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Why is const int fine for char brace init?

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11

I thought brace initialization doesn't allow narrowing. But why is int const allowed for char brace initialization?

int value1 = 12;

char c1{value1};  // error! no narrowing



const int value2 = 12;

char c2{value2};   // why is this fine?

See it on Godbolt.

c++ c++11 const narrowing

share|improve this question

edited 11 hours ago

Boann

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BK C.BK C.

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11

I thought brace initialization doesn't allow narrowing. But why is int const allowed for char brace initialization?

int value1 = 12;

char c1{value1};  // error! no narrowing



const int value2 = 12;

char c2{value2};   // why is this fine?

See it on Godbolt.

c++ c++11 const narrowing

share|improve this question

edited 11 hours ago

Boann

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asked 12 hours ago

BK C.BK C.

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add a comment | 

I thought brace initialization doesn't allow narrowing. But why is int const allowed for char brace initialization?

int value1 = 12;

char c1{value1};  // error! no narrowing



const int value2 = 12;

char c2{value2};   // why is this fine?

See it on Godbolt.

c++ c++11 const narrowing

share|improve this question

edited 11 hours ago

Boann

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asked 12 hours ago

BK C.BK C.

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int value1 = 12;

char c1{value1};  // error! no narrowing



const int value2 = 12;

char c2{value2};   // why is this fine?

share|improve this question

edited 11 hours ago

Boann

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asked 12 hours ago

BK C.BK C.

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share|improve this question

edited 11 hours ago

Boann

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asked 12 hours ago

BK C.BK C.

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edited 11 hours ago

Boann

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edited 11 hours ago

Boann

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asked 12 hours ago

BK C.BK C.

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asked 12 hours ago

BK C.BK C.

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2 Answers
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13

const int value2 = 12;

value2 is a compile-time constant. The compiler can easily (and has to) prove that the value is 12 which happens to be within the range of values representable by char.

int value1 = 12;

value1 is not a compile-time constant. The value of the variable could change at runtime.

The exact wording of the standard rule (quoting latest draft, emphasis added):

[dcl.init.list]/7

A narrowing conversion is an implicit conversion

• from an integer type or unscoped enumeration type to an integer type that cannot represent all the values of the original type, except where the source is a constant expression whose value after integral promotions will fit into the target type.

share|improve this answer

edited 11 hours ago

Max Langhof

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answered 11 hours ago

eerorikaeerorika

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add a comment | 

-4

#include <iostream>

#include <vector>





int main() {

    int value1 = 12;

    char c1{value1};  // error! no narrowing



    const int value2 = 12;

    char c2{value2};   // why is this fine?

}

Warning thrown by your program.

test.cpp: In function ‘int main()’:

test.cpp:7:19: warning: narrowing conversion of ‘value1’ from ‘int’ to ‘char’ inside { } [-Wnarrowing]

     char c1{value1};  // error! no narrowing

                   ^

Its happens because const int 12 is optimized by compiler to literal 12. While int value1 is not compile time constant(although known at compile time in this case).

Note: For c++11 or greater, contrexpr is recommended.

share|improve this answer

edited 11 hours ago

answered 11 hours ago

v78v78

1,96699 silver badges2121 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This reasoning is wrong. Compiler optimizations have not happened at the point where the compiler decides whether the code is valid or not. And you get the same error even without any optimizations. Further, value1 is definitely not opaque to compiler optimizations. It's just that value2 is a constant expression while value1 is not.
  
  – Max Langhof
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Corrected, @MaxLanghof
  
  – v78
  11 hours ago
  

  
  
  
  

add a comment | 

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13

const int value2 = 12;

value2 is a compile-time constant. The compiler can easily (and has to) prove that the value is 12 which happens to be within the range of values representable by char.

int value1 = 12;

value1 is not a compile-time constant. The value of the variable could change at runtime.

The exact wording of the standard rule (quoting latest draft, emphasis added):

[dcl.init.list]/7

A narrowing conversion is an implicit conversion

• from an integer type or unscoped enumeration type to an integer type that cannot represent all the values of the original type, except where the source is a constant expression whose value after integral promotions will fit into the target type.

share|improve this answer

edited 11 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 11 hours ago

eerorikaeerorika

105k66 gold badges8686 silver badges162162 bronze badges

add a comment | 

13

const int value2 = 12;

value2 is a compile-time constant. The compiler can easily (and has to) prove that the value is 12 which happens to be within the range of values representable by char.

int value1 = 12;

value1 is not a compile-time constant. The value of the variable could change at runtime.

The exact wording of the standard rule (quoting latest draft, emphasis added):

[dcl.init.list]/7

A narrowing conversion is an implicit conversion

• from an integer type or unscoped enumeration type to an integer type that cannot represent all the values of the original type, except where the source is a constant expression whose value after integral promotions will fit into the target type.

share|improve this answer

edited 11 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 11 hours ago

eerorikaeerorika

105k66 gold badges8686 silver badges162162 bronze badges

add a comment | 

const int value2 = 12;

value2 is a compile-time constant. The compiler can easily (and has to) prove that the value is 12 which happens to be within the range of values representable by char.

int value1 = 12;

value1 is not a compile-time constant. The value of the variable could change at runtime.

