Re: Two dimensional array Question

pete <pfiland@[EMAIL PROTECTED]
> writes:

> mdh wrote:
>>>
>>>> The exercise associated with this, used the construct
>>>> *p = arr[1 or 0 ] to point to either the "first" or "second" row of
>>>> the array. Does the compiler "know" where to point to because it has
>>>> been given this information by the declaration of "13" in arr[2][13].
>>>> Hopefully this makes sense too!
>>> Yes and yes.
>>>
>>  I was in a hurry to get to work this am...so forgot to ask this.
>> In a multidimensional array, one can simply drop the second []?  as in
>> *p=arr[i] and this is legal?
>> Thanks.
>
> It's legal regardless of whether or not
> you understand what you're writing.
> But it's better if you realize that
>
>     *p = arr[i];
>
> means the same thing as
>
>     *p = &arr[i][0];

The trouble is this phrase "means the same as".  Both right hand sides
are certainly closely related expressions, but they have different
types, different sizes (sizeof arr[i] != sizeof &arr[i][0]) and they
can't be used in the same places (you can, for example take the
address of one but not of the other).

Sometimes it helps to gloss over these differences and sometimes it
does not.  I am not sure which is the case here.

> because
>
>     &arr[i][0] means &*(*(arr + i) + 0)
>
> which can be simplified to
>
>     (*(arr + i) + 0)
>      *(arr + i)
>        arr[i]

C is not good for equational reasoning.  The OP might find this
helpful:

#include <stdio.h>

#define PSZ(exp) printf("sizeof " #exp " = %d\n", (int)sizeof exp)

int main(void)
{
     int i = 0;
     int arr[2][13] = {0};

     PSZ(&arr[i][0]);
     PSZ(&*(*(arr + i) + 0));
     PSZ((*(arr + i) + 0));
     PSZ(*(arr + i));
     PSZ(arr[i]);

     return 0;
}

-- 
Ben.