Hi: Thank you all very much. 2012/2/21 Bernd Petrovitsch <bernd@xxxxxxxxxxxxxxxxxxx>: > On Die, 2012-02-21 at 20:30 +0800, Tao Jiang wrote: > [...] >> Now I know in the most modern machine there is no difference between BE and LE >> at so called 'bit order' level. >> Right? > > One main difference between *byte* order and *bit* order is: > > What are the means to address individual *bits*? > a) Bit shift and masking as in "1 << bit-number": > This has a mathematical background and - implicitly - the > least-significant bit has - thus - the number 0. > I can't even think of an insane reason (let alone a sane one) to > break the "shift left is for unsigned numbers equivalent to > doubling" property - apart from the fact that it is defined in that > way by C - and all other languages I came across. And the same holds > for all CPUs/assembler instruction sets .... > b) use a bit-field as in "unsigned char b0:1, b1:1, b2:1, b3:1, b4:1, > b5:1, b6:1, b7:1;": > It is not defined by any C-standard and is - thus - up to the > compiler, if b0 == (1 << 0) or b0 == (1 << 7) or anything else. > c) bit-test/st/clr assembler instructions in the architecture: Go read > *if* your CPU has such stuff and how it relates to the "bit-shift and > mask" method. > I would be greatly surprised if it is different (on i386, it is equal > since ages BTW) mainly because it makes absolutely no sense. > d) There is hardware with bit-addressable memory out there. Go read the > manual and the same as c) > I doubt that it is different even for really old machines .... > > Bernd > -- > Bernd Petrovitsch Email : bernd@xxxxxxxxxxxxxxxxxxx > LUGA : http://www.luga.at > _______________________________________________ Kernelnewbies mailing list Kernelnewbies@xxxxxxxxxxxxxxxxx http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
