Returning function pointer type

Question

Often I find the need to write functions which return function pointers. Whenever I do, the basic format I use is:

typedef int (*function_type)(int,int);

fu         


        

Answer 1

While wrapping some C code in C++ classes, I had the same desire as the original poster: return a function pointer from a function without resorting to typedef'ing the function pointer prototype. I hit a problem with C++ const correctness which I thought was worth sharing, even if it's a little off-topic (C++) but it does relate directly to the original question: the syntax for returning a C function pointer without resorting to a typedef.

The code below defines a class A which stores a function pointer and exposes it to the outside world through the get_f() call. This is the function that should return a function pointer without a typedef.

The point (which stumped me for some time) was how to declare that get_f() was a const function, i.e. it wouldn't alter A.

The code contains 2 variants: the first uses a typedef for the function pointer prototype, whilst the second writes everything out in full. The #if switches between the two.

#include <iostream>

int my_f(int i)
{
  return i + 1;
}

#if 0 // The version using a typedef'ed function pointer

typedef int (*func_t)(int);

class A
{
public:
  A(func_t f) : m_f(f) {}
  func_t get_f() const { return m_f; }

private:
  func_t m_f;
};

int main(int argc, char *argv[])
{
  const A a(my_f);
  std::cout << "result = " << a.get_f()(2) << std::endl;
}

#else // The version using explicitly prototyped function pointer    

class A
{
public:
  A(int (*f)(int)) : m_f(f) {}
  int (*get_f() const)(int) { return m_f; }

private:
  int (*m_f)(int);
};

int main(int argc, char *argv[])
{
  const A a(my_f);
  std::cout << "result = " << a.get_f()(2) << std::endl;
}

#endif

The expected/desired output is:

result = 3

The key point is the position of the const qualifier in the line:

int (*get_f() const)(int) { return m_f; }