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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
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Data science time! April 2019 and salary with experience
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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}







17















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




















  • 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
    11 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
    11 hours ago











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

    – Barry
    11 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
    11 hours ago




















17















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




















  • 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
    11 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
    11 hours ago











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

    – Barry
    11 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
    11 hours ago
















17












17








17


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
















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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 11 hours ago









Barry

187k21331612




187k21331612










asked 11 hours ago









AlexanderAlexander

915414




915414








  • 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
    11 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
    11 hours ago











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

    – Barry
    11 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
    11 hours ago
















  • 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
    11 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
    11 hours ago











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

    – Barry
    11 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
    11 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
11 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
11 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
11 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
11 hours ago













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

– Barry
11 hours ago





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

– Barry
11 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
11 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
11 hours ago














1 Answer
1






active

oldest

votes


















16














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
























  • 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
    9 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
    9 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
    9 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 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
    9 hours ago














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1 Answer
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active

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16














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
























  • 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
    9 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
    9 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
    9 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 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
    9 hours ago


















16














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
























  • 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
    9 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
    9 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
    9 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 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
    9 hours ago
















16












16








16







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













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












share|improve this answer



share|improve this answer










answered 11 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 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
    9 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
    9 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
    9 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 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
    9 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
    9 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
    9 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
    9 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 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
    9 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
9 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
9 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
9 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
9 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
9 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
9 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
9 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
9 hours ago






















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