views:

185

answers:

3

I'd like to set up a function pointer as a member of a class that is a pointer to another function in the same class. The reasons why I'm doing this are complicated.

In this example, I would like the output to be "1"

class A {
public:
 int f();
 int (*x)();
}

int A::f() {
 return 1;
}


int main() {
 A a;
 a.x = a.f;
 printf("%d\n",a.x())
}

But this fails at compiling. Why?

+10  A: 

The syntax is wrong. A member pointer is a different type category from a ordinary pointer. The member pointer will have to be used together with an object of its class:

class A {
public:
 int f();
 int (A::*x)(); // <- declare by saying what class it is a pointer to
};

int A::f() {
 return 1;
}


int main() {
 A a;
 a.x = &A::f; // use the :: syntax
 printf("%d\n",(a.*a.x)()); // use together with an object of its class
}

a.x does not yet say on what object the function is to be called on. It just says that you want to use the pointer stored in the object a. Prepending a another time as the left operand to the .* operator will tell the compiler on what object to call the function on.

Johannes Schaub - litb
+1  A: 

You need to use a pointer to a member function, not just a pointer to a function.

class A { 
    int f() { return 1; }
public:
    int (A::*x)();

    A() : x(&A::f) {}
};

int main() { 
   A a;
   std::cout << (a.*a.x)();
   return 0;
}
Jerry Coffin
+2  A: 

int (*x)() is not a pointer to member function. A pointer to member function is written like this: int (A::*x)(void) = &A::f;.

Bertrand Marron