References and memory

[fungivore]

Hi, I have a few questions about references.

Consider the following code:

#include <iostream>
using namespace std;
int main()
{
int x = 10; // line 1
int& y = x; // line 2
cout << &x << endl; // line 3
cout << &y << endl; // line 4
}

When running this program, lines 3 and 4 print out the same memory address. My impression was that the memory allocation in this program, starting from line 1, went like this:

line 1: allocate memory on the stack for x, copy the value 5 into the bucket
line 2: allocate memory on the stack for y, copy the memory address of x into the bucket
line 3: print out the memory location of the bucket for x
line 4: print out the memory location of the bucket for y

Since line 4 prints out what appears to be the memory location of the bucket for x, does this mean that the & operator works differently on reference types (return the memory address contained in y rather than the memory address of the bucket for y), or that y is sharing the same memory location as x, or something else?

Thanks.

[Gaminic]

The behaviour you describe would be correct if y was an int*. There are a few juicy topics on References floating around the forum right now, if you really want to read more, but prepare for severe headaches.

[clanmjc]

You can think of references like this, they are just a "label" for the value in memory, no additional memory is used. Instead of referring to a variable by it's memory address 0x0F032010 we can name that address in your case y

In other words, a reference is just a name, another name for a variable or any object for that matter.

[Moschops]

In other words, a reference is just a name, another name for a variable or any object for that matter.

Bingo. It's another name for the exact same thing in the exact same piece of memory.

Unless you want to end up in a screaming match, go no further :)

How C++ is to implement references is not specified in the standard, and as such opinions on what a reference "is" vary significantly.

[viliml]

if you declare int& y=x, y will have the same adress as x, and therefore line 3 and 4 print tha same thing, and whaever you do to x, the same is done with y.

[fungivore]

Thanks guys.