passing object by reference in C++

Question

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?

Answer 1

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