Re: Two dimensional array Question

Ben Bacarisse wrote:
> 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;
> }
> 
All true on the face of it but sizeof was not what was asked, it was 
assignment. Given..

   int *p;
   p = arr[i];

Yes, arr[i] is an array 13 of int but expressing an array as above 
yields the address of its first element, in this case int*.

   p = &arr[i][0];

The address of the int yields the same int*.

-- 
Joe Wright
"Everything should be made as simple as possible, but not simpler."
                     --- Albert Einstein ---