achan1989
I want tidy code that creates an opcode from two pieces, and encodes it as a little endian binary. Where “the OGF occupies the upper 6 bits of the Opcode, and the OCF occupies the remaining 10 bits.” This code is almost what I want, except that the resulting binary is big endian:
def opcode_bin(ogf, ocf), do: <<ogf::6, ocf::10>>
<<16, 1>> = opcode_bin(4, 1)
# I want <<1, 16>>
The following works, but the intermediate step isn’t very elegant:
def opcode_bin(ogf, ocf) do
<<opcode::16>> = <<ogf::6, ocf::10>>
<<opcode::little-16>>
end
Is there a nicer way to accomplish this, maybe by specifying that I want little-endian output directly?
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NobbZ
Perhaps using
Binary.<<</2.Something like
x = ogf <<< 10 ||| ocf; <<x::little-16>>.That way you don’t have to pack just for immediate unpacking.
Maybe you can even inline
xinto the binary, I’m never sure what’s possible within a binary literal, I use them to rarely for creating values.aziz
Careful with the
Bitwiseoperators.ogf <<< 10would only shift and not mask like<<ogf::6>>. Masking cuts off the extra bits. So as an expression it can be:I guess you’d agree that’s hardly more readable. Alternatively you could create a handy
swap16function that swaps its bytes. Then you can have everything in one line and the intent is clear as day.NobbZ
Trimming would have happened in the
little-16part. At least forogf, but indeed I’ve forgotten aboutocf(or did I just assume that data has been sanitizer already?)achan1989
Interesting. Where are the docs for that? I had no idea that was a thing.
NobbZ
Sorry, my fault, it was
Bitwise, notBinary… That is the kind of stuff that happens when one replies to technical topics while waiting for the train home…