cppcheck sept 18 redux

[Mike Kaplinskiy]

On Tue, Sep 22, 2009 at 10:48 AM, Luke Benstead <kazade at gmail.com> wrote:
> 2009/9/22 Ben Klein <shacklein at gmail.com>:
>> 2009/9/23 Luke Benstead <kazade at gmail.com>:
>>> If it IS the case that this doesn't cause a crash and is perfectly
>>> valid, can someone explain to me how/why this works? Or point me (no
>>> pun intended) to the bit in the C spec that explains it? Coz the way I
>>> read it, it has to dereference dmW, otherwise how would the compiler
>>> find the address of the array? ... so confused :)
>>
>> I believe it's because the array (as a pointer) is at the same
>> location as start of the struct (as a pointer). Compiler then applies
>> pointer arithmetic without dereferencing.
>>
>
> Ah, I see.. but in that case, how is an array different to using a
> pointer, like in Vitaliy's example? Surely that's the same thing
> essentially?
>
> Luke.
>
>
>

It basically has to do with semantics. If you have ``char arr[5];'',
the ``arr`` will refer to the address of the first element, ie
``arr==&arr''. The same applies for fixed size arrays in structs -
because the array is allocated as part of the struct,
``dmW->dmFormName'' refers to the address of the 0th element (which as
the compiler knows is ``dmW+__offset of dmFormName''), or as above
``dmW->dmFormName=&dmW->dmFormName''.

For the pointer case (``char *arr;''), ``arr'' would be an actual
pointer - that is, ``&arr != arr'', since it stores an address and has
a location of its own (whereas fixed length arrays just have a
location).

This would make a nice interview question.

Mike.