Why does "if constexpr" not behave as expected in such a case?

Asked Oct 29 '18 at 11:33

Active Oct 29 '18 at 11:33

Viewed 90 times

#include <type_traits>
#include <iostream>

void f()
{}

#define M1(...) \
if constexpr (std::is_void_v<decltype(__VA_ARGS__)>)\
{ std::cout << "is_void" << std::endl; }\
else\
{ std::cout << "is_not_void" << std::endl; }

#define M2(...) \
if constexpr (std::is_void_v<decltype(__VA_ARGS__)>)\
{ __VA_ARGS__; }\
else\
{ auto tmp = __VA_ARGS__; }

int main()
{
    M1(f()); // OK! output "is_void"

    M2(f());
    // compiler error:
    // variable has incomplete type 'void' at line: auto tmp = __VA_ARGS__;
}

My compiler is Clang 7.0.

In M2, I know auto tmp = __VA_ARGS__; is illegal because decltype(__VA_ARGS__) is void. However, I have used if constexpr, so to me, the compiler should ignore the illegal branch.

Is this comforming to the C++17?

asked Oct 29 '18 at 11:33

xmllmx

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Related, possibly duplciate https://stackoverflow.com/questions/46512248/why-doesnt-an-if-constexpr-make-this-core-constant-expression-error-dissappear – StoryTeller - Unslander Monica Oct 29 '18 at 11:39
Simpler mcve: https://godbolt.org/z/JdfpWk – YSC Oct 29 '18 at 11:42
auto temp = f(); is supposed to be illegal regardless of its placement. Even though branch is not instantiated syntax is still checked and needs to be valid. – user7860670 Oct 29 '18 at 11:43

Why does "if constexpr" not behave as expected in such a case?

0 Answers0