overflowing bitfields

[.rhavin grobert]

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?

[Jack Klein]

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.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.club.cc.cmu.edu/~ajo/docs/FAQ-acllc.html

[James Kanze]

On Oct 18, 4:30 am, Jack Klein <[e-mail]:

On Wed, 17 Oct 2007 07:41:08 -0700, ".rhavin grobert" <[e-mail]>
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.

--
James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]