i'm learning c pointer , please explain the difference between-->

Question

1.it is c pointer for swaping two numbers
#include <stdio.h>
void swap (int *num1, int *num2);
int main() {
int x = 25;
int y = 100;
printf("x is %d, y is %d\n", x, y);
swap(&x, &y);
printf("x is %d, y is %d\n", x, y);
return 0;
}
void swap (int *num1, int *num2) {
int temp;
temp = *num1;
*num1 = *num2;
*num2 = temp;
}
2. why this code will not work-->
#include <stdio.h>
void swap (int num1, int num2);
int main() {
int x = 25;
int y = 100;
printf("x is %d, y is %d\n", x, y);
swap(&x, &y);
printf("x is %d, y is %d\n", x, y);
return 0;
}
void swap (int *num1, int *num2) {
int temp;
temp = num1;
num1 = num2;
num2 = temp;
}

Answer 1

Directly addresses your problem (2 hours ago -- please search first!)
https://www.sololearn.com/Discuss/1582970/what-is-the-problem-in-the-code

Answer 2

the second will not work bcos the variables declared inside the swap function is not a pointer even if it is the address of x and y that is passed to it.using a pointer allows you to dereference the x and y variable