Pointer outputs in C

Question

I am trying to understand why the following piece of code produces the output of 7.

int main()
{
    int a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int b[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int c = 5;
    int *p = a;

    printf("%d", (c[b])[p]);

    return 0;
}

Sourav Ghosh · Accepted Answer · 2015-03-16T10:54:11.373

6

This is pure pointer arithmatic. Read more about pointer arithmatic here and here

Just FYI, a[5] == 5[a] == *(5 + a) == *(a + 5). Related Reading.

So, in your code, c[b] == b[c] == b[5] == 6.

Then, p being equal to a, the base address of the a array, (c[b])[p] == 6[p] == 6[a] == a[6] == 7.

edited Mar 16 '15 at 10:54

answered Mar 16 '15 at 10:46

Sourav Ghosh

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Thank you, you are always very helpful. – Promisek3u Mar 16 '15 at 10:48
1

Nice answer, more info at: http://c-faq.com/aryptr/joke.html – David Ranieri Mar 16 '15 at 10:53
1

@AlterMann Thanks for the link. Vey useful. Added in my answer. :-) – Sourav Ghosh Mar 16 '15 at 10:55
[With C arrays, why is it the case that `a[5] == 5[a]`?](http://stackoverflow.com/questions/381542/with-c-arrays-why-is-it-the-case-that-a5-5a?rq=1) – phuclv Mar 16 '15 at 11:27

score 2 · Answer 2 · answered Mar 16 '15 at 10:49

2

Well, since c is 5, c[b] is the same as b[5], which is 6. Since p points to a, 6[p] is the same as a[6], which is 7.

answered Mar 16 '15 at 10:49

David Schwartz

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Pointer outputs in C

2 Answers2