Pointer Confusion with C++

Question

I am very confused with c++ pointers and reference operators. My main confusion is the following (simple ) code:

#include <iostream>

using namespace std;

void changeInt(int &a)
{
    a *= 3;
}

int main()
{
    int n = 3;
    changeInt(n);
    cout << n << endl; 

    return 0;
}

Mainly, I am confused as to why changing the address (&a) changes the actual variable (n). When I first attempted this problem this was my code:

#include <iostream>

using namespace std;

void changeInt(int &a)
{
    *a *= 3;
}

int main()
{
    int n = 3;
    changeInt(n);
    cout << n << endl; 

    return 0;
}

But this gives me an error. Why is it that when I change the address it changes the variable, but when I change the value pointed by the address I get an error?

Answer 1

Your second example is not valid C++, you can only dereference a pointer (or an object whose type overload operator* , which is not your case).

Your first example pass the parameter by reference ( int &a is not "the address of a", it is a reference to a), which is why a change to a really is a change to the object being passed by the function (in you case, n )

Answer 2

The ampersand ( & ) in that context means a reference , not the "address". Eg:

int some_int;

int & a = some_int; // Declare 'a', a reference to 'some_int'

int * p = &some_int; // '&' in this context is "the address of" 'some_int'

A reference is equivalent to a pointer in many ways, but it behaves like a value type. See this thread and the wikipedia entry to learn more.

Answer 3

The ampersand indicates that a variable is passed by reference to your function -- but inside the function the variable is treated as if it were passed by value. This is syntactic sugar , to make writing code that accepts references simpler to understand.