CODEWITHSUNDEEP
carrayspointers
Krishna sundar
Krishna sundar

Reputation: 59

array indexing method a tricky case

#include<stdio.h>
int main() {
int buff[] = {1,2,3,4,5,6,9,10};
char c = (buff+1)[5];
printf("%d\n",c);//output is 9
return 0;
}

Can someone explain it clearly how this is happening and why

Upvotes: 2

Views: 109

Answers (2)

Jacob Pollack
Jacob Pollack

Reputation: 3751

Recall:

In C the square braces [ ] are implicitly *( ... ).

What is going on in the snippet of code you provided is not obvious pointer arithmetic. This line:

char c = (buff+1)[5];

... is equivalent to the following (by the C standard):

char c = *( ( buff + 1 ) + 5 );

... which points to the 7th element in the array (6th position) and dereferences it. It should output 9, not 19.

Remark:

Following the note about square braces, it's important to see that the following is equivalent.

arr[ n ] <=> n[ arr ]

... where arr is an array and n is a numerical value. A more complicated example:

' '[ "]; i < 0; i++; while ( 1 ); do something awesome (y)." ];

... of entirely valid pointer arithmetic.

Upvotes: 8

P0W
P0W

Reputation: 47784

{1,    2,  3,  4,  5,  6,  9,  10};
 |     |
 buff  buff+1 = {2, 3, 4, 5, 6, 9, 10} (say buff_1)
                                |
 buff_1[5]    =                 9

Upvotes: 6

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