How to convert to left justified bits?

Quote:

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Originally Posted by Subsonics

Having thought this over I think that what is refered to as unused bits would be to the right of 1 only, otherwise I can´t understand how the data could be understood by the recieving end. In this case the value 1 would be encoded as: 00000000 00000000 00100000 (16bit). I have discussed this earlier here perhaps the older solution was the correct one. That involved copying the short to a byte array and performing right shifts to spread out the data over three bytes.

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Based on the specs, I would agree with you. I don't know that you have to use a byte array for the original data really; masking would seem to be the way to go. (Although a byte array for the output would work.)

I think my confusion came from that I wan´t to encode 12bit samples. Right now I can only assume that it´s encoded just like 16bit but with the highest legal sample value being 4095. I guess that I will have to try and see. :-) I don´t remember exactly how I ended up solving it the last time but, I think masking was used to get the value of the last byte with the the 2 least significant bits.

If it happens to be encoded into two bytes I think I could use something like this:

Where 00001111 11111111 will become 01111111 01111100

Code:

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        UInt16 sample = 0xfff;
        UInt8  outputBuf[2];
        sample <<= 3;
        memcpy(&outputBuf[0], &sample, 2);
        outputBuf[0] >>= 1;

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Subsonics, based on the spec for MIDI you posted, it looks like you can't just shift 16-bits... Instead you need to split up into groups of 7 bits because the spec calls for the high bit to be zero in each byte. Only 7 bits are used. So not only shifting the data left but spreading it out some as you cross the byte boundaries.

Quote:

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Originally Posted by nonoob

Subsonics, based on the spec for MIDI you posted, it looks like you can't just shift 16-bits... Instead you need to split up into groups of 7 bits because the spec calls for the high bit to be zero in each byte. Only 7 bits are used. So not only shifting the data left but spreading it out some as you cross the byte boundaries.

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I´m not sure what you mean now. In the post above I have done that, the MSB is padded with zero in both bytes. First I shift three bits to the left which leaves me with one leading zero, then I copy the short across two bytes in the output buffer and make a right shift to the last byte. If you refer to the original post then that´s only one part in solving this that I couldn´t figure out how to do. But I don´t think I will need it now as I´m pretty sure that the left justifying that is mentioned in the doc only is relevant for the least significant byte.