How can I swap two numbers by address?

Question

#include<stdio.h>
void fun(int **a,int **b)
{
    int **k;
    k=a;
    a=b;
    b=k;
}
int main()
{
    int a=3,b=6,*x=&a,*y=&b;
    fun(&x,&y);
    printf("%d %d",a,b);
    return 0;
}

I swap the address,why it still outputs 3 6. So what's the correct way to swap the two address?

Answer 1

p =... never persists across the function call. Like everything else in C, if you want to change the caller's data, dereference the pointers you're given. k should be int * , and the assignments should be int *k = *a; *a = *b; *b = k; int *k = *a; *a = *b; *b = k; . that's it. Note also this changes nothing about the original a and b . All your swapping at this point are pointer values.

This code swaps two integers

#include<stdio.h>

void fun(int *a,int *b)
{
    int k = *a;
    *a = *b;
    *b = k;
}

int main()
{
    int a=3,b=6;
    printf("%d %d\n",a,b);
    fun(&a,&b);
    printf("%d %d\n",a,b);
    return 0;
}

Output

3 6
6 3

Whereas, this code swaps two pointers. The integers they point to (a and b) remain as-is.

#include<stdio.h>

void fun(int **a,int **b)
{
    int *k = *a;
    *a = *b;
    *b = k;
}

int main()
{
    int a=3,b=6,*x=&a,*y=&b;
    printf("%d %d\n",a,b);
    printf("%d %d\n",*x,*y); // note using x and y with deref
    fun(&x,&y);
    printf("%d %d\n",a,b);
    printf("%d %d\n",*x,*y); // note using x and y with deref
    return 0;
}

Output

3 6
3 6
3 6
6 3

Answer 2

Lets make it clear from the first expression.

int a=3,b=6,*x=&a,*y=&b;

Now breaking it down.

int a = 3, b = 6 

Here a and b are two memory locations. Say a is 10000000 and b is 10010000 . So, after we write a = 3, b = 6 the value 3 takes the place in memory location 10000000 and 6 in 10010000 . Now,

*x=&a,*y=&b;

means, x is an integer pointer that has the address of integer variable a which is 10000000 and y has the address of b which is 10010000 . Note that, both x and y are also variable which can store the address as value of any integer variable. x and y has locations in memory let say 11000000 and 11011111 . Now,

fun(&x,&y);

You are passing the address of x and y to the function fun . It looks like fun(11000000, 11010000) . Note that, you are not passing the address of a and b ! In function fun these addresses are in a ( pa for clarity ) and b ( pb for clarity),

void fun(int **a,int **b)
{
    int **k;
    k = a;
    a = b;
    b = k;
}

Here, a ( pa ), b ( pb ) and k are pointers of pointer. They can hold the address of an integer pointer. a ( pa ) and b ( pb ) are holding the address of x and y respectively. Now when you swapping a ( pa ) and b ( pb ) you actually swapping the addresses ( of x and y ) in a ( pa ) and b ( pb ). So, after the function fun finishes it's execution a ( pa ) is pointing y and b ( pb ) is pointing x . In main function, nothing actually happened. When you do the following,

void fun(int **a, int **b)
{
    int *k;
    k=*a;
    *a = *b;
    *b = k;
}

it swaps the address in x and y and they now pointing to b and a (in main function).

Now the final version comes, If you would do the following, it actually make you happy ,

void fun(int **a, int **b)
{
    int k;
    k=**a;
    **a = **b;
    **b = k;
}

I hope you understand it. Happy coding !

Answer 3

Try this to swap the values to which *a and *b of fun() points

void fun(int **a, int **b)
{
int k;
k = **a;
**a = **b;
**b = k;
}

Answer 4

Instead of this:

int **k;
k=a;
a=b;
b=k;

you should have tried this:

int *k;
k=*a;
*a=*b;
*b=k;

And yeah, you should print *x and *y , because those are the variables you swap, not a and b .