Thread: a simple C++ question

wrong geuss again!

no i was wrong. I read the call as (A,B,C) for some reason when it was (B,C,A). You were right and i apologize for misleading you.

How dose the 'address of' operator function when put with a variable decleration like that?

Ive only see it used to send a pointer of a variable to a function.

like.

void do_something(int *p_a) {
*p_a=0; //modifys the variable declared elsewhere
}



int a;
do_something(&a);

Its a reference. It provides an alias for another variable. You can imagine a reference as an already dereferenced pointer. i.e.

int a=10; // an int
int& ref =a; // now ref is just another name for a so any changes to ref actually change a.
ref=30;
cout<<a;

now lets look at using this as a function parameter.

void func(int byValue,int& byRef);

now suppose we call this func with ints a and b like this....
func(a,b);
now any changes made to a in the func are local to the func and the value of a in the calling function never changes. You have in fact passed a copy of a to func. Now b is different. Any changes to b make changes to b in the calling function because instead of passing a copy you have passed a reference which is basically an address like a pointer but it behaves as if it is already dereferenced.

Thanks