Using function pointers?

Question

I'm not yet very familiar with these but maybe someone can shine light on this example.

Imagine I have class CFoo and it will have a function to add a handler and a function which is a function pointer.

So something like this:

class CFoo {

int *pointedFunc(int a, int b) = 0;

void setFunc(int *func(int a, int b))
{
    pointedFunc = func;
}
};

Given the above context, I want to know the proper way of doing this. I don't think I have done it properly. Also, how would I go about calling pointedFunc?

Thanks

Answer 1

In your example, pointedFunc is a member function that returns an int *. To make it a function pointer, you need parens around pointedFunc, like:

int (*pointedFunc)( int a, int b );

A typedef might make it clearer:

class CFoo {
    CFoo() : pointedFunc( NULL ) {}
    typedef int (*funcType)(int, int);
    funcType pointedFunc;

    void setFunc( funcType f ) {
         pointedFunc = f;
    }
};

To call the function, you can use either pointedFunc( 1, 2 ) or (*pointedFunc)(1, 2). I tend to use the latter to make it clear that you are going through a function pointer, but either will work.

Answer 2

Right now you have a member function returning int *, not a pointer to a function returning int. A set of parenthesis would fix that.

int (*pointedFunc)(int a, int b);

void setFunc(int (*pfunc)(int a, int b))
{
    pointedFunc = pfunc;
}

Also, member variables get initialized in the constructor ctor-initializer-list, like

CFoo::CFoo() : pointedFunc(0) {}

not like you did. Your = 0 was actually a pure-specifier (and it won't work unless the member function is virtual), when you fix the pointer-return-type vs pointer-to-function issue you'll find that the compiler also complains about your attempt to initialize it.

Using a typedef as Graeme suggests is the easiest and best way to keep track of function-pointer types.