What is 0[p] doing?

Question

What is the following C code doing?

int i;
int* p = &i;
0[p] = 42;

I would have though that this would not event compile. But it even executes without a segmentation fault. So I wonder what strange use of the [] operator I have missed.

score 13 · Accepted Answer · edited Dec 27 '13 at 16:00

13

The C Standard defined the operator [] this way:

Whatever a and b are a[b] is considred as *((a)+(b))

And that's why 0[p] == *(0 + p) == *(p + 0) == p[0] which is the first element of the array.

edited Dec 27 '13 at 16:00

Danvil

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answered Dec 27 '13 at 13:13

rullof

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score 9 · Answer 2 · answered Dec 27 '13 at 13:05

9

0[p] is equivalent to p[0]. Both are converted as

0[p] = *(0+p) and p[0] = *(p+0)

From above statements both are equal.

answered Dec 27 '13 at 13:05

Chinna

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score 7 · Answer 3 · answered Dec 27 '13 at 13:04

7

0[p]

in 0[p] = 42;

is equivalent to p[0]

+ operation is commutative and we have:

p[0] == *(p + 0) == *(0 + p) == 0[p]

answered Dec 27 '13 at 13:04

ouah

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What is 0[p] doing?

3 Answers3