How does the swap function work (pointer is used)?

Question

#include <stdio.h>

void swap(int *a, int *b);

int main(void)
{
    int x = 1;
    int y = 2;

    printf("x is %i, y is %i\n", x, y);
    swap(&x, &y);
    printf("x is %i, y is %i\n", x, y);
}

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

In this code (I didn't write it btw) I don't understand how x and y are being swapped when we are actually swapping a and b in the function below. i do not know what the relationship between 'x and y' and 'a and b' is.

I know that * is a pointer, and that '&' gives the address to things, but idk how it's pointing to x and y when they're ( x and y) not mentioned in the actual function.

i don't know what 'placeholder' is being used.

Please explain this if you can!

Re "*nut idk how it's pointing to x and y*", You effectively did `a = &x; b = &y;` by calling the sub using `swap(&x, &y)`. You literally made `a` point to `x` and `b` point to `y`. — ikegami, Jul 11 '20 at 09:15

score 0 · Answer 1 · answered Jul 11 '20 at 09:12

0

When you pass the address of x and y to swap function. The parameters of this function point to new address and swap the content of two address as the function does.

answered Jul 11 '20 at 09:12

Farhad Sarvari

1,051
7
13

RobertS supports Monica Cellio · Accepted Answer · 2020-07-11T10:57:30.183

It is because a and b are pointers and you actually pass the addresses of x and y in main() as arguments to the function swap() by reference:

swap(&x, &y);

If you use the & operator it gains the address of an object.

The parameter a gets assigned by the address of x and the parameter b gets assigned by the address of y.

If you dereference the pointers a and b properly and have a correct algorithm (as it is in this case), you interchange the values of x and y inside of main() and not the values of the pointers a and b.

To speak about the algorithm itself:

int tmp = *a;
*a = *b;
*b = tmp;

int tmp = *a; - The value of x is assigned to a temporary object. This is needed since we change the value of x at the next step.
*a = *b; - The value of y is assigned to x.
*b = tmp; - The value which was stored previously in x and was stored as place between in tmp (a temporary object) is now assigned to y.

Note that at all of these steps, the * operator doesn't mean to declare a pointer twice. It is used to dereference the pointer.

For more information about "dereferencing", take a look at:

What does "dereferencing" a pointer mean?

If the values of x and y have been passed by value (without the &) and the parameters a and b were just of type int, not int * then you would only change a and b inside of swap().

I recommend you to learn more about pointers. For example in a free copy of Modern C made by community member @JensGustedt.

How does the swap function work (pointer is used)?

2 Answers2