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The code below, is it ill-formed NDR or is it well formed?
Announcing the arrival of Valued Associate #679: Cesar Manara
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The Ask Question Wizard is Live!Where in C++14 Standard does it say that a non-constexpr function cannot be used in a definition of a constexpr function?Inheriting constructorsWhy does an overridden function in the derived class hide other overloads of the base class?x[0] == 1 constant expression in C++11 when x is const int[]?constexpr bug in clang but not in gcc?Rationale for [dcl.constexpr]p5 in the c++ standardWhy can't lambda, when cast to function pointer, be used in constexpr context?Ill-Formed, No Diagnostic Required (NDR): ConstExpr Function Throw in C++14constexpr reference to non-const objectWhy is this constexpr function ill-formed?Constexpr constructor fails to satisfy the requirements, but still constexpr. Why?
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}
Clang accepts the following code, but gcc rejects it.
void h() { }
constexpr int f() {
return 1;
h();
}
int main() {
constexpr int i = f();
}
Here is the error message:
g++ -std=c++17 -O2 -Wall -pedantic -pthread main.cpp && ./a.out
main.cpp: In function 'constexpr int f()':
main.cpp:5:6: error: call to non-'constexpr' function 'void h()'
h();
~^~
main.cpp: In function 'int main()':
main.cpp:9:24: error: 'constexpr int f()' called in a constant expression
constexpr int i = f();
~^~
main.cpp:9:19: warning: unused variable 'i' [-Wunused-variable]
constexpr int i = f();
This could well be the case where both compilers are correct, once we consider [dcl.constexpr]/5, given that f() is not a constant expression, as it doesn't satisfy [expr.const]/(4.2), as it calls a non-constexpr function h. That is, the code is ill-formed, but no diagnostic is required.
One other possibility is that the code is well formed, as [expr.const]/(4.2) doesn't apply in this case because the call to h in f is not evaluated. If this is the case, gcc is wrong and clang is correct.
c++ language-lawyer c++17 constexpr
edited 5 hours ago
Barry
187k21331612
asked 6 hours ago
AlexanderAlexander
915414
add a comment |
Clang accepts the following code, but gcc rejects it.
void h() { }
constexpr int f() {
return 1;
h();
}
int main() {
constexpr int i = f();
}
Here is the error message:
g++ -std=c++17 -O2 -Wall -pedantic -pthread main.cpp && ./a.out
main.cpp: In function 'constexpr int f()':
main.cpp:5:6: error: call to non-'constexpr' function 'void h()'
h();
~^~
main.cpp: In function 'int main()':
main.cpp:9:24: error: 'constexpr int f()' called in a constant expression
constexpr int i = f();
~^~
main.cpp:9:19: warning: unused variable 'i' [-Wunused-variable]
constexpr int i = f();
This could well be the case where both compilers are correct, once we consider [dcl.constexpr]/5, given that f() is not a constant expression, as it doesn't satisfy [expr.const]/(4.2), as it calls a non-constexpr function h. That is, the code is ill-formed, but no diagnostic is required.
One other possibility is that the code is well formed, as [expr.const]/(4.2) doesn't apply in this case because the call to h in f is not evaluated. If this is the case, gcc is wrong and clang is correct.
c++ language-lawyer c++17 constexpr
edited 5 hours ago
Barry
187k21331612
asked 6 hours ago
AlexanderAlexander
915414
4
Clang does not allow callingh()before returning, so the real question here is: Is a compiler allowed to ignore dead ill-formed code?
– idmean
6 hours ago
"as it calls a non-constexpr functionh". But it doesn't actually callh. I'd say that gcc is wrong here.
– geza
6 hours ago
I'm adding the C++14 tag since on C++11 there's no question that it's ill-formed.
– Barry
5 hours ago
This code works in GCC's trunk.. godbolt.org/z/f04MCq and godbolt.org/z/bAyE8a .. So it's already fixed.. Free upvotes. If compiled with GCC 8.3 it will fail but compile with Trunk and it works fine.
