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Home/ Questions/Q 5988839
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T23:00:57+00:00 2026-05-22T23:00:57+00:00

This is just an experiment code. struct B { virtual B* operator -> ()

  • 0

This is just an experiment code.

struct B
{
  virtual B* operator -> () { return this; }
  void foo () {} // edit: intentionally NOT virtual
};

struct D : B
{
  virtual D* operator -> () { return this; }
  void foo () {}
};

int main ()
{
  B &pB = *new D;
  pB->foo();  // calls B::foo() !
}

I know that operator has to be called using object or reference; thus in above case does reference pB still resolute to the object of B ?
Though it will not be practical, but for curiosity, is there any way to invoke D::operator -> through pB ?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T23:00:57+00:00Added an answer on May 22, 2026 at 11:00 pm

    I think it is invoking D::operator->, but the return value is being treated as a B*, so B::foo() is being called.

    This is an artifact of how covariant return types behave.

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