How to pass a generic function pointer as parameter

Question

I have several functions that having similiar declarations:

int foo(int a);
int bar(int a);
int test(int a);

And the logics of my message handlers are exactly the same:

void HandleFoo(int a) {
    process(a);
    int ret = foo(a);
    if (ret) 
        print(a);
}

void HandleBar(int a) {
    process(a);
    int ret = bar(a);
    if (ret) 
        print(a);
}

void HandleTest(int a) {
    process(a);
    int ret = test(a);
    if (ret) 
        print(a);
}

So I am wondering if it is possible to write a general function:

void Handle(int a, func_pointer fn) {
    process(a);
    int ret = fn(a);
    if (ret) 
        print(a);
}

The fn is like a generic function pointer that can accept foo, bar and test

Is this possible?

Btw, currently no C++11 and boost in my project, only using TR1.

Answer 1

You can use a template

Example (not your code exactly) :

int add1(int n) { return n + 1; }
int add2(int n) { return n + 2; }

template<typename Adder>
void AddHandler(int n, Adder adder)
{
    int r = adder(n);
    std::cout << r << std::endl;
}

int main(void)
{
    AddHandler(1, add1);
    AddHandler(3, add2);

    return 0;
}

This outputs as expected :

2
5

You can see it live here http://ideone.com/q3FyI5

Answer 2

Convert the functions to functors, i.e.

struct foo
{
    int operator()(int a) {}
};
struct bar
{
    int operator()(int a) {}
};
struct test
{
    int operator()(int a) {}
};

then the previous approach works, with a slight tweak

template <typename Functor>
void Handle(int a)
{
    process(a);
    int ret = Functor()(a); // <- tweak, instantiate functor and then call
    if (ret) 
        print(a);
}

You can take this approach or @undu's where you pass the function to call as an argument.