Why doesn't std::is_invocable_r_v<void, TFun> return false if the function does not return void

See the test case

#include <iostream>
#include <type_traits>


int main(){
    const char* hw = "hello world";
    auto task = [hw] ()->const char * { return hw ; };

    std::cout << std::is_invocable_r_v<const char *, decltype(task)> << std::endl;
    std::cout << std::is_invocable_r_v<void, decltype(task)> << std::endl;
    std::cout << std::is_invocable_r_v<const char, decltype(task)> << std::endl;
}

Result is

1
1
0

but I would expect

1
0
0

See

https://godbolt.org/z/d1cbfbsEY

>Solution :

Why doesn’t std::is_invocable_r_v<void, TFun> return false if the function does not return void

From https://en.cppreference.com/w/cpp/types/is_invocable I see that:

2. Determines whether INVOKE<R>(std::declval<Fn>(), std::declval<ArgTypes>()...) is well formed when treated as an unevaluated operand.

Then I click on "INVOKE" and jump to https://en.cppreference.com/w/cpp/utility/functional which has:

The exposition-only operation INVOKE<R>(f, t1, t2, ..., tN) is defined as static_cast<void>(INVOKE(f, t1, t2, ..., tN)) if R is (possibly cv-qualified) void, otherwise INVOKE(f, t1, t2, …, tN) implicitly converted to R.

Surely static_cast<void>(task()) is fine, void just discard the value. std::is_invocable_r_v<void, decltype(task)> should be true.