CODEWITHSUNDEEP
c
John Nash
John Nash

Reputation: 77

Multidimensional arrays and Addresses

I wrote the following code In c just to check whether the code would break or not:

int main(void)
{
    int A [5] [2] [3];
    printf("%d\n\n", A[6]);
    printf("%d\n\n", &A[6][0][0]);
    system("pause");
}

Now, the code does not break which was something I was not expecting. When we declare a multidimensional array: int A [5][2][3], doesn't that conceptually mean that A in its first level is a one-dimensional array of 5 elements ( 0 - 4 ) and every element of that array is itself a one-dimensional array of 2 elements and every element of that array is a one-dimensional array of 3 elements? If that concept is correct, how can A[6][0][0] even exist - since in the first level we only have 5 elements ( 0 based ) .

Any help, would be greatly appreciated.

Upvotes: 0

Views: 114

Answers (2)

K-ballo
K-ballo

Reputation: 81349

You are accessing a position outside the array, there is no A[6]. That's undefined behavior, and anything could happen.

Note that A[5] is a well defined location (past the end of the array), so getting a pointer to it is legal but trying to access that pointer is not. However, getting a pointer to A[6] or any other greater index is completely undefined.

Upvotes: 5

pb2q
pb2q

Reputation: 59617

C doesn't do array bounds checking. For int A[4], think of A[5] as like pointer arithmetic: *(A + 5).

Even if A has only been declared with 5 elements (0-4), C will allow you to reference memory outside the declared bounds of your array, using A.

The result of this access is undefined. If you read A[5], you might get garbage; if you write A[5], i.e. A[5] = 13, you may corrupt another allocated block of memory.

The wiki article is a pretty good start for understanding this, but I recommend that you pick up a copy of K&R, read it, and do all the exercises.

Upvotes: 0

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