int a[5]={1,2,3,4,5}; int *p= (int*) (&a+1); printf ("%d", * (p-1)); Why is the answer 5?

A is an array name, which is the first address of an array.
Take the address operator to a, get a pointer to the array!!! This sentence is particularly important!
It is equivalent to
int (*p) [5] = &a;
P is a pointer to an array containing 5 int elements!!

Then after executing p+1, the offset of P is equivalent to P + sizeof (int) * 5!!

And the program is forced to convert the pointer p to a int* so p-1 is actually p-sizeof (int)
So, p-1 points to the last element in the array, which is 5.

int a[5]={1,2,3,4,5}; int *p= (int*) (&a+1); printf ("%d", * (p-1)); Why is the answer 5?