– Brandon
5 hours ago
add a comment |
Clang accepts the following code, but gcc rejects it.
void h() { }
constexpr int f() {
return 1;
h();
}
int main() {
constexpr int i = f();
}
Here is the error message:
g++ -std=c++17 -O2 -Wall -pedantic -pthread main.cpp && ./a.out
main.cpp: In function 'constexpr int f()':
main.cpp:5:6: error: call to non-'constexpr' function 'void h()'
h();
~^~
main.cpp: In function 'int main()':
main.cpp:9:24: error: 'constexpr int f()' called in a constant expression
constexpr int i = f();
~^~
main.cpp:9:19: warning: unused variable 'i' [-Wunused-variable]
constexpr int i = f();
This could well be the case where both compilers are correct, once we consider [dcl.constexpr]/5, given that f() is not a constant expression, as it doesn't satisfy [expr.const]/(4.2), as it calls a non-constexpr function h. That is, the code is ill-formed, but no diagnostic is required.
One other possibility is that the code is well formed, as [expr.const]/(4.2) doesn't apply in this case because the call to h in f is not evaluated. If this is the case, gcc is wrong and clang is correct.
c++ language-lawyer c++17 constexpr
edited 5 hours ago
Barry
187k21331612
asked 6 hours ago
AlexanderAlexander
915414
Clang accepts the following code, but gcc rejects it.
void h() { }
constexpr int f() {
return 1;
h();
}
int main() {
constexpr int i = f();
}
Here is the error message:
g++ -std=c++17 -O2 -Wall -pedantic -pthread main.cpp && ./a.out
main.cpp: In function 'constexpr int f()':
main.cpp:5:6: error: call to non-'constexpr' function 'void h()'
h();
~^~
main.cpp: In function 'int main()':
main.cpp:9:24: error: 'constexpr int f()' called in a constant expression
constexpr int i = f();
~^~
main.cpp:9:19: warning: unused variable 'i' [-Wunused-variable]
constexpr int i = f();
This could well be the case where both compilers are correct, once we consider [dcl.constexpr]/5, given that f() is not a constant expression, as it doesn't satisfy [expr.const]/(4.2), as it calls a non-constexpr function h. That is, the code is ill-formed, but no diagnostic is required.
One other possibility is that the code is well formed, as [expr.const]/(4.2) doesn't apply in this case because the call to h in f is not evaluated. If this is the case, gcc is wrong and clang is correct.
c++ language-lawyer c++17 constexpr
c++ language-lawyer c++17 constexpr
edited 5 hours ago
Barry
187k21331612
asked 6 hours ago
AlexanderAlexander
915414
edited 5 hours ago
Barry
187k21331612
asked 6 hours ago
AlexanderAlexander
915414
edited 5 hours ago
Barry
187k21331612
edited 5 hours ago
Barry
187k21331612
edited 5 hours ago
Barry
187k21331612
187k21331612
asked 6 hours ago
AlexanderAlexander
915414
asked 6 hours ago
AlexanderAlexander
915414
asked 6 hours ago
AlexanderAlexander
915414
915414
4
Clang does not allow callingh()before returning, so the real question here is: Is a compiler allowed to ignore dead ill-formed code?
– idmean
6 hours ago
"as it calls a non-constexpr functionh". But it doesn't actually callh. I'd say that gcc is wrong here.
– geza
6 hours ago
I'm adding the C++14 tag since on C++11 there's no question that it's ill-formed.
– Barry
5 hours ago
This code works in GCC's trunk.. godbolt.org/z/f04MCq and godbolt.org/z/bAyE8a .. So it's already fixed.. Free upvotes. If compiled with GCC 8.3 it will fail but compile with Trunk and it works fine.
– Brandon
5 hours ago
add a comment |
4
Clang does not allow callingh()before returning, so the real question here is: Is a compiler allowed to ignore dead ill-formed code?
– idmean
6 hours ago
"as it calls a non-constexpr functionh". But it doesn't actually callh. I'd say that gcc is wrong here.
– geza
6 hours ago
I'm adding the C++14 tag since on C++11 there's no question that it's ill-formed.
