How this code is working please explain the output

Question

#include <stdio.h>
void f(char**);
int main() {
char *argv[] = {"ab", "cd", "ef", "gh", "ij", "kl"};
f(argv);
return 0;
}
void f(char **p){
char *t;
t = (p += sizeof(int))[-1];
printf("%s\n", t);
}

Answer 1

To understand this code, we should first realize that

a[n] is equivalent to *(a+n)

Assuming sizeof(int) returns 4, we can handtrace f(argv) easily.

t = (p += sizeof(int))[-1]

is equivalent to

t = *((p += 4) - 1)

(p += 4) returns a pointer which points to the fifth element in p. Deducting it by 1 causes the pointer to point to the fourth element of p. This value is retrieved and stored in t. The value of t is then printed.

"gh" is printed.

Answer 2

#include <stdio.h>

int main(void) {
char x[5] = {1, 2, 3, 4, 5};
char *ptr = (char*)(&x+1);
printf("%d%d\n", *(x+1), *(ptr-1));
return 0;
}
How's this works