c++ std::function type cheking correct?

Question

I wish to use std::function type from so that function signature is checked on assignment. But I fail to understand what is happening in this case

//g++  5.4.0
#include <iostream>
#include <functional>
int f(int x)      { std::cout << "//f:int->int\n";    return x; }
int g(double x)   { std::cout << "//g:double->int\n"; return 1; }
int main()
{
    std::function<int(int)> fct;
    fct = f; fct(1);
    fct = g; fct(1);
}
//trace
//
//f:int->int
//g:double->int

The behavior for f is what I want, but I thought that "fct=g;" would cause a compile time error.

Any light on this case please ?

Answer 1

std::function is flexible and uses type erasure underneath, so if you have std::function object with with signature std::function<R(Args...)>, it will accept any Callable that can be called with type Args... and returns type R,

So in your case for std::function<int(int)>, function of type int g(double); can be called with type int arguments compiler will just promote int to double,

If you run this code

#include <iostream>
#include <functional>
int f(int x)      { std::cout << x << " " << "//f:int->int\n";    return x; }
int g(double x)   { std::cout << x << " " << "//g:double->int\n"; return 1; }
int main()
{
    std::function<int(int)> fct;
    fct = f; fct(1);
    fct = g; fct(2.5);
}

You can see that fct will only accept int and afterwards compiler promote it to double, So In the output of fct(2.5); it will print 2 //g:double->int and not 2.5 //g:double->int