Is the copy constructor called when you dereferencing a pointer when passing by reference to a function

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dereferencing a pointer when passing by reference

Is the copy constructor called when you dereferencing a pointer when passing by reference to a function?

Here is a simple example

int& returnSame( int &example ) { return example: }

int main()
{
  int inum = 3;
  int *pinum = & inum;

  std::cout << "pinum: " <<  returnSame(*pinum) << std::endl;

  return 0;          

}

My Guess what happeing:

When we dereference pinum we would expect the copy constructor to be called, but since the function is passed by value this will not be called?

If pinum copy constructor was called then a temporary object would be produced and the reference of that would be used, which would be very bad news in the form of undefined behaivour...

So what does happen...undefined behavior?

@ Oli Charlesworth that question was closed due to multiple typos — MWright, Jul 05 '12 at 15:21
@MWright That question was closed because it was "ambiguous, vague, incomplete, overly broad, or rhetorical". — Etienne de Martel, Jul 05 '12 at 15:22
@MWright - closure is not permanent. Votes can reopen questions our you can flag and explain that you've fixed it for a moderator to reopen it — Flexo, Jul 05 '12 at 15:23

score 10 · Accepted Answer · answered Jul 05 '12 at 15:20

10

No, the copy constructor is not called.

The dereference operator creates an lvalue referring to the existing object. The reference parameter is bound to this existing object.

answered Jul 05 '12 at 15:20

Ben Voigt

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Is the copy constructor called when you dereferencing a pointer when passing by reference to a function

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