Why use dereferencing when function takes arguments in C++

Question

#include <iostream>
using namespace std;

class Dummy {
  public:
    bool isitme (Dummy& param);
};

bool Dummy::isitme (Dummy& param)
{
  if (&param == this) return true;
  else return false;
}

int main () {
  Dummy a;
  Dummy* b = &a;
  if ( b->isitme(a) )
    cout << "yes, &a is b\n";
  return 0;
}

I was looking at this C++ example and I don't understand why bool isitme (Dummy& param); uses the dereferencing sign '&'. The argument is an Dummy object itself right, why it is the object's address?

It seems you need to pick up [a good book](https://stackoverflow.com/q/388242/2878796) — UnholySheep, Jun 07 '18 at 06:59

score 3 · Answer 1 · answered Jun 07 '18 at 06:59

Ampersand is not the "dereferencing sign". It is used here in two different ways which I will explain below.

The "dereference sign" is asterisk (*).

bool Dummy::isitme (Dummy& param) Here, Dummy& param means that param is a reference to a Dummy object.
if (&param == this) return true; Here, &param denotes address of param.

score 0 · Answer 2 · answered Jun 07 '18 at 07:04

0

The isitme takes its argument by reference that is it takes the address of the object instead of copying it. The point is not to copy the object and to work on the original one.

answered Jun 07 '18 at 07:04

navyblue

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Why use dereferencing when function takes arguments in C++

2 Answers2