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Home/ Questions/Q 3270698
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T18:39:47+00:00 2026-05-17T18:39:47+00:00

I saw the following code: class NullClass { public: template<class T> operator T*() const

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I saw the following code:

class NullClass {
public:
    template<class T> operator T*() const { return 0; }
};

const NullClass NULL;

void f(int x);
void f(string *p);

f(NULL); // converts NULL to string*, then calls f(string*)

Q1> I have problems to understand the following statement

template<class T> operator T*() const { return 0; }

Specially, what is the meaning of operator T*()?

Q2> Why f(NULL) is finally triggering the f(string*)?

Thank you

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  1. Editorial Team
    Editorial Team
    2026-05-17T18:39:48+00:00Added an answer on May 17, 2026 at 6:39 pm

    what is the meaning of operator T*()?

    It is a user-defined conversion operator. It allows an object of type NullClass to be converted to any pointer type.

    Such conversion operators can often lead to subtle, unexpected, and pernicious problems, so they are best avoided in most cases (they are, of course, occasionally useful).

    Why f(NULL) is finally triggering the f(string*)?

    NULL is of type NullClass. It can’t be converted to an int, but the user-defined conversion NullClass -> T* can be used to convert it to a string*, so void f(string*) is selected.

    Note that this works because there is only one overload of f that takes a pointer. If you had two overloads,

    void f(int*);
    void f(float*);
    

    the call would be ambiguous because the NullClass -> T* conversion can be converted to both int* and float*.

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