In C, why does this "backwards" indexing work?

Question

Consider the following:

int main() {
        int a[] = {42, 9001, -1337};
        printf("%d, %d\n", a[0], 0[a]);
        return 0;
}

I didn't see 0[a] before, but it appears to do the same as a[0] (the same appears to be true for other numbers as well, not just for 0).

Is that my compiler's fault (GCC)? Is there any documentation on this behavior? What purpose does it serve?

score 16 · Accepted Answer · answered Dec 13 '14 at 13:45

16

a[0] is the same as *(a+0)
0[a] is the same as *(0+a)

We know that a+0 is same as 0+a.So *(a+0)==*(0+a) which would mean that a[0]==0[a].

answered Dec 13 '14 at 13:45

Spikatrix

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It was referenced in [r/programminghorror](https://www.reddit.com/r/programminghorror/comments/v7pyc8/comment/ibm1i77/?utm_source=bob&utm_medium=nokia5110), but in a positive way. – kdb Jun 09 '22 at 11:18

score 5 · Answer 2 · answered Dec 13 '14 at 13:43

5

a[0] is another form of *(a + 0) , which can be rewritten as *(0 + a), which can be rewritten as 0[a]

So, essentially, a[0] and 0[a] represents the same.

There is no fault or error from your compiler.

answered Dec 13 '14 at 13:43

Sourav Ghosh

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score 0 · Answer 3 · answered Dec 13 '14 at 13:47

0

If you give a[0]. compiler will make it to *(a+0).

So you are giving the 0[a]. It will make that as *(0+a). So it is working.

answered Dec 13 '14 at 13:47

Karthikeyan.R.S

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In C, why does this "backwards" indexing work?

3 Answers3