Question

I am trying to understand functions returning a reference. For that I have written a simple program:

#include<iostream>
using namespace std;

class test
{
   int i;
   friend test& func();

public:
   test(int j){i=j;}
   void show(){cout<<i<<endl;}
};

test& func()
{
   test temp(10);
   return temp; //// Address of temp=0xbfcb2874
}

int main()
{
   test obj1(50); // Address of obj1=0xbfcb28a0
   func()=obj1; <= Problem:The address of obj1 is not changing
   obj1.show(); // // Address of obj1=0xbfcb28a0
   return 0;
}

I ran the program using gdb and observed that the address of obj1 still remains same, but I expect it to get changed to 0xbfcb2874. I am not clear with the concept. Please help.

Answer 1

There are several problems in your code:

(1) That is not how you want to return a reference. temp(10) is an automatic (i.e. resides in stack) variable and will be destroyed once your program goes out of scope of the test function. A better way to show this would be to return a reference to a variable passed (e.g. for chaining of calls):

Test& func(Test& some_param) {
  // Do something with some_param...

  // Return it as a reference.
  return some_param;
}

(2) You are assigning the value of obj1 to func(), while what you want would be to assign the return value of func() to a variable. Try this:

Test obj1(50);
Test& obj2 = func(obj1);  // Address of obj2 should now be the same as obj1.

(3) func() need not be a friend of Test. In fact, it should not. A friend class/functions allow the class/functions to access private members of Test. That is not what you want to do too often.