Does std::is_constructible work with arguments that are convertible to parameters?

Question

Does

std::is_constructible<T, Arg1>

work if

Arg1

is a type that is convertible to a valid one-parameter constructor for

T

? It appears to work when the type has a non-templated conversion operator, but does not work (under some compilers) if the conversion operator is templated.

In the below example, the final

static_assert

fails under GCC 7.2 and clang 5.0, but passes under MSVC 19. Is there undefined behavior in here, or is one of the compilers misbehaving?

#include <type_traits>



struct foo

{

    foo(int) {}

    foo(int, int) {}

};



struct converts

{

    template <class T>

    operator T(){}

};



int main()

{

    // These compile

    foo f1(converts());

    foo f2(converts(), converts());

    static_assert(std::is_constructible<foo, converts, converts>::value, "foo(converts(), converts())");

    // This line doesn't

    static_assert(std::is_constructible<foo, converts>::value, "foo(converts())");

}

Live example: https://godbolt.org/g/EcFqMP

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Answer 1

This is just a Most Vexing Parse issue. Since the line foo f1(converts()) can be seen as a declaration of a function f1, the compiler must treat it as a declaration. If you switch the parentheses for braces, the first line stops compiling:

int main()
{
    foo f1{converts()}; // Now broken
    foo f2{converts(), converts()};
}

^{Live on godbolt}

The error message helpfully tells us that the compiler can't determine whether to call foo(int), or foo(const foo&), or foo(foo&&), as converts could also be converted into foo with its templated conversion operator.

When you use the parentheses, we can see by using decltype that the compiler sees f1 as a declaration of a function of type foo(converts(*)()) – a function returning foo, which takes one argument which is a function pointer.

Does std::is_constructible work with arguments that are convertible to parameters?

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