duanchanguo7603 2015-07-15 15:23
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有效地将压缩值写入流

Currently, I'm trying to translate some compression algorithm from existing C code.

Encoding and decoding doesn't appear difficult to me. It's more about the serialization to and from a stream (be it a file or a socket).

The input is 12 bit and the compressed output is 7 bits. But writing something to a stream always involves writing entire 8 bits.

So as there is always 1 bit remaining for each value, does that mean I would have to buffer 7 bytes just to be able to write 8 values? Which would give the following bytes (whereas all 1s belong to the first value, all 2's to the second one, etc.)

11111112
22222233
33333444
44445555
55566666
66777777
78888888

The real codec or the language being used both don't really matter (actually: the codec is G.711 and the language is Golang). So maybe the go-Tag is inappropriate.

Any clue on this?

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  • douhoushou8385 2015-07-15 16:31
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    Use binary shift operators to load seven bits at a time into a bit buffer, and whenever the bit buffer has eight bits, output that. At the end, if there are any bits leftover, output a final byte with what's in the buffer.

    So something like (don't know Go, but this should be close):

    bits = 0
    bitbuf = 0
    ... some loop ...
        ...make your seven bits ...
        bitbuf |= sevenbits << bits
        bits += 7
        if bits >= 8 {
            output(bitbuf & 0xff)
            bitbuf >>= 8;
            bits -= 8;
        }
    ...
    if bits > 0 {
        output(bitbuf)
    }
    
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