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Home/ Questions/Q 6986771
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T18:51:06+00:00 2026-05-27T18:51:06+00:00

I have this declaration struct Z { void operator ()( int a ) {

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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?

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  1. Editorial Team
    Editorial Team
    2026-05-27T18:51:06+00:00Added an answer on May 27, 2026 at 6:51 pm

    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);
    
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