What is ->* operator in C++?

Question

C++ continues to surprise me. Today i found out about the ->* operator. It is overloadable but i have no idea how to invoke it. I manage to overload it in my class but i have no clue how to call it.

struct B { int a; };

struct A
{
    typedef int (A::*a_func)(void);
    B *p;
    int a,b,c;
    A() { a=0; }
    A(int bb) { b=b; c=b; }
    int operator + (int a) { return 2; }
    int operator ->* (a_func a) { return 99; }
    int operator ->* (int a) { return 94; }
    int operator * (int a) { return 2; }
    B* operator -> () { return p; }


    int ff() { return 4; }
};


void main()
{
    A a;
    A*p = &a;
    a + 2;
}

edit:

Thanks to the answer. To call the overloaded function i write

void main()
{
    A a;
    A*p = &a;
    a + 2;
    a->a;
    A::a_func f = &A::ff;
    (&a->*f)();
    (a->*f); //this
}

No, you don't call the overloaded operator in this case. What you call is the built-in one. Your overloaded operator requires an int as the second parameter. — AnT stands with Russia
– AnT stands with Russia, Commented Nov 22, 2009 at 20:00
@acidzombie24: OK, now you finally calling the overloaded version. Again, you could do it without any updates by doing a->*42, as sellibitze suggested. — AnT stands with Russia
– AnT stands with Russia, Commented Nov 22, 2009 at 20:10

Cat Plus Plus · Accepted Answer · 2009-11-22 19:29:33Z

17

Just like .*, ->* is used with pointers to members. There's an entire section on C++ FAQ LITE dedicated to pointers-to-members.

#include <iostream>

struct foo {
    void bar(void) { std::cout << "foo::bar" << std::endl; }
    void baz(void) { std::cout << "foo::baz" << std::endl; }
};

int main(void) {
    foo *obj = new foo;
    void (foo::*ptr)(void);

    ptr = &foo::bar;
    (obj->*ptr)();
    ptr = &foo::baz;
    (obj->*ptr)();
    return 0;
}

answered Nov 22, 2009 at 19:29

Cat Plus Plus

131k27 gold badges205 silver badges226 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

AnT stands with Russia · Accepted Answer · 2009-11-22 19:57:23Z

11

The overloaded ->* operator is a binary operator (while .* is not overloadable). It is interpreted as an ordinary binary operator, so in you original case in order to call that operator you have to do something like

A a;
B* p = a->*2; // calls A::operator->*(int)

What you read in the Piotr's answer applies to the built-in operators, not to your overloaded one. What you call in your added example is also the built-in operator, not your overloaded one. In order to call the overloaded operator you have to do what I do in my example above.

answered Nov 22, 2009 at 19:57

AnT stands with Russia

322k44 gold badges548 silver badges793 bronze badges

1 Comment

user34537 Over a year ago

Your right about piotr answer but with it i figured out how to call the overloaded operator as shown in my edited question

Birdman · Accepted Answer · 2009-11-22 22:06:19Z

1

Like any other opperator, you can also call it explicitly:

a.operator->*(2);

answered Nov 22, 2009 at 22:06

Birdman

8692 gold badges10 silver badges14 bronze badges

Collectives™ on Stack Overflow

What is ->* operator in C++?

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related