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Why does the below code not generate an error and print the value stored in arr then followed by a junk value ?

int main() {
    int arr[1]={10};
    printf("%d %d\n",0[arr], 1[arr] );
    return 0;
}
ArtKorchagin
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Rahul Gautam
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2 Answers2

2

In C, 0[arr] == arr[0].

So,

0[arr]==arr[0]==10.

and

1[arr]==arr[1]==Junk value

Check this SO question and it's answers.

The C standard defines the [] operator as follows:

a[b] == *(a + b)

Therefore a[5] will evaluate to:

*(a + 5) and 5[a] will evaluate to:

*(5 + a) and from elementary school math we know those are equal. (Addition is commutative.)

This is the direct artifact of arrays behaving as pointers, "a" is a memory address. "a[5]" is the value that's 5 elements further from "a". The address of this element is "a + 5". This is equal to offset "a" from "5" elements at the beginning of the address space (5 + a).

Community
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Keshava GN
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  • Since you knew a duplicate question existed, next time just close it as a duplicate, instead of posting an identical answer. – Lundin Dec 11 '15 at 10:42
0

1[arr] is the same as arr[1].

This is possible because E1[E2] is equivalent to (*((E1) + (E2)) in C by definition of [] operator.

Note that if you declare:

int arr[1]={10};

There is only element in the array, arr[0] and there is no arr[1] element.

ouah
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