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POHow to convert 8 17-bit integers into 17 8-bit integers efficiently
 

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  1. POHow to convert 8 17-bit integers into 17 8-bit integers efficiently
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    <p>Okay, I have the following problem: I have a set of 8 (unsigned) numbers that are all 17bit (a.k.a. none of them are any bigger than 131071). Since 17bit numbers are annoying work work with (keeping them in a 32-bit int is a waste of space), I would like to turn these into 17 8-bit numbers, like so:</p> <p>If I have these 8 17-bit integers:</p> <pre><code>[25409, 23885, 24721, 23159, 25409, 23885, 24721, 23159] </code></pre> <p>I would turn them into a base 2 representationL</p> <pre><code>["00110001101000001", "00101110101001101", "00110000010010001", "00101101001110111", "00110001101000001", "00101110101001101", "00110000010010001", "00101101001110111"] </code></pre> <p>Then join that into one big string:</p> <pre><code>"0011000110100000100101110101001101001100000100100010010110100111011100110001101000001001011101010011010011000001001000100101101001110111" </code></pre> <p>Then split that into 17 strings, each with 8 chars:</p> <pre><code>["00110001", "10100000", "10010111", "01010011", "01001100", "00010010", "00100101", "10100111", "01110011", "00011010", "00001001", "01110101", "00110100", "11000001", "00100010", "01011010", "01110111"] </code></pre> <p>And, finally, convert the binary representations back into integers</p> <pre><code>[49, 160, 151, 83, 76, 18, 37, 167, 115, 26, 9, 117, 52, 193, 34, 90, 119] </code></pre> <p>This method works, but it's not very efficient, I am looking for something more efficient than this, preferrably coded in C++, since that's the language I am working with. I just can't think of any way to do this more efficient, and 17-bit numbers aren't exactly easy to work with (16-bit numbers would be much nicer to work with). </p> <p>Thanks in advance, xfbs</p>
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    1. COWhat platform are you working on that you're worried about wasting space by storing these in 32-bit integers?
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    2. CO@Praetorian: presumably any platform that has less memory available than double the size of the data.
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    3. CO@Steve Jessop Well, it's not exactly double, just 88% more space :-) But seriously, if those numbers are expected as 17-bit integers by some other part of the program, instead of an array of bytes squeezed next to each other, the OP would have re-convert these to the 17-bit representation. It's arguable whether he's saving memory with all the format massaging code.
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