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).

Think harder.

     *p = arr[i];
means the same thing as
     *p = (arr[i] + 0);

     sizeof arr[i]
does not mean the same thing as
     sizeof (arr[i] + 0)

My claim was about two expression statements.
You're talking about array type expressions
in a context which is irrelevant to this.

Consider new.c:
Do you think new.c is missing a cast somewhere?
Do you have a compiler that thinks
that new.c is missing a cast somewhere?
Do you know under what cir***stances
cast operators are required in C?

/* BEGIN new.c */

int main(void)
{
     int arr[2][13] = {0};
     int **p1;
     int **p2;

     *p1 = arr[1];
     *p2 = &arr[1][0];
     return *p1 != *p2;
}

/* END new.c */


-- 
pete