Explain the code

Question

ary[-1]

Answer 1

int ary[3]={1,2,3,4};
int *k=ary +3;
Printf("%d",k[-1]);

Answer 2

pointer arithmetic says that given a pointer p and an integer i, p + i the pointer p advanced by i * sizeof(*p) bytes. which means that the pointer "k" will point to the 3th elem ( in your case k will point to the last elem since ary has a size of 4) .

+-----+-----+-----+-----+-----+
| 1 | 2 | 3 | 4 |
+-----+-----+-----+-----+-----+
^ ^
| |
ary k == ary + 3

the syntax "array[i]" is equivalent to " *(array + i )". // the compiler converts the array operation in pointers before accessing the array elements.

so k[-1] == *(k + -1) == *(k - 1). // k is a pointer to the 3th elem, subtraction of 1 resulted in an address with decrement of 1*sizeof(*k) and by using the dereferencing operator we get the value 3.

+-----+-----+-----+-----+-----+
| 1 | 2 | 3 | 4 |
+-----+-----+-----+-----+-----+
^ ^ ^
| | |
| | |
ary k[-1] ary + 3 == k

Answer 3

Thank you