C++ pass another method as argument

Question

In a simple code shown below, there is a function run7 which receives a function as an argument. In main function, a function test is passed to it and it works fine. However, my method2 cannot pass the method1 to this function. It causes error:

main.cpp:24:15: error: cannot convert ‘A::method1’ from type ‘void (A::)(int)’ to type ‘void (*)(int)’
   run7(method1);
               ^

I want to call pass method1 to run7 without changing the structure of run7. How to fix method2?

#include <iostream>

using namespace std;

void run7 ( void (*f)(int) )
{
    f(7);
}

void test(int a)
{
    cout<<a<<endl;
}

class A
{
public:

    int m=4;

    void method1(int a)
    {
        cout<< a*m <<endl;
    }

    void method2()
    {
        run7(method1);
    }
};

int main()
{
    run7(test);
    return 0;
}

Answer 1

If you look at the error closely:

error: cannot convert ‘A::method1’ from type ‘void (A::)(int)’ to type ‘void (*)(int)’

You'll see that the types are different. That's because class methods do not have the same type as raw function - they need that extra object to get called on. There is no way to get that code to compile since calling method1 requires an A which requires storage which is impossible to pass in as a raw function pointer.

What you can do instead is change run to take a type-erased functor:

void run7 ( std::function<void(int)> f ) {
    f(7);
}

And then pass in a functor which also passes in this:

void method2()
{
    run7(std::bind(&A::method1, this,           // option 1
                   std::placeholders::_1)); 
    run7([this](int x){ this->method1(x); });   // option 2
}