Re: How to figure out the byteorder only with one byte number?
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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
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