Question

I’ve never used it before and just stumbled upon it in an article… I thought it would be the equivalent to *x->y but apparently it isn’t.

Here’s what I tried, and gave me an error:

struct cake {
 int * yogurt;
} * pie;

int main(void) {
 pie = new cake;
 pie->yogurt = new int;
 return pie->*yogurt = 4;
}

Answer 1

Its used when you have pointers to member functions.

When you have a pointer to a function of a class, you call it in much the same way you would call any member function

object.membername( … )

or

objectptr->membername( … )

but when you have a member function pointer, an extra * is needed after the . or -> in order that the compiler understand that what comes next is a variable, not the actual name of the function to call.

Here’s an example of how its used.

class Duck
{
public:

  void quack() { cout << "quack" << endl; }
  void waddle() { cout << "waddle" << endl; }
};

typedef void (Duck::*ActionPointer)();

ActionPointer myaction = &Duck::quack;

void takeDuckAction()
{    
    Duck myduck;
    Duck *myduckptr = &myduck;

    (myduck.*myaction)();
    (myduckptr->*myaction)();
}