C/C++: with int i[3], why i[2] is the same as 2[i]?

Question

I just know this trick.

In C++:

#include <iostream>
using namespace std;

int main() {
    int i[3]={5,7,11};
    cout << 2[i];
    return 0;
}

In C:

#include <stdio.h>
int main(void) 
{  
    int i[3]={5,7,11};
    printf("%d", 2[i]);
    return 0; 
}

yes, these are allowed, both output are 11.

why i[2] can be expressed as 2[i]?

http://madebyevan.com/obscure-cpp-features/#what-square-brackets-really-mean — Biffen, Dec 01 '14 at 12:50
Mostly because arrays in C/C++ aren't "real" but are just a different way to deal with pointers. — Hot Licks, Dec 01 '14 at 12:50
Well, this is something I didn't know :) Everyday you learn something new. — XCS, Dec 01 '14 at 12:52
@HotLicks That's not true. Arrays are 100% real. They're just treated differently to other objects (they decay to pointers and aren't copyable). — Joseph Mansfield, Dec 01 '14 at 12:53
While this may be something you don't know, it is also something you don't _need_ to know... — Lundin, Dec 01 '14 at 13:30

score 11 · Accepted Answer · answered Dec 01 '14 at 12:49

11

a[i] = i[a]

*(a+i) = *(i+a)

Hence both are same and is valid.

answered Dec 01 '14 at 12:49

Gopi

1 Answers1