Calling non-static constexpr member function in template definition

Question

#include <type_traits>

template<size_t S> struct A
{
    constexpr size_t size() const noexcept { return S; } // Not static on purpose!
};

struct B : public A<123> {};

template <class T>
typename std::enable_if<std::is_base_of_v<A<T().size()>, T>, bool>::type // (*)
f(const T&, const T&) noexcept { return true; }

int main() {
    B b1, b2;
    f(b1, b2);
}

In my original question in (*) line I mistakenly used T()::size() , which is obviously incorrect since size() is not static. The code works with T().size() and std::declval<T>().size() . So the question now is what is the difference and if any of these ways are more correct or better?

Answer 1

You didn't specify which compiler you're using, but gcc's error message offers a big honking clue:

t.C:12:52: error: cannot call member function ‘constexpr size_t A<S>::size() const
[with long unsigned int S = 123; size_t = long unsigned int]’ without object

After adjusting the declaration of the method accordingly:

static constexpr size_t size() noexcept { return S; }

gcc then compiled the shown code without issues.

If your intent is really to have size() be a regular class method, then you'll need to use std::declval instead of invoking it as a static method, in your template.

Answer 2

size is a non-static function and requires an object to call. Make it static and remove const .