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The code below, is it ill-formed NDR or is it well formed?

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The code below, is it ill-formed NDR or is it well formed?

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13

1

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

share|improve this question

edited 1 hour ago

Barry

187k21331612

asked 1 hour ago

AlexanderAlexander

895414

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm adding the C++14 tag since on C++11 there's no question that it's ill-formed.
  
  – Barry
  1 hour 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
  1 hour ago
  
  

  
  
  
  

add a comment | 

13

1

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

share|improve this question

edited 1 hour ago

Barry

187k21331612

asked 1 hour ago

AlexanderAlexander

895414

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm adding the C++14 tag since on C++11 there's no question that it's ill-formed.
  
  – Barry
  1 hour 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
  1 hour 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

share|improve this question

edited 1 hour ago

Barry

187k21331612

asked 1 hour ago

AlexanderAlexander

895414

void h() 

constexpr int f() 
 return 1;
 h();


int main() 
 constexpr int i = f();

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();

share|improve this question

edited 1 hour ago

Barry

187k21331612

asked 1 hour ago

AlexanderAlexander

895414

share|improve this question

edited 1 hour ago

Barry

187k21331612

asked 1 hour ago

AlexanderAlexander

895414

edited 1 hour ago

Barry

187k21331612

edited 1 hour ago

Barry

187k21331612

asked 1 hour ago

AlexanderAlexander

895414

asked 1 hour ago

AlexanderAlexander

895414

1 Answer
1

active

oldest

votes

12

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;

share|improve this answer

answered 1 hour 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 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
  23 mins 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
  21 mins 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
  15 mins 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
  11 mins ago
  
  

  
  
  
  

add a comment | 

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12

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;

share|improve this answer

answered 1 hour 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 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
  23 mins 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
  21 mins 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
  15 mins 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
  11 mins ago
  
  

  
  
  
  

add a comment | 

12

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;

share|improve this answer

answered 1 hour 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 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
  23 mins 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
  21 mins 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
  15 mins 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
  11 mins 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;

share|improve this answer

answered 1 hour ago

BrianBrian

67k799192

constexpr int g(int x) 
 if (x%2 == 0) h();
 return 0;

share|improve this answer

answered 1 hour ago

BrianBrian

67k799192

answered 1 hour ago

BrianBrian

67k799192

answered 1 hour ago

BrianBrian

67k799192