Compound Types

[Pebble]

I've ran the following program:

#include <iostream>
int main()
{
int aval = 6;
int valu = 3;
int &six = aval;
int &three = valu;
--aval;
std::cout << &six << " " << &three;
return 0;

}

And the output was 0x28fef4 0x28fef0

I was expecting to see 6 3

What is causing this output?

Using Codeblocks.

[vlad from moscow]

You are outputing addresses of references. Instead of

std::cout << &six << " " << &three;

use

std::cout << six << " " << three;

to get the desired result.

[jaded7]

You declare two ints, aval and valu
You then declare two references to integers. The references, "six" and "three" refer to the integers "aval" and "valu" respectively. A reference is just an "alias", which means that any changes to the reference also change the data it refers to.

Then you try to do output of "&six". The ampersand is the "address-of" operator, which means when you do this...

cout << &six

You are not printing the referenced data (aval), but the address-of the reference. Hence why you have an address printed.

[Pebble]

Finally got it. If it was described in the book (C++ Primer) the same way you described it, I would'nt have had a problem.

Thank's for the replies :)