Re: [C or C++] Is this legal? sizeof *p

From: Jeff Schwab (jeffplus_at_comcast.net)
Date: 01/04/04 

Date: Sun, 04 Jan 2004 01:38:42 -0500

Jack Klein wrote:
> On Sat, 03 Jan 2004 17:17:30 -0500, Jeff Schwab <jeffplus@comcast.net>
> wrote in alt.comp.lang.learn.c-c++:
>
>
>>Jason wrote:
>>
>>>{
>>>struct a_struct *p;
>>>
>>>p = malloc( sizeof *p );
>>>}
>>>
>>>That should be legal shouldn't it? Despite the fact that p doesn't point to
>>>anything.
>>>
>>>* For c++ p = static_cast<struct a_struct*> malloc(sizeof *p );
>>>
>>>
>>
>>
>>Nope. It usually works fine, but don't count on it. The pointer might
>>not hold a valid address, and the simple act of dereferencing it might
>>cause a bus error. Sorry.
>
>
> Jeff, you are completely wrong about this one.
>
> In C++, and in C except for one special case in C99, the sizeof
> operator is guaranteed not to evaluate its operand, merely use the
> operand to identify the type and hence the size.
>
> So this code is perfectly legal in both C and C++.
>
> The special case in C99 that I mentioned above does not apply here.
> That is for the a VLA (variable length array) which was added to C in
> 1999, where the size of an automatic array is defined at run time. In
> that case if sizeof is applied to a VLA, it must obviously be
> evaluated at run time. But applying sizeof to a VLA is applying
> sizeof to an array, not a pointer, and using the name of an array as
> an operand of the sizeof operator is one of the few situations when
> the array name is NOT silently converted to a pointer.
>

D'oh! Thank you for correcting me. Of course you are entirely right.

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