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Home/ Questions/Q 9030009
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T07:23:04+00:00 2026-06-16T07:23:04+00:00

I’m trying to learn C++ and ran into this confusing example in the tutorial

  • 0

I’m trying to learn C++ and ran into this confusing example in the tutorial I’m using perhaps someone can clarify this for me.

The following code is from the tutorial. I’ve put a “tag” (if you will // <--- )indicating the parts that are confusing to me.

#include <iostream>
using namespace std;

class CDummy {
  public:
    // signature: accepts an address 
    int isitme (CDummy& param);
};

// actual function that accepts an address passed in from main set to
// the param variable
int CDummy::isitme (CDummy& param) // <--- 1
{

Now here is where the confusing part comes. I’m taking the address of
an address variable??? Wasn’t the address already passed in?

  if (&param == this) return true; // <--- 2
  else return false;
}

int main () {
  CDummy a;
  CDummy* b = &a;
  // passes in the class 
  if ( b->isitme(a) )
    cout << "yes, &a is b";
  return 0;
}

Now below in the code that makes sense to me

#include <iostream>
using namespace std;

class CDummy {
  public:
    int isitme (CDummy *param);
};

int CDummy::isitme (CDummy *param) // <--- 1
{

this part makes perfect sense. param is a pointer and I’m comparing
the pointer of class a with the pointer of class b.

  if (param == this) return true; // <--- 2
  else return false;
}

int main () {
  CDummy a;
  CDummy *b = &a;
  // pass in the address.
  if ( b->isitme(&a) )
    cout << "yes, &a is b";
  return 0;
}

Which one of the code samples is correct? Correct in the sense that this is the preferred way to do it, because they both work. Also, why am I taking an address of an address in the first example?

Apologies if this has been answered before but I couldn’t find it.
Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-16T07:23:06+00:00Added an answer on June 16, 2026 at 7:23 am

    You need to take you favorite C++ book and read about references.

    In this declaration

    int CDummy::isitme (CDummy& param)
    

    param is not an address, as you seem to believe. It is a reference.

    You are not the first one to be misled by the & character used in reference declarations. This question has been asked many times before on StackOverflow

    What is a reference variable in C++?

    What are the differences between a pointer variable and a reference variable in C++?

    but a book/guide/tutorial will make a better starting point, in my opinion.

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