Pointer value difference not as expected

Why does the following code display 4-5 as 4 and not -1?

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int fun(int *a, int *b){
  *a = *a+*b;
  *b = *a-*b;
}

void main(){
  int a=4;
  int b=5;
  int *p=&a, *q=&b;
  fun(p,q);
  printf("%d\t%d",a,b);
}

The sum comes as 9 (as expected), but why does the difference come as 4?

This assigns the result of *a + *b to *a, so after

*a = *a + *b; // 4 + 5

then *a is 9, so

*b = *a - *b;

makes it *b = 9 - 5, which is 4.

The problem has nothing to do with pointers. You would have the same problem if you did it without:

int a = 4;
int b = 5;
a = a + b; // assign the result of 4 + 5 (9) to a
b = a - b; // assign the result of 9 - 5 (4) to b

Also note that the program has undefined behavior since fun is declared to return an int but it doesn't return anything.

For starters all these headers

#include <string.h>
#include <math.h>
#include <stdlib.h>

are redundant because neither declaration from the headers is used in the program.

Further according to the C Standard the function main without parameters shall be declared like

int main( void )

The function fun has return type int but returns nothing.

int fun(int *a, int *b){

It should be declared like

void fun(int *a, int *b){

As for your problem then after the first statement of the function

*a = *a+*b;

the object *a was changed and in the next statement

*b = *a-*b;

there is used already a new value of the object.

Instead you could write for example

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

    *a = *a + *b;
    *b = tmp - *b;
}

If a=4 and b=5:

int fun(int *a, int *b){
   *a = *a+*b;
   *b = *a-*b;
}

This function would:

1. [a=4 and b=5]: the first line sets a to 9 = (4+5)
2. [a=9 and b=5]: the second line sets b to 4 = (9-5)

Try this instead:

void fun(int *a, int *b){
   int temp = *a+*b;
   *b = *a-*b;
   *a = temp;
}