– Barry
5 hours ago
This code works in GCC's trunk.. godbolt.org/z/f04MCq and godbolt.org/z/bAyE8a .. So it's already fixed.. Free upvotes. If compiled with GCC 8.3 it will fail but compile with Trunk and it works fine.
– Brandon
5 hours ago
4
4
Clang does not allow calling
h() before returning, so the real question here is: Is a compiler allowed to ignore dead ill-formed code?– idmean
6 hours ago
Clang does not allow calling
h() before returning, so the real question here is: Is a compiler allowed to ignore dead ill-formed code?– idmean
6 hours ago
"as it calls a non-constexpr function
h". But it doesn't actually call h. I'd say that gcc is wrong here.– geza
6 hours ago
"as it calls a non-constexpr function
h". But it doesn't actually call h. I'd say that gcc is wrong here.– geza
6 hours ago
I'm adding the C++14 tag since on C++11 there's no question that it's ill-formed.
– Barry
5 hours ago
I'm adding the C++14 tag since on C++11 there's no question that it's ill-formed.
– Barry
5 hours ago
This code works in GCC's trunk.. godbolt.org/z/f04MCq and godbolt.org/z/bAyE8a .. So it's already fixed.. Free upvotes. If compiled with GCC 8.3 it will fail but compile with Trunk and it works fine.
– Brandon
5 hours ago
This code works in GCC's trunk.. godbolt.org/z/f04MCq and godbolt.org/z/bAyE8a .. So it's already fixed.. Free upvotes. If compiled with GCC 8.3 it will fail but compile with Trunk and it works fine.
– Brandon
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
Clang is correct. A call to f() is a constant expression since the call to h() is never evaluated, so [dcl.constexpr]/5 doesn't apply. The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions. Indeed, a function like the following is well-formed because a call to it can be a constant expression when x is odd:
constexpr int g(int x) {
if (x%2 == 0) h();
return 0;
}
answered 6 hours ago
BrianBrian
67k799192
You answer seems to be correct, but I don't agree with your reasoning. Consider this:constexpr int f() { h(); return 1; }. In this case the code is ill-formed NDR according to [dcl.constexpr]/5 because the functionfinvokes a non-constexpr functionh, and this is not allowed according to [expr.const]/(4.2). See also this question in SO.
– Alexander
4 hours ago
@Alexander In the case where the call toh()is before the return statement, every call tofwill fail to be a constant expression, and [dcl.constexpr]/5 applies. In the case where the call toh()is after the return statement, every call tofwill be a constant expression, so [dcl.constexpr]/5 does not apply.
– Brian
4 hours ago
Ok, but why did you say "The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions."? This is what is confusing in your answer.
– Alexander
4 hours ago
1
@Alexander I'm not sure what part of that was unclear. I linked a section of the standard that lists the constructs that are forbidden from appearing in aconstexprfunction's body. A call to a non-constexprfunction is not one of the forbidden constructs. However, if the call to the non-constexprfunction becomes inevitable (i.e., occurs along all paths) then [dcl.constexpr]/5 becomes violated.
– Brian
4 hours ago
add a comment |
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Clang is correct. A call to f() is a constant expression since the call to h() is never evaluated, so [dcl.constexpr]/5 doesn't apply. The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions. Indeed, a function like the following is well-formed because a call to it can be a constant expression when x is odd:
constexpr int g(int x) {
if (x%2 == 0) h();
return 0;
}
answered 6 hours ago
BrianBrian
67k799192
You answer seems to be correct, but I don't agree with your reasoning. Consider this:constexpr int f() { h(); return 1; }. In this case the code is ill-formed NDR according to [dcl.constexpr]/5 because the functionfinvokes a non-constexpr functionh, and this is not allowed according to [expr.const]/(4.2). See also this question in SO.
– Alexander
4 hours ago
@Alexander In the case where the call toh()is before the return statement, every call tofwill fail to be a constant expression, and [dcl.constexpr]/5 applies. In the case where the call toh()is after the return statement, every call tofwill be a constant expression, so [dcl.constexpr]/5 does not apply.
– Brian
4 hours ago
Ok, but why did you say "The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions."? This is what is confusing in your answer.
