How to swap pointers

[Thiengineer]

I don't know how to approach this problem, please help!

The following code is in main:

main()

        {  int x=1,y=2;
           int *xp,*yp;
           .
           .
           .
           xp = &x;
           yp = &y;
           intpswap(<arg1>,<arg2>);
           x=3;y=4;
           printf("*xp = %d, *yp = %d\n",*xp,*yp); 
        }

Question:

Write a function called intpswap that swaps what xp and yp point at,
and thus printf prints "*xp = 4, *yp =3".

[horace1]

post what you have done so far

[Thiengineer]

This is what I have so far...am I on the right track? Please help!!

void intpswap(int *xp, int *yp)
{
int t = *xp;
*xp = *yp;
*yp = t;
}

[willakawill]

This is what I have so far...am I on the right track? Please help!!

void intpswap(int *xp, int *yp)
{
int t = *xp;
*xp = *yp;
*yp = t;
}

On the right track with some distance still to cover :)
with

int t = *xp;

You are assigning the value '1' to the int 't'
with

*xp = *yp;

you are assigning the value that yp points to, 2, to the memory space that xp points to, x. So x now equals 2
with

*yp = t;

you are assigning the value at t, 1, to where yp points to, y. So now y = 1
The original pointers xp and yp have not changed. In order to change them we have to pass a reference to them:

void intpswap(int **xp, int **yp)
{
   int *t = *xp;
   *xp = *yp;
   *yp = t;
}

int main() {
	int x=1,y=2;
	int *xp,*yp;

	xp = &x;
	yp = &y;
	intpswap(&xp, &yp);

	x=3;y=4;
	printf("*xp = %d, *yp = %d\n",*xp,*yp); 

	getchar();
	return 0;
}

see if you can tell us what is happening here