFFFFFFFFFFFFFFn"); // nope, obviously
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFF (!!!)n");
#endif
}
With this output:
1 - 2u > 0
0 < 0xFFFFFFFFFFFFFFFF
-1 < 0
-1 == 0xFFFFFFFFFFFFFFFF (!!!)
TLDR dumbed down version:
l
and ll
are effectively (not literally!) ignored by the preprocessor conditionals (basically, everything is treated as if it had a ll
suffix), however u
is considered (normally, as for every C integer constant)!
After reading all the marvelous answers, I created some more examples that reveal some expected but yet interesting behavior:
#include <stdio.h>
int main()
{
#if (1 - 2u > 0) // If one operand is unsigned, the result is unsigned.
// Usual implicit type conversion.
printf("1 - 2u > 0n");
#endif
#if (0 < 0xFFFFFFFFFFFFFFFF)
printf("0 < 0xFFFFFFFFFFFFFFFFn");
#endif
#if (-1 < 0)
printf("-1 < 0n");
#endif
#if (-1 < 0xFFFFFFFFFFFFFFFF)
printf("-1 < 0xFFFFFFFFFFFFFFFFn"); // nope
#elif (-1 > 0xFFFFFFFFFFFFFFFF)
printf("-1 > 0xFFFFFFFFFFFFFFFFn"); // nope, obviously
#endif
#if (-1 == 0xFFFFFFFFFFFFFFFF)
printf("-1 == 0xFFFFFFFFFFFFFFFF (!!!)n");
#endif
}
With this output:
1 - 2u > 0
0 < 0xFFFFFFFFFFFFFFFF
-1 < 0
-1 == 0xFFFFFFFFFFFFFFFF (!!!)
answered 5 hours ago
MichiMichi
414317
414317
add a comment |
add a comment |
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8
C11 §6.10.1 Conditional inclusion ¶4: … For the purposes of this token conversion and evaluation, all signed integer types and all unsigned integer types act as if they have the same representation as, respectively, the types
intmax_t
anduintmax_t
defined in the header<stdint.h>
. …– Jonathan Leffler
9 hours ago