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Talk:Bitwise IO: Difference between revisions
→Real intention
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:::Endianness is an issue here because you have to specify how a tuple of N bits is to be packed into an integral unit called byte, since your I/O deals with these units, not with bits. Note, this has nothing to do with the endianness of the machine. If N=8, then, to illustrate the case, let B<sub>i</sub>=(1,0,0,0,1,0,0,1), then B is 137<sub>10</sub> using big endian and 145<sub>10</sub> using little endian encoding, and there are N! variants of packing in total. The code you provided looks like big endian. If you don't like the term ''endian'', no problem. You can simply provide a formula, like byte = Σ B<sub>i</sub>*2<sup>i-1</sup> (=little endian) or Σ B<sub>i</sub>*2<sup>8-i</sup> (=big endian).
::::As far as I know, packing into bytes is as expected in common mathematics (that is, if we want to talk about endianness, big endian): less significative bits are on the right, most significative bits are on the left, so that being A<sub>i</sub> with i ∈ [0,7] the binary digits, A<sub>0</sub> is the ''unity'' bit, which is so to say A<sub>0</sub>⋅2<sup>0</sup>. So having the binary number 10010001, this is ''packed'' into a byte as 10010001 simply. When we are interested in less than 8 bits, they should be ''counted'' (always so to say) from right to left; e.g. if I want to write the 4 bits 1001, these must be right-aligned into the ''variable'' holding the bits. But these are conventions the user of the functions must know, they are not mandatory to do the way I did. I've implemented all the stuff so that you must always right-align the bits, then the functions will shift to left so that the intended most significative bit of the ''bits datum'' becomes the leftmost bit in the ''container'' (unsigned int in the C implementation). It is enough the user of the functions gets knowledge about how your functions extract bits (in which order) from the data they want to pass; then, it is ''intended'' that the first (being it the leftmost or the rightmost according to your convention) must be the first of the output stream. So that it will be the first bit you read when you ask for a single bit from the so output stream. Maybe your misunderstanding comes from the fact that you handle single bits (in your expl, still not looked your code) as unit held by an array. In C it would be like having char bitarray[N], where each ''char'' can be only 0 or 1. Doing this way, you can decide your preferred convention, i.e. if bitarray[0] is the first bit to be output or it is instead bitarray[N-1]. In C this implementation would be hard to use; if I want to write integers less than 16, which can be stored in just 4 bits, it is enough I pass to the function the number, like 12 (binary 1100), and tell the function I want just (the first) 4 bits; otherwise, I (as user of the functions) should split the integer into chars, each rapresenting a bit, pack it into an array in the right ''endianness'', and then pass the array to the function. Maybe this is easier in Ada (I don't know), but it would be harder in C. Consider e.g. the integers output array of the LZW task; in C, it is enough to take one integer per time, and say to the function to output (e.g.) 11 bits of that integer (of course, 11 bits must be enough to hold the value!); you are outputting 11-bits long ''words'' (which are the ''words'' of the LZW dictionary). Hope I've understood well what you've said and expressed well my explanation (too long as usual, sorry). --[[User:ShinTakezou|ShinTakezou]] 21:53, 20 December 2008 (UTC)
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