On Tue, Sep 22, 2009 at 10:48 AM, Luke Benstead <kaz...@gmail.com> wrote: > 2009/9/22 Ben Klein <shackl...@gmail.com>: >> 2009/9/23 Luke Benstead <kaz...@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.
