16-bit to 10-bit conversion code explanation

Question

I came across the following code to convert 16-bit numbers to 10-bit numbers and store it inside an integer. Could anyone maybe explain to me what exactly is happening with the AND 0x03?

// Convert the data to 10-bits    
    int xAccl = (((data[1] & 0x03) * 256) + data[0]);    
    if(xAccl > 511) {    
      xAccl -= 1024;  
    }

Link to where I got the code: https://www.instructables.com/id/Measurement-of-Acceleration-Using-ADXL345-and-Ardu/

Take a random 16-bit value, and do the operations one by one *on paper*. See what you get at each step of the way. — Some programmer dude, Oct 20 '18 at 14:29
`data[1] & 0x03` means to take the number in `data` at index 1 and compare the binary representation of it to the binary representation of the `0x03` (this is hex). See [this](https://stackoverflow.com/questions/4757447/understanding-the-behavior-of-a-single-ampersand-operator-on-integers) for how the `&` comparison works. — CodingYoshi, Oct 20 '18 at 14:34
Also please don't spam with unrelated language tags. Judging from the context of the link its code from an Arduino project, which means it's C++. — Some programmer dude, Oct 20 '18 at 14:36
Sorry about that, its code taken from an Arduino project yes, but my code is written in a combination of C, C++ and C# for the different aspects of the project, but will definitely keep that in mind, thanks again! — Zandré van Heerden, Oct 20 '18 at 14:44

score 1 · Answer 1 · answered Oct 20 '18 at 14:29

1

The bitwise operator & will make a mask, so in this case, it voids the 6 highest bits of the integer.

Basically, this code does a modulo % 1024 (for unsigned values).

answered Oct 20 '18 at 14:29

Matthieu Brucher

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Oh, the solution is so simple, for some reason I just did not get it, thank! – Zandré van Heerden Oct 20 '18 at 14:34
This is not quite a modulo operation, because it result is n the [-512,511] range. – 1201ProgramAlarm Oct 20 '18 at 21:11
Hence the comment in the parentheses. – Matthieu Brucher Oct 20 '18 at 21:54

score 0 · Answer 2 · answered Oct 20 '18 at 14:33

data[1] takes the 2nd byte; & 0x03 masks that byte with binary 11 - so: takes 2 bits; * 256 is the same as << 8 - i.e. pushes those 2 bits into the 9th and 10th positions; adding data[0] to data combines these two bytes (personally I'd have used |, not +).

So; xAccl is now the first 10 bits, using big-endian ordering.

The > 511 seems to be a sign check; essentially, it is saying "if the 10th bit is set, treat the entire thing as a negative integer as though we'd used 10-bit twos complement rules".

Thank you for the clear explanation, really appreciate it. Was wondering about the + myself, but might just edit that to |. Thank you! — Zandré van Heerden, Oct 20 '18 at 14:38

16-bit to 10-bit conversion code explanation

2 Answers2