is there any defined behaviour when doing this:

__________________________
// lets assume you defined 'BYTE'
// to whatever is a 8-bit-byte on your sys;-)

struct x {
BYTE nTreeBit : 3;
BYTE nFiveBit : 5;
}

x.nTreeBit = 42;
__________________________


my i assume nFiveBit stayes untouched?

On Wed, 17 Oct 2007 07:41:08 -0700, ".rhavin grobert" <clqrq@yahoo.de>
wrote in comp.lang.c++:

> is there any defined behaviour when doing this:
>
> __________________________
> // lets assume you defined 'BYTE'
> // to whatever is a 8-bit-byte on your sys;-)
>
> struct x {
> BYTE nTreeBit : 3;
> BYTE nFiveBit : 5;
> }
>
> x.nTreeBit = 42;
> __________________________
>
>
> my i assume nFiveBit stayes untouched?

Yes, you may. And the value in nTreeBit is implementation-defined.

--
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On Oct 18, 4:30 am, Jack Klein <jackkl...@spamcop.net> wrote:
> On Wed, 17 Oct 2007 07:41:08 -0700, ".rhavin grobert" <cl...@yahoo.de>
> wrote in comp.lang.c++:

> > is there any defined behaviour when doing this:

> > __________________________
> > // lets assume you defined 'BYTE'
> > // to whatever is a 8-bit-byte on your sys;-)

> > struct x {
> > BYTE nTreeBit : 3;
> > BYTE nFiveBit : 5;
> > }

> > x.nTreeBit = 42;
> > __________________________

> > my i assume nFiveBit stayes untouched?

> Yes, you may. And the value in nTreeBit is implementation-defined.

Not if BYTE is an unsigned type. If BYTE is unsigned, nTreeBit
is guaranteed to be 2.

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