# Why is a[i] == i[a] in C/C++ arrays?

The definition of [] subscript operator operator in C, according to (C99, 6.5.2.1p2), is that

E1[E2] is identical to (*((E1)+(E2)))

Compilers use pointer arithmetic internally to access array elements. And because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th element of E1 (counting from zero).

Therefore, **a[b]** is defined as :

a[b] == *(a + b)

So will evaluate to

a[8] == *(a + 8)

Here, a is a pointer to the first element of the array and a[8] is the value of an elements which is 8 elements further from a, which is the same as *(a + 8) and **8[a]** will evaluate to following which means both are same.

8[a] == *(8 + a)

So by addition commutative property, **a[8] == 8[a]**

Sample program showing above results :

`// C program to illustrate` `// a[i] == i[a] in arrays ` `#include <stdio.h>` `int` `main()` `{` ` ` `int` `a[] = {1, 2, 3, 4, 5, 6, 7};` ` ` `printf` `(` `"a[5] is %d\n"` `, a[5]);` ` ` `printf` `(` `"5[a] is %d\n"` `, 5[a]);` ` ` `return` `0;` `}` |

Output:

a[5] is 6 5[a] is 6

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