what does & do as argument?

[ballabash]

I stumbled on this piece of code earlier (don't remember where, so I duplicated it here). It outputs 6. What I don't understand is how. Can anyone explain please? Thank you. I realize a pointer must be created somewhere, but it can't be "r" because we're doing r*=2.

#include <iostream>

using std::cout;
using std::endl;

void f(int &r);

int main()
{
	int i = 3;
	f(i);
	cout << i << endl;
	return 0;
}

void f(int &r)
{
	r = r*2;
}

[Skillless]

Its a reference, because im bad at explaining stuff,
read this : http://cplusplus.com/doc/tutorial/pointers/ (Reference operator (&))

[Bazzy]

Actually the right tutorial is this: http://cplusplus.com/doc/tutorial/functions2/

[ballabash]

From what I read the tutorial just explains what happens when you use it, but not why it happens. Is it something like (from the example above) in function f, the value 3 with the alias i also gets the alias r, instead of a new variable being created for r in pass by value?

[Kyon]

For standard types the overhead of copying the value is negligible. Though, your impression seems to be right, indeed. A reference is made to the inputted variable. This means that the function gets access to the variable you pass into it. It also means that you don't have to copy a lot of values (which is specifically helpful for vectors, strings and other containers or classes). When you understand how to use references and pass by reference, I suggest you check out the part in the tutorial where constant references are explained.