CODEWITHSUNDEEP
c++
dead programmer
dead programmer

Reputation: 4365

Wrong output of reference to pointer to int

I am trying to use a reference to pointer to int like in below program. But I am not getting the expected output.

Output:

9 5
5 9

Expecting:

9 5
9 5

Code:

#include <iostream>
using namespace std;

void swap (int *& a, int *&b)
{
    int *t = a;
    a = b;
    b = t;
}

int main()
{
    int a = 5, b = 9;       
    int *p = &a;
    int *q = &b;

    swap (p, q);
    cout << *p << " " << *q << endl;
    cout << a << " " << b << endl;

    return 0;
}

Why is my expectation wrong? I head that reference is nothing just an other name of the target variable.

Upvotes: 0

Views: 104

Answers (5)

molbdnilo
molbdnilo

Reputation: 66371

You're swapping the values of the pointers.

Look at this illustration:

First, p is pointing at a, q is pointing at b:

  p         a
+---+     +---+
+ ------> | 5 |
+---+     +---+

  q         b
+---+     +---+
+ ------> | 9 |
+---+     +---+

After you swap p and q, q is pointing at a, and p is pointing at b: 

  q         a
+---+     +---+
+ ------> | 5 |
+---+     +---+

  p         b
+---+     +---+
+ ------> | 9 |
+---+     +---+

But both a and b still have their old values.

Upvotes: 1

masoud
masoud

Reputation: 56479

I head that reference is nothing just a other name of the target variable

Yes, and target variable in your case is a pointer of the variable, not the variable. Because you're using * in the function declaration. Then, you're swaping pointers not values.

Use one of these ways:

void swap(int &a, int &b)
{
    int t=a;
    a=b;
    b=t;
}
// ...

int a=5,b=9;        
swap(a,b);

or

void swap(int *a, int *b)
{
    int t=*a;
    *a=*b;
    *b=t;
}
// ...

int a=5,b=9;        
swap(&a,&b);

Upvotes: 0

CashCow
CashCow

Reputation: 31435

In general, you are swapping two pointers and not what they point to.

Maybe your function is confusing you as you are calling the values a and b. Your function swaps two pointers such that the first one now points where the second one was pointing, and the second points to where the first was pointing.

p was previously pointing to a and q to b. Now p points to b and q points to a thus your output is 9 5 5 9

There are references in your swap function. These are references to the pointers, so a and b in the function scope become aliases to the parameters passed in, i.e. p and q and thus modify p and q (to point elsewhere).

Upvotes: 0

user1357649
user1357649

Reputation:

In your case, you're swapping the pointers (and not the values). Since you're swapping the pointers, the actual values inside a and b remain unchanged. Hencewhy when you print the value of a, you get 5 and when you print b you get 9.

Upvotes: 0

Benedikt Waldvogel
Benedikt Waldvogel

Reputation: 12866

You swap the pointers, not the values. Your expectation is wrong.

Upvotes: 2

Related Questions