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Home/ Questions/Q 3348286
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T01:31:27+00:00 2026-05-18T01:31:27+00:00

class Test { public: operator Test * () { return NULL; }; }; int

  • 0
class Test
{
public:
    operator Test * () { return NULL; };
};

int main()
{
    Test test;
    if (test == NULL)
        printf("Wtf happened here?\n");

    return 0;
}

How is it that this code compiles? How did Test get a comparison operator? Is there some implicit casting going around? What does that overloaded operator even mean (and do)?

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  1. Editorial Team
    Editorial Team
    2026-05-18T01:31:28+00:00Added an answer on May 18, 2026 at 1:31 am

    The overloaded operator adds a conversion from Test to Test *. Since there is no comparision operator defined that takes Test and NULL as arguments, any conversion operators that exists are tried. operator Test * returns a type which is comparable with NULL, so it is used.

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