How does this array program in C give the result 10?

Question

I am using Ubuntu 12.04lts with the GCC compiler. This program gives the result 10. Could you anybody please describe why this program gives the result like this?

 #include <stdio.h>
 void main(void)
 {
     int arr[1] = {10};
     printf("\n%d\n\n", 0[arr]);
 }

digital_revenant · Accepted Answer · 2013-10-19T18:19:08.913

5

arr[0] gets internally expanded to *(arr+0). Similarly 0[arr] gets expanded to *(0+arr) which points to the same thing. Hence you see 10.

In general for an array or a pointer a, a[b] always means *(a+b) where a is the starting address of the array or pointer and b is the offset. Thus, a[b] and b[a] are equivalent.

edited Oct 19 '13 at 18:19

answered Oct 19 '13 at 18:13

digital_revenant

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In other words: the very definition of `a[b]` is `*(a+b)`, thus the subscript operator, despite its asymmetrical appearance, is commutative. – Matteo Italia Oct 19 '13 at 18:15
1

And for a pointer, `p` and integer `i`, `p[i]` and `i[p]` reference the same object too. (Not just arrays, but pointers too.) – Jonathan Leffler Oct 19 '13 at 18:20
@JonathanLeffler yes, of course. I will mention that. – digital_revenant Oct 19 '13 at 18:21

score 0 · Answer 2 · answered Oct 19 '13 at 18:17

0

below line means arr is int type array and it has size 1 and it is initialized with 10 ie index 0 has 10

int arr[1] = {10};

then next line printf statement printing the value of arr at index 0.

printf("\n%d\n\n",0[arr]);

answered Oct 19 '13 at 18:17

Rocker

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How does this array program in C give the result 10?

2 Answers2