– Alexander
4 hours ago
1
@Alexander I'm not sure what part of that was unclear. I linked a section of the standard that lists the constructs that are forbidden from appearing in aconstexprfunction's body. A call to a non-constexprfunction is not one of the forbidden constructs. However, if the call to the non-constexprfunction becomes inevitable (i.e., occurs along all paths) then [dcl.constexpr]/5 becomes violated.
– Brian
4 hours ago
add a comment |
Clang is correct. A call to f() is a constant expression since the call to h() is never evaluated, so [dcl.constexpr]/5 doesn't apply. The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions. Indeed, a function like the following is well-formed because a call to it can be a constant expression when x is odd:
constexpr int g(int x) {
if (x%2 == 0) h();
return 0;
}
answered 6 hours ago
BrianBrian
67k799192
You answer seems to be correct, but I don't agree with your reasoning. Consider this:constexpr int f() { h(); return 1; }. In this case the code is ill-formed NDR according to [dcl.constexpr]/5 because the functionfinvokes a non-constexpr functionh, and this is not allowed according to [expr.const]/(4.2). See also this question in SO.
– Alexander
4 hours ago
@Alexander In the case where the call toh()is before the return statement, every call tofwill fail to be a constant expression, and [dcl.constexpr]/5 applies. In the case where the call toh()is after the return statement, every call tofwill be a constant expression, so [dcl.constexpr]/5 does not apply.
– Brian
4 hours ago
Ok, but why did you say "The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions."? This is what is confusing in your answer.
– Alexander
4 hours ago
1
@Alexander I'm not sure what part of that was unclear. I linked a section of the standard that lists the constructs that are forbidden from appearing in aconstexprfunction's body. A call to a non-constexprfunction is not one of the forbidden constructs. However, if the call to the non-constexprfunction becomes inevitable (i.e., occurs along all paths) then [dcl.constexpr]/5 becomes violated.
– Brian
4 hours ago
add a comment |
Clang is correct. A call to f() is a constant expression since the call to h() is never evaluated, so [dcl.constexpr]/5 doesn't apply. The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions. Indeed, a function like the following is well-formed because a call to it can be a constant expression when x is odd:
constexpr int g(int x) {
if (x%2 == 0) h();
return 0;
}
answered 6 hours ago
BrianBrian
67k799192
Clang is correct. A call to f() is a constant expression since the call to h() is never evaluated, so [dcl.constexpr]/5 doesn't apply. The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions. Indeed, a function like the following is well-formed because a call to it can be a constant expression when x is odd:
constexpr int g(int x) {
if (x%2 == 0) h();
return 0;
}
answered 6 hours ago
BrianBrian
67k799192
answered 6 hours ago
BrianBrian
67k799192
answered 6 hours ago
BrianBrian
67k799192
answered 6 hours ago
BrianBrian
67k799192
67k799192
You answer seems to be correct, but I don't agree with your reasoning. Consider this:constexpr int f() { h(); return 1; }. In this case the code is ill-formed NDR according to [dcl.constexpr]/5 because the functionfinvokes a non-constexpr functionh, and this is not allowed according to [expr.const]/(4.2). See also this question in SO.
– Alexander
4 hours ago
@Alexander In the case where the call toh()is before the return statement, every call tofwill fail to be a constant expression, and [dcl.constexpr]/5 applies. In the case where the call toh()is after the return statement, every call tofwill be a constant expression, so [dcl.constexpr]/5 does not apply.
– Brian
4 hours ago
Ok, but why did you say "The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions."? This is what is confusing in your answer.
– Alexander
4 hours ago
1
@Alexander I'm not sure what part of that was unclear. I linked a section of the standard that lists the constructs that are forbidden from appearing in aconstexprfunction's body. A call to a non-constexprfunction is not one of the forbidden constructs. However, if the call to the non-constexprfunction becomes inevitable (i.e., occurs along all paths) then [dcl.constexpr]/5 becomes violated.
