Why is array indexing symmetric in index and name

Question

Looking at some weird obfuscation contest code today I realized that array indexing is symmetric, in other words, x[n] is the same as n[x]. For example, consider the code below:

#include <iostream>

int main()
{
    int x[] = {0, 1, 2, 3, 4};
    std::cout << x[3] << ' ' << 3[x]; // both display 3
}

Live on Coliru

Is this indeed standard compliant, and if yes, is there any good reason why? And a bonus if you can provide a standard reference/quote.

PS: the code compiles fine with both gcc and clang

@NathanOliver Please close it if you find it, I was almost sure that there should be another question like it, but couldn't find it. — vsoftco, Sep 27 '16 at 18:42
That's indeed a good find! Although it should probably be also tagged as C++. — vsoftco, Sep 27 '16 at 18:45
@NathanOliver No problem. (cc vsoftco) A user couple days back asked the same question. — Khalil Khalaf, Sep 27 '16 at 18:48
@NathanOliver I've done the same now, though _@vsoftco's_ question title sounds way better. — πάντα ῥεῖ, Sep 27 '16 at 18:57
That is why we need to keep it. It should be a good sign post. — NathanOliver, Sep 27 '16 at 19:00
@NathanOliver _Upvoting_ always marks good dupe sign posts :) — πάντα ῥεῖ, Sep 27 '16 at 19:10

Innot Kauker · Accepted Answer · 2016-09-27T18:51:36.867

2

The reason is that in C (and C++) both expressions are equal to *(x + 3) == *(3 + x).

edited Sep 27 '16 at 18:51

answered Sep 27 '16 at 18:43

Innot Kauker

1,333
1
18
30

With pointers yes, but I see `[ ]` more like an operator (although it's clearly not for PODs). – vsoftco Sep 27 '16 at 18:43
C-style arrays decay to pointers in many contexts. This is one of these I believe. – Innot Kauker Sep 27 '16 at 18:45

Why is array indexing symmetric in index and name

1 Answers1