Why is a[i] == i[a] in C/C++ arrays?
The definition of  subscript operator operator in C, according to (C99, 18.104.22.168p2), 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 == *(a + 8)
Here, a is a pointer to the first element of the array and a 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[a]
Sample program showing above results :
a 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 email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Please Login to comment...