Swap integers using pointers - c++

Question

Consider the function :
void swap (int *p,int *q)
{
int temp;
temp=*p;
*p=*q;
*q=temp;
}
swap(4 , 5)
Output:
5 4
I am aware that this is the right way to swap two integers using pointers, as mentioned in the books. But what is actually going on in there? It's very confusing. Consider this function , where the step *p=*q is replaced by p=q i.e., with the indirection operator * removed .
void swap (int *p,int *q)
{
int temp;
temp=*p;
p=q;
*q=temp;
}
Now ,what would happen? In this case the output i get for
swap(4 , 5)
Output:
4 4

Answer 1

Think of an int as a box with a number in it, and think of a pointer as an entry on your clipboard telling you which box to look in.
When you do
*p = *q;
you're looking on your list to find out which box you should use for p, and which box you should use for q; then you're looking what number is in the q box, and copying it into the p box. You still have two separate boxes, and two separate entries on your clipboard; it's just that the two boxes have the same number in.
But when you do
p = q;
you're not touching the boxes at all. You're only changing your clipboard. You're looking up which box you should use for q, and then you're scrubbing out the entry for p on your clipboard and replacing it with the same entry that you found under q. Now you have two boxes, with different numbers in; but your clipboard contains two entries that are the same, and both direct you to the same box. One of your boxes is now unreachable (you don't have an entry on your clipboard pointing you to it), and whenever you look up p or q, you'll end up looking in the same box.
(It's worth noting that this clipboard is only used for executing this particular function, and it gets thrown away afterwards. Changes that you make to the pointers that you pass in as arguments to the function won't have any effect after the function has finished executing. But I'd get your head round the other concepts first off, before you start worrying about scope.)

Answer 2

Because you've removed the step where the int 5 is moved to *p, so you're only moving the 4 across to *q via temp.

Answer 3

Note that int *p and int *q are pointers (represent memory addresses!!). So, when you write p=q you end-up with two pointers that point to the same address!!!
void swap (int *p,int *q)
{
int temp;
temp=*p; // temp = 4
p=q; // p points at the same address with q (where 5 is stored)
*q=temp; // store 4 where q points
}
You now understand that both pointers point to a memory location where integer 4 has been stored! Both pointers point to the same memory location.

Answer 4

In general, the best method for understanding what happens in code with pointers is to bring out a pencil and some paper and start drawing.
I used my computer for this, which is much, much slower.
Since swap(4,5) doesn't actually work, I'll assume that we have two variables in the code that calls swap:
int x = 4;
int y = 5;
swap(&x, &y);
The first two lines of swap are identical in both versions, so the first two pictures apply to both.
As you enter swap, it will look something like this:
As you can see, p points at x, and q points at y.
The value of temp is indeterminate.
Now we have followed the p pointer (*p) and copied the value of x into temp.
Version one:
Here, we follow the pointer q (*q), fetching the value of y (i.e. 5) and then we follow the pointer p (*p) and store that value into x.
And lastly, we follow q so we can store the value of temp, i.e. the value that x used to have, in y.
Version two
Here, we copy the value of q into p.
Since p and q are pointers, they now point to the same thing (y).
As you can see, the variable x hasn't changed; we just changed p so it points somewhere else.
This step is exactly the same as in the first version: store the value of temp in y by following q.
Hope this makes it clearer.

Answer 5

your program will crash because you are passing constant value swap(4,5); and there is no address for 4 and 5what p and q gonna point?
void swap(int *p,int *q){
......
}
expecting two variable to whom pointer p and q can point.

Answer 6

Copy this code to your code editor and test run it. Try to understand what this code is doing using comments.
//Start from main() function
#include <iostream>
using namespace std;
// We are here from main function
/*
This function takes two arguments
both are pointer of type int. When declaring
a pointer of type int we write "int *x". Which means
that the vaiable x is pointer and it can only hold
an address of a varible instead of it's value.
However, We now have addresses of p in x and
q in y. We want to exchange values but taking
addresses of them instead. This is a bit confusing.
*/
void swap(int *x, int *y)
{
int temp = 0;
/*
Things are more confusing in the line below.
" * " is a multipurpose operator.
1. When you put this operator in the middle
of two variable (like a*b ) then it works as a
multipliyer.
2. When you put it when declaring a variable
( like int *x ) then it works as if you are
declaring a pointer of type int.
3. When you put it before a pointer variable name but not
declaring the pointer variable then it gives the
value the pointer pointing to.
In our case the third statement is happening. As x has
the address of p, *x means the value of p.
*/
temp = *x; // Here, *x = 10 (because p = 10) so temp = 10.
/*
Here, we are replacing the value of *x (or p, which is 10)
by *y (or q which is 20).
So, after the execution of the line below
we will have *x = 20 or p = 20. It means we are losing
the previous value of p. This is the reason we kept the value
of p in temp.
*/
*x = *y; // Now p = 20 and q = 20
/*
The line below sets *y (or q ) to the value of temp
(which is 10).
*/
*y = temp; //Now p = 20 and q = 10
}
int main()
{
int p = 0;
int q = 0;
cout << "Enter p : ";
cin >> p; // Say you have entered 10. Now p = 10
cout << "Enter q : ";
cin >> q; // Say you have entered 20. Now q = 20
/*
This line takes you to the swap function above.
We are not sending p, q here. We are sending pointer
or addresses of p and q.
*/
swap(&p, &q); // This line takes you to the swap function above.
cout << "p : " << p << endl;
cout << "q : " << q << endl;
return 0;
}

Answer 7

The function which works correctly:
void swap (int *p,int *q)
{
int temp; //You take an integer variable
temp=*p; //You store the value pointed to by p to temp which is 4
*p=*q; //You copy the value pointed to by q to p which is 5, hence p points to 5
*q=temp; //You store the value in temp as the value pointed by q, hence q points to 4 as temp holds 4
}
Now the code with your changes:
void swap (int *p,int *q)
{
int temp;
temp=*p; //You store the value pointed to by p to temp which is 4
p=q; //p and q are addresses, you equate them so the value pointed becomes equal, if you make change in one the other changes automatically. So basically the value of 'p' in the calling function doesn't change, the change is a local change as the arguments to the called function are passed by reference. So the calue pointed to by p is not altered in 'swap' function.
*q=temp; //Now to make the value pointed to by q as 4
}

Answer 8

You can use the xor swap algorithm:
#include <stdio.h>
int main(int argc, char** argv)
{
int i = 1;
int j = 2;
int *p = &i;
int *q = &j;
printf("BEFORE: %d - %d\n", *p, *q);
if (*p != *q)
{
/* swap */
*p ^= *q;
*q ^= *p;
*p ^= *q;
}
printf("AFTER: %d - %d\n", *p, *q);
return 0;
}
reference:
http://en.wikipedia.org/wiki/XOR_swap_algorithm