why is the output 1 in this case and when i am doing t-p then its giving me -1

Question

As you can see it is giving me 1 but why? When i do t-p then it gives me -1 . Why ?

int f=4, o=8;
int *p, *t;
p = &f;
t = &o;

printf("Difference between the two pointer is %d\n",p-t); 

Answer 1

p-t is not well-defined behavior. You cannot do pointer arithmetic unless the pointers point at the same array. Any kind of result is possible.

Furthermore the result of subtracting two pointers is a large integer type called ptrdiff_t. To print that one with printf you need to use %tu.

Although it is quite possible that the two variables are allocated adjacently in memory with 1 sizeof(int) bytes in between. Do this instead:

#include <stdint.h>
printf("%llu\n", (uintptr_t)p);
printf("%llu\n", (uintptr_t)t);

On a gcc x86 Linux PC I get:

140720377482824
140720377482828

So the result -1 or 1 aren't very strange despite invoking undefined behavior.