The exact wording of the standard rule (quoting latest draft, emphasis added):

[dcl.init.list]/7

A narrowing conversion is an implicit conversion

• from an integer type or unscoped enumeration type to an integer type that cannot represent all the values of the original type, except where the source is a constant expression whose value after integral promotions will fit into the target type.

share|improve this answer

edited 11 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 11 hours ago

eerorikaeerorika

105k66 gold badges8686 silver badges162162 bronze badges

share|improve this answer

edited 11 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 11 hours ago

eerorikaeerorika

105k66 gold badges8686 silver badges162162 bronze badges

edited 11 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

edited 11 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 11 hours ago

eerorikaeerorika

105k66 gold badges8686 silver badges162162 bronze badges

answered 11 hours ago

eerorikaeerorika

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-4

#include <iostream>

#include <vector>





int main() {

    int value1 = 12;

    char c1{value1};  // error! no narrowing



    const int value2 = 12;

    char c2{value2};   // why is this fine?

}

Warning thrown by your program.

test.cpp: In function ‘int main()’:

test.cpp:7:19: warning: narrowing conversion of ‘value1’ from ‘int’ to ‘char’ inside { } [-Wnarrowing]

     char c1{value1};  // error! no narrowing

                   ^

Its happens because const int 12 is optimized by compiler to literal 12. While int value1 is not compile time constant(although known at compile time in this case).

Note: For c++11 or greater, contrexpr is recommended.

share|improve this answer

edited 11 hours ago

answered 11 hours ago

v78v78

1,96699 silver badges2121 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This reasoning is wrong. Compiler optimizations have not happened at the point where the compiler decides whether the code is valid or not. And you get the same error even without any optimizations. Further, value1 is definitely not opaque to compiler optimizations. It's just that value2 is a constant expression while value1 is not.
  
  – Max Langhof
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Corrected, @MaxLanghof
  
  – v78
  11 hours ago
  

  
  
  
  

add a comment | 

-4

#include <iostream>

#include <vector>





int main() {

    int value1 = 12;

    char c1{value1};  // error! no narrowing



    const int value2 = 12;

    char c2{value2};   // why is this fine?

}

Warning thrown by your program.

test.cpp: In function ‘int main()’:

test.cpp:7:19: warning: narrowing conversion of ‘value1’ from ‘int’ to ‘char’ inside { } [-Wnarrowing]

     char c1{value1};  // error! no narrowing

                   ^

Its happens because const int 12 is optimized by compiler to literal 12. While int value1 is not compile time constant(although known at compile time in this case).

Note: For c++11 or greater, contrexpr is recommended.

share|improve this answer

edited 11 hours ago

answered 11 hours ago

v78v78

1,96699 silver badges2121 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This reasoning is wrong. Compiler optimizations have not happened at the point where the compiler decides whether the code is valid or not. And you get the same error even without any optimizations. Further, value1 is definitely not opaque to compiler optimizations. It's just that value2 is a constant expression while value1 is not.
  
  – Max Langhof
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Corrected, @MaxLanghof
  
  – v78
  11 hours ago
  

  
  
  
  

add a comment | 

#include <iostream>

#include <vector>





int main() {

    int value1 = 12;

    char c1{value1};  // error! no narrowing



    const int value2 = 12;

    char c2{value2};   // why is this fine?

}

Warning thrown by your program.

test.cpp: In function ‘int main()’:

test.cpp:7:19: warning: narrowing conversion of ‘value1’ from ‘int’ to ‘char’ inside { } [-Wnarrowing]

     char c1{value1};  // error! no narrowing

                   ^

Its happens because const int 12 is optimized by compiler to literal 12. While int value1 is not compile time constant(although known at compile time in this case).

Note: For c++11 or greater, contrexpr is recommended.

share|improve this answer

edited 11 hours ago

answered 11 hours ago

v78v78

1,96699 silver badges2121 bronze badges

#include <iostream>

#include <vector>





int main() {

    int value1 = 12;

    char c1{value1};  // error! no narrowing



    const int value2 = 12;

    char c2{value2};   // why is this fine?

}

test.cpp: In function ‘int main()’:

test.cpp:7:19: warning: narrowing conversion of ‘value1’ from ‘int’ to ‘char’ inside { } [-Wnarrowing]

     char c1{value1};  // error! no narrowing

                   ^

share|improve this answer

edited 11 hours ago

answered 11 hours ago

v78v78

1,96699 silver badges2121 bronze badges

answered 11 hours ago

v78v78

1,96699 silver badges2121 bronze badges

answered 11 hours ago

v78v78

1,96699 silver badges2121 bronze badges