Parameterizing a function using a struct

Question

The code below does not compile. The function foo takes a function pointer f as an argument, and f strictly takes one int as argument and returns an int. In this example, foo always invokes f with the argument 3. I would like to pass a function to foo so that when f(3) is evaluated, it's evaluated in conjunction with some other parameter; however, I can't pass a function with 2 ints as arguments to foo (this problem is an analogy for a real problem).

#include <iostream>

void foo(int(*f)(int))
{
  std::cout << f(3) << std::endl;
}

struct plusX
{
  plusX(int x_) : x(x_) {}
  int x;
  int plus(int y)
  {
    return x + y;
  }
};

int main()
{
  plusX px4(4);
  foo(&px4.plus);  // ERROR!
}

ISO C++ forbids taking the address of a bound member function to form a pointer to member function. Say '&plusX::plus'

Answer 1

Two solutions. The first solution doesn't require refactoring code if existing code uses functions.

1) Using functors:

#include <iostream>

template<typename F>
void foo(F&& f)
{
  std::cout << f(3) << std::endl;
}

struct my_functor
{
  my_functor(int x_) : x(x_) {}
  int operator()(int y)
  {
    return x + y;
  }
  int x;
};

int main()
{
  my_functor f(4);
  foo(f);
}

2) Using std::function and std::bind

#include <iostream>
#include <functional>

void foo(const std::function<int(int)>& f)
{
  std::cout << f(3) << std::endl;
}

int add(int x, int y)
{
  return x + y;
}

int main()
{
  std::function<int(int)> f_binded = std::bind(add, 4, std::placeholders::_1);
  foo(f_binded);
}