Confused with C pointers - c

Question

#include<stdio.h>
// Function to swap two values but does not work
void swapDoesNotWork (int *ptrX, int *ptrY);
// Function to swap two values and works fine
void swap (int *ptrX, int *ptrY);
void swap (int *px, int *py) {
int temp;
temp = *px;
*px = *py;
*py = temp;
}
void swapDoesNotWork (int *px, int *py) {
printf("\n\n");
int temp;
temp = *px;
px = py;
py = &temp;
}
int main() {
int x = 5;
int y = 10;
swapDoesNotWork(&x, &y);
printf("++++++++++++++++++\n");
printf("pre x:%d\n", x);
printf("pre y:%d\n", y);
printf("\n");
printf("After calling swapDoesNotWork(&x, &y)...\n");
printf("post x:%d\n", x);
printf("post y:%d\n", y);
printf("++++++++++++++++++\n\n");
x = 5;
y = 10;
printf("= = = = = = = = =\n\n");
printf("pre x:%d\n", x);
printf("pre y:%d\n", y);
swap(&x, &y);
printf("\n");
printf("After calling swap(&x, &y)...\n");
printf("post x:%d\n", x);
printf("post y:%d\n", y);
printf("= = = = = = = = =\n\n");
return 0;
The output of the above program when compiled and executed is:
infi#linux% ./swap_test.o
++++++++++++++++++
pre x:5
pre y:10
After calling swapDoesNotWork(&x, &y)...
post x:5
post y:10
++++++++++++++++++
= = = = = = = = =
= = = = = = = = =
pre x:5
pre y:10
After calling swap(&x, &y)...
post x:10
post y:5
As can be seen, the swapDoesNotWork function does not seem to change the values as is the case with the swap function.
I am new to C language, coming from mostly scripting background. Can someone help me why swapDoesNotWork functon is not changing the values?

Answer 1

Here are all the modifications performed by swapDoesNotWork:
temp = ...
px = ...
py = ...
All of these are assignments to local variables. Local variables are destroyed when the function returns, so swapDoesNotWork has no lasting effects. It only changes variables that are about to stop existing anyway.
On the other hand, swap contains these lines:
*px = ...
*py = ...
These are assignments to locations pointed to by px and py, respectively. Even though px and py are local variables themselves, they can point to variables outside of the current function (in this case the function is called as swap(&x, &y), so they point to the x and y variables in main).

Answer 2

With what you are doing you just cancelled the whole point of using pointers.
What you are doing is basically the same as this:
temp = a;
a = b;
b = temp;
which means that you are acting on a copy of the original variables. So what you do although it is changing the values of the pointers (if you try a printf inside the function the results should be correct) after the function is finished the results disappear. In order for that function to work you would have to pass as an argument a double pointer. Hope I helped