Using std::invoke when a function is overloaded

Question

I am trying to use std::invoke() with an overloaded function:

#include <iostream>
#include <functional>

struct S {
    void foo(int) { }
    void foo(int, int) { }
};

int main()
{
    S s;
    std::invoke(&S::foo, s, 1);
}

but I get an error: 'std::invoke': no matching overloaded function found. It works fine, when there is the only one function with name foo(). Is it possible to use std::invoke() when a function is overloaded?

You can help compiler to choose the proper overload by casting: `static_cast(&S::foo)` — rafix07, Apr 09 '21 at 11:23
You can also eliminate the problem by wrapping the call in a lambda. `std::invoke([&]() { s.foo(1); })`. — super, Apr 09 '21 at 11:26

Michal · Accepted Answer · 2021-04-09T11:32:09.870

3

There are two example solutions:

std::invoke(static_cast<void(S::*)(int)>(&S::foo), s, 1);
std::invoke([&s]() { s.foo(1); });

edited Apr 09 '21 at 11:32

answered Apr 09 '21 at 11:29

Michal

627
3
13

2

why do you cast `1` to an `int` ? – 463035818_is_not_an_ai Apr 09 '21 at 11:29

Using std::invoke when a function is overloaded

1 Answers1