Question

#include <stdio.h>

int main(void)
{
    int a[5]={1,2,3,4,5};
    int *ptr=(int*)(&a+1);
    printf("%d %d\n",*(a+1),*(ptr-1));
    return 0;
}

Output:

2,5

I could not understand how the *(ptr-1) evaluated to 5 (correct output). But when I manually
did it was 1. My understanding is *(ptr-1) will get evaluated to *(&a+1-1) which would be
*(&a) which is 1.

Please help me to understand this concept.

Answer 1

int *ptr=(int*)(&a+1);

makes &a + 1 makes &a + sizeof (a) as a is type of int [5] which makes ptr to point to actually a[5] (invalid/beyond the defined limit)

(ptr - 1) points thus points to a[4] and *(ptr - 1) will print 5.