This is interesting only for C programmers, and probably not useful even to them.
But you’re still reading, so I guess you’re really jobless. So here it is:
Anyone familiar with pointers in C knows the output of this code:
int arr[3] = {5,6,7};
printf("%d",arr[2]);
(OUTPUT : 7)
But did you know that even this code works :
int arr[3] = {5,6,7};
printf("%d",2[arr]);
(OUTPUT : 7)
Amazing, huh?
(Nah, not really… I just write these articles to show people that they’re geeks! But do read on…)
Well, here’s how it works:
You probably know that arr[2] is equivalent to *(arr + 2)
Actually, arr[2] converts internally to *(arr + 2) before evaluation
So, if
arr[2] => *(arr + 2)
Isn’t it quite reasonable to expect that
2[arr] => *(2 + arr)
Well, reasonable or not, thats how it works! And I’m sure you know *(x + 2) is equivalent to *(2 + x), so there!
Makes perfect sense doesn’t it!
arr[2] is equivalent to 2[arr] !!!!!!!!!!!!!!!!
other tongue 