std::is_invocable is false but std::invoke works

Question

The following program's output seems to contradict itself:

#include <type_traits>
#include <iostream>
#include <functional>

void foo(int&){ std::cout << "called\n"; }

int main() {
    int a;
    foo(a);
    std::cout << std::is_invocable_v<decltype(foo), decltype(a)> << std::endl;
    std::invoke(foo, a);
}

The output is:

called
0
called

Which seems to me to be invoking a function that is not invocable? What is going on here?

Answer 1

decltype(a) is int . This corresponds to invoking f with an int prvalue -- something like f(7) . That one indeed doesn't compile, because a non- const lvalue reference cannot bind to a prvalue.

What you're doing instead in main is calling f with an lvalue , a , to which the reference can bind just fine.

To get the correct result from std::is_invocable , use the expression form of decltype by adding parentheses:

std::is_invocable_v<decltype(foo), decltype((a))>
//                                          ^ ^