– Brian
4 hours ago
add a comment |
You answer seems to be correct, but I don't agree with your reasoning. Consider this:constexpr int f() { h(); return 1; }. In this case the code is ill-formed NDR according to [dcl.constexpr]/5 because the functionfinvokes a non-constexpr functionh, and this is not allowed according to [expr.const]/(4.2). See also this question in SO.
– Alexander
4 hours ago
@Alexander In the case where the call toh()is before the return statement, every call tofwill fail to be a constant expression, and [dcl.constexpr]/5 applies. In the case where the call toh()is after the return statement, every call tofwill be a constant expression, so [dcl.constexpr]/5 does not apply.
– Brian
4 hours ago
Ok, but why did you say "The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions."? This is what is confusing in your answer.
– Alexander
4 hours ago
1
@Alexander I'm not sure what part of that was unclear. I linked a section of the standard that lists the constructs that are forbidden from appearing in aconstexprfunction's body. A call to a non-constexprfunction is not one of the forbidden constructs. However, if the call to the non-constexprfunction becomes inevitable (i.e., occurs along all paths) then [dcl.constexpr]/5 becomes violated.
– Brian
4 hours ago
You answer seems to be correct, but I don't agree with your reasoning. Consider this:
constexpr int f() { h(); return 1; }. In this case the code is ill-formed NDR according to [dcl.constexpr]/5 because the function f invokes a non-constexpr function h, and this is not allowed according to [expr.const]/(4.2). See also this question in SO.– Alexander
4 hours ago
You answer seems to be correct, but I don't agree with your reasoning. Consider this:
constexpr int f() { h(); return 1; }. In this case the code is ill-formed NDR according to [dcl.constexpr]/5 because the function f invokes a non-constexpr function h, and this is not allowed according to [expr.const]/(4.2). See also this question in SO.– Alexander
4 hours ago
@Alexander In the case where the call to
h() is before the return statement, every call to f will fail to be a constant expression, and [dcl.constexpr]/5 applies. In the case where the call to h() is after the return statement, every call to f will be a constant expression, so [dcl.constexpr]/5 does not apply.– Brian
4 hours ago
@Alexander In the case where the call to
h() is before the return statement, every call to f will fail to be a constant expression, and [dcl.constexpr]/5 applies. In the case where the call to h() is after the return statement, every call to f will be a constant expression, so [dcl.constexpr]/5 does not apply.– Brian
4 hours ago
Ok, but why did you say "The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions."? This is what is confusing in your answer.
– Alexander
4 hours ago
Ok, but why did you say "The call to h() in the body of f() is not ill-formed because the constraints on constexpr functions don't say anything about not being allowed to call non-constexpr functions."? This is what is confusing in your answer.
– Alexander
4 hours ago
1
1
@Alexander I'm not sure what part of that was unclear. I linked a section of the standard that lists the constructs that are forbidden from appearing in a
constexpr function's body. A call to a non-constexpr function is not one of the forbidden constructs. However, if the call to the non-constexpr function becomes inevitable (i.e., occurs along all paths) then [dcl.constexpr]/5 becomes violated.– Brian
4 hours ago
@Alexander I'm not sure what part of that was unclear. I linked a section of the standard that lists the constructs that are forbidden from appearing in a
constexpr function's body. A call to a non-constexpr function is not one of the forbidden constructs. However, if the call to the non-constexpr function becomes inevitable (i.e., occurs along all paths) then [dcl.constexpr]/5 becomes violated.– Brian
4 hours ago
add a comment |
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4
Clang does not allow calling
h()before returning, so the real question here is: Is a compiler allowed to ignore dead ill-formed code?– idmean
6 hours ago
"as it calls a non-constexpr function
h". But it doesn't actually callh. I'd say that gcc is wrong here.– geza
6 hours ago
I'm adding the C++14 tag since on C++11 there's no question that it's ill-formed.
– Barry
5 hours ago
This code works in GCC's trunk.. godbolt.org/z/f04MCq and godbolt.org/z/bAyE8a .. So it's already fixed.. Free upvotes. If compiled with GCC 8.3 it will fail but compile with Trunk and it works fine.
– Brandon
5 hours ago