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class - passing object by reference in C++

The usual way to pass a variable by reference in C++(also C) is as follows:

void _someFunction(dataType *name){ // dataType e.g int,char,float etc.
/****
definition
*/
}

int main(){
    dataType v;
    _somefunction(&v);  //address of variable v being passed
    return 0;
}

But to my surprise, I noticed when passing an object by reference the name of object itself serves the purpose(no & symbol required) and that during declaration/definition of function no * symbol is required before the argument. The following example should make it clear:

// this
#include <iostream>
using namespace std;

class CDummy {
  public:
    int isitme (CDummy& param);     //why not (CDummy* param);
};

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

int main () {
  CDummy a;
  CDummy* b = &a;
  if ( b->isitme(a) )               //why not isitme(&a)
    cout << "yes, &a is b";
  return 0;
}

I have problem understanding why is this special treatment done with class . Even structures which are almost like a class are not used this way. Is object name treated as address as in case of arrays?

question from:https://stackoverflow.com/questions/18147038/passing-object-by-reference-in-c

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What seems to be confusing you is the fact that functions that are declared to be pass-by-reference (using the &) aren't called using actual addresses, i.e. &a.

The simple answer is that declaring a function as pass-by-reference:

void foo(int& x);

is all we need. It's then passed by reference automatically.

You now call this function like so:

int y = 5;
foo(y);

and y will be passed by reference.

You could also do it like this (but why would you? The mantra is: Use references when possible, pointers when needed) :

#include <iostream>
using namespace std;

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


int CDummy::isitme (CDummy* param)
{
    if (param == this) return true;
    else return false;
}

int main () {
    CDummy a;
    CDummy* b = &a;             // assigning address of a to b
    if ( b->isitme(&a) )        // Called with &a (address of a) instead of a
        cout << "yes, &a is b";
    return 0;
}

Output:

yes, &a is b

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