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?

score 3 · Accepted Answer · answered Sep 24 '14 at 03:17

3

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)

answered Sep 24 '14 at 03:17

quantdev

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@Asad-ullahKhan : no it doesn't : the page talks about the `&` operator, but your are not using it (you are declaring a reference, with `int &`) – quantdev Sep 24 '14 at 03:19
OHHH that makes much more sense! Thanks you! – Asad-ullah Khan Sep 24 '14 at 03:20
@Asad-ullahKhan in code: `int a = b & c;` you do not think that address of `c` is doing something to b, do you? – Slava Sep 24 '14 at 03:29

score 2 · Answer 2 · edited May 23 '17 at 11:57

2

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

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.

edited May 23 '17 at 11:57

Community

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answered Sep 24 '14 at 03:16

glampert

4,371
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score 1 · Answer 3 · answered Sep 24 '14 at 03:18

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.

Pointer Confusion with C++

3 Answers3