How can I pass a member function where a free function is expected?

Question

The question is the following: consider this piece of code:

#include 


class aClass
{
public:
    void aTest(int a, int b)
    {
        prin         


        

Answer 1

@Pete Becker's answer is fine but you can also do it without passing the class instance as an explicit parameter to function1 in C++ 11:

#include <functional>
using namespace std::placeholders;

void function1(std::function<void(int, int)> fun)
{
    fun(1, 1);
}

int main (int argc, const char * argv[])
{
   ...

   aClass a;
   auto fp = std::bind(&aClass::test, a, _1, _2);
   function1(fp);

   return 0;
}

Answer 2

Not sure why this incredibly simple solution has been passed up:

#include <stdio.h>

class aClass
{
public:
    void aTest(int a, int b)
    {
        printf("%d + %d = %d\n", a, b, a + b);
    }
};

template<class C>
void function1(void (C::*function)(int, int), C& c)
{
    (c.*function)(1, 1);
}
void function1(void (*function)(int, int)) {
  function(1, 1);
}

void test(int a,int b)
{
    printf("%d - %d = %d\n", a , b , a - b);
}

int main (int argc, const char* argv[])
{
    aClass a;

    function1(&test);
    function1<aClass>(&aClass::aTest, a);
    return 0;
}

Output:

1 - 1 = 0
1 + 1 = 2

Answer 3

A pointer to member function is different from a pointer to function. In order to use a member function through a pointer you need a pointer to it (obviously ) and an object to apply it to. So the appropriate version of function1 would be

void function1(void (aClass::*function)(int, int), aClass& a) {
    (a.*function)(1, 1);
}

and to call it:

aClass a; // note: no parentheses; with parentheses it's a function declaration
function1(&aClass::test, a);

Answer 4

Since 2011, if you can change function1, do so, like this:

#include <functional>
#include <cstdio>

using namespace std;

class aClass
{
public:
    void aTest(int a, int b)
    {
        printf("%d + %d = %d", a, b, a + b);
    }
};

template <typename Callable>
void function1(Callable f)
{
    f(1, 1);
}

void test(int a,int b)
{
    printf("%d - %d = %d", a , b , a - b);
}

int main()
{
    aClass obj;

    // Free function
    function1(&test);

    // Bound member function
    using namespace std::placeholders;
    function1(std::bind(&aClass::aTest, obj, _1, _2));

    // Lambda
    function1([&](int a, int b) {
        obj.aTest(a, b);
    });
}

(live demo)

Notice also that I fixed your broken object definition (aClass a(); declares a function).

Answer 5

I made the member function as static and all works:

#include <iostream>

class aClass
{
public:
    static void aTest(int a, int b)
    {
        printf("%d + %d = %d\n", a, b, a + b);
    }
};

void function1(int a,int b,void function(int, int))
{
    function(a, b);
}

void test(int a,int b)
{
    printf("%d - %d = %d\n", a , b , a - b);
}

int main (int argc, const char* argv[])
{
    aClass a;

    function1(10,12,test);
    function1(10,12,a.aTest); // <-- How should I point to a's aClass::test function?

    getchar();return 0;
}

Answer 6

If you actually don't need to use the instance a (i.e. you can make it static like @mathengineer 's answer) you can simply pass in a non-capture lambda. (which decay to function pointer)

---

#include <iostream>

class aClass
{
public:
   void aTest(int a, int b)
   {
      printf("%d + %d = %d", a, b, a + b);
   }
};

void function1(void (*function)(int, int))
{
    function(1, 1);
}

int main()
{
   //note: you don't need the `+`
   function1(+[](int a,int b){return aClass{}.aTest(a,b);}); 
}

Wandbox

---

note: if aClass is costly to construct or has side effect, this may not be a good way.