overloading operator ()

Question

I have this declaration

struct Z {
    void operator ()( int a ) {
        cout << "operator()() " << a << endl;
    }
};

Z oz, *zp = &oz;

oz(1); //ok
(*zp)(2); //ok
zp(3); //"error: 'zp' cannot be used as a function"

Is there a way to modify struct declaration, so a call to No. 3 would succeed?

I do not think this is possible, I am fairly sure you cannot overload pointers to your type. You could however write a wrapper class around your pointers that does allow this syntax. — , Dec 22 '11 at 04:36
If some thought was put into it, I'm sure some guru could come up with a way using the preprocessor to achieve your goal. But I sure hope that person wouldn't waste their time, since what you ask for is completely useless, and of no benefit to anyone, including you. — Benjamin Lindley, Dec 22 '11 at 04:37
Not that exact syntax, but you could add things so that `zp->call(3)` and/or `call(zp, 3)` work. — aschepler, Dec 22 '11 at 04:46

score 5 · Accepted Answer · answered Dec 22 '11 at 04:33

5

That's expected behavior. zp is a pointer (a Z *), and operator() is overloaded for Z, not Z *. When you deference the pointer with *zp, you get a Z &, for which operator() is overloaded.

Unfortunately, you can't overload an operator for a pointer type (I think it has something to do with the fact that pointers are not user-defined types, but I don't have the Standard in front of me). You could simplify the syntax with references:

Z & r = *zp;
r(3);

answered Dec 22 '11 at 04:33

Etienne de Martel

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no :) you destroyed the whole meaning of it with the last edit – Ulterior Dec 22 '11 at 04:44
your last edit introduced an irrelevant addition, something that was not intended to be used in the first place. – Ulterior Dec 22 '11 at 05:20
Would you kindly tell me what irrelevant addition? – Etienne de Martel Dec 22 '11 at 05:25
the addition reference variable, it's already used as a call Nr.2 – Ulterior Dec 22 '11 at 05:30
@Ulterior Yeah, but notice how I say "You could simplify the syntax". – Etienne de Martel Dec 22 '11 at 06:04

score 0 · Answer 2 · answered Dec 22 '11 at 04:54

Provided your Z type has no state you can change it into a function.

typedef void Y(int);
void y( int a ) {
  cout << "y() " << a << endl;
}

Y& oy = y;
Y* yp = &oy;

Now both oy(1) and yp(1) are legal, because function pointers are implicitly dereferenced when called.

overloading operator ()

2 Answers2