Re: Two dimensional array Question

Ben Bacarisse wrote:
> pete <pfiland@[EMAIL PROTECTED]
> writes:
> 
>> Ben Bacarisse wrote:
>>> pete <pfiland@[EMAIL PROTECTED]
> writes:
> <snip>
>>>> 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.
> 
> OK, but you'll have to help me if you think I've missed something.  I
> really do think I get it.
> 
> I know you meant "in most expression contexts"

I didn't mean that.
I meant that given

     int arr[2][13], **p, i;

then

     *p = arr[i];

means exactly the same thing as

     *p = &arr[i][0];


-- 
pete