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

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  • Difficulty Level : Medium
  • Last Updated : 01 Jul, 2017
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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

This article is contributed by Mandeep Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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