如何将成员函数作为参数传递给不期望它的函数？

Question

Say I have a function foo :

void foo(void (*ftn)(int x))
{
  ftn(5);
}

It needs as a parameter a void function that accepts an int as a parameter. Consider

void func1(int x) {}

class X {
public:
  void func2(int x) {}
};

Now foo(&func1) is ok.

But foo(&X::func2) isn't ok because X::func2 isn't static and needs a context object and its function pointer type is different.

I tried foo(std::bind(&X:func2, this)) from inside X but that raises a type mismatch too.

What is the right way of doing this?

Answer 1

Based on the comments, if you cannot change the signature of foo to take anything but a raw function pointer... then you'll have to do something like this:

struct XFunc2Wrapper {
    static X* x;

    static void func2(int v) {
        x->func2(v);
    }
};

And then just do foo(&XFunc2Wrapper::func2) once you set XFunc2Wrapper::x to be your X . It doesn't have to be nested in a struct, it can just be some global pointer, but nesting helps establish the intent behind the code better.

But this should definitely be last resort after (as per Captain Obvlious) trying to do foo(std::function<void(int)> ) .