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Why is const int fine for char brace init?

Why can't I convert 'char**' to a 'const char* const*' in C?Why use static_cast<int>(x) instead of (int)x?How to convert a std::string to const char* or char*?What is the difference between char * const and const char *?What is the difference between const int*, const int * const, and int const *?assigning char to int reference and const int reference in C++Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsInitialize vector <char> with int valuesCannot initialize a vector of const char*/string array with an initializer-list on declaration

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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 c1value1; // error! no narrowing

const int value2 = 12;
char c2value2; // why is this fine?

See it on Godbolt.

c++ c++11 const narrowing

share|improve this question

edited 8 hours ago

Boann

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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 c1value1; // error! no narrowing

const int value2 = 12;
char c2value2; // why is this fine?

See it on Godbolt.

c++ c++11 const narrowing

share|improve this question

edited 8 hours ago

Boann

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asked 9 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 c1value1; // error! no narrowing

const int value2 = 12;
char c2value2; // why is this fine?

See it on Godbolt.

c++ c++11 const narrowing

share|improve this question

edited 8 hours ago

Boann

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asked 9 hours ago

BK C.BK C.

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int value1 = 12;
char c1value1; // error! no narrowing

const int value2 = 12;
char c2value2; // why is this fine?

share|improve this question

edited 8 hours ago

Boann

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asked 9 hours ago

BK C.BK C.

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share|improve this question

edited 8 hours ago

Boann

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asked 9 hours ago

BK C.BK C.

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edited 8 hours ago

Boann

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edited 8 hours ago

Boann

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asked 9 hours ago

BK C.BK C.

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asked 9 hours ago

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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 8 hours ago

Max Langhof

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answered 9 hours ago

eerorikaeerorika

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add a comment | 

-4

#include <iostream>
#include <vector>


int main() 
 int value1 = 12;
 char c1value1; // error! no narrowing

 const int value2 = 12;
 char c2value2; // 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 c1value1; // 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 9 hours ago

answered 9 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
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Corrected, @MaxLanghof
  
  – v78
  8 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 8 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 9 hours ago

eerorikaeerorika

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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 8 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 9 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 8 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 9 hours ago

eerorikaeerorika

105k66 gold badges8686 silver badges162162 bronze badges

share|improve this answer

edited 8 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 9 hours ago

eerorikaeerorika

105k66 gold badges8686 silver badges162162 bronze badges

edited 8 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

edited 8 hours ago

Max Langhof

15.9k33 gold badges2626 silver badges4848 bronze badges

answered 9 hours ago

eerorikaeerorika

105k66 gold badges8686 silver badges162162 bronze badges

answered 9 hours ago

eerorikaeerorika

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-4

#include <iostream>
#include <vector>


int main() 
 int value1 = 12;
 char c1value1; // error! no narrowing

 const int value2 = 12;
 char c2value2; // 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 c1value1; // 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 9 hours ago

answered 9 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
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Corrected, @MaxLanghof
  
  – v78
  8 hours ago
  

  
  
  
  

add a comment | 

-4

#include <iostream>
#include <vector>


int main() 
 int value1 = 12;
 char c1value1; // error! no narrowing

 const int value2 = 12;
 char c2value2; // 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 c1value1; // 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 9 hours ago

answered 9 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
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Corrected, @MaxLanghof
  
  – v78
  8 hours ago
  

  
  
  
  

add a comment | 

#include <iostream>
#include <vector>


int main() 
 int value1 = 12;
 char c1value1; // error! no narrowing

 const int value2 = 12;
 char c2value2; // 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 c1value1; // 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 9 hours ago

answered 9 hours ago

v78v78

1,96699 silver badges2121 bronze badges

#include <iostream>
#include <vector>


int main() 
 int value1 = 12;
 char c1value1; // error! no narrowing

 const int value2 = 12;
 char c2value2; // 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 c1value1; // error! no narrowing
 ^

share|improve this answer

edited 9 hours ago

answered 9 hours ago

v78v78

1,96699 silver badges2121 bronze badges

answered 9 hours ago

v78v78

1,96699 silver badges2121 bronze badges

answered 9 hours ago

v78v78

1,96699 silver badges2121 bronze badges