2 ms·

This looks magical :-) Are you sure you're decoding bits from an octet stream (normal 8-bit bytes) and not using logical booleans? I ask this because I don't s

• fleitz 16y ago
  Yes very sure. nthbit will extract the nth bit from an integer type. The bitwise logic is the &&& and <<< in F# you use triples for bitwise. But there is a bug in there, i forgot to times sz by 8 http://msdn.microsoft.com/en-us/library/dd469495.aspx http://msdn.microsoft.com/en-us/library/dd469495.aspx bitstream turns a sequence of integer types into a sequence of bits. So if you wanted to turn a stream into bits you'd do something like the following. new System.IO.FileStream('foo.txt') |> Seq.unfold (fun s -> (s , match s.read with | -1 -> None | x -> Some(x))) |> bitstream I just realized that the bitstream function is unnecessary and you could instead write let bits (bitContainer:'T) : seq<bool> = [0..(sizeof<'T>*8)] |> Seq.ofArray |> Seq.map (fun nth -> (bitContainer &&& ((1 :> 'T) <<<nth)) And change the above code to: new System.IO.FileStream('foo.txt') |> Seq.unfold (fun s -> (s , match s.read with | -1 -> None | x -> Some(x))) |> Seq.map bits |> Seq.concat
  • fleitz 16y ago
    Just for fun here's the reverse to turn a sequence of booleans into a sequence of integer types (byte,int,long,etc). Note that the length of sequence s must be a multiple of sizeof<'T> times 8 let bitsToBitContainer (s:seq<bool>) : seq<'T> = let sz = sizeof<'T>*8 s |> Seq.scan (fun (i,n) x -> (i++,(n<<<1)|||(x&&&1)) (0,0) |> Seq.filter (fun (i,n) -> i = sz) |> Seq.map (fun (i,n) -> n)
    • mahmud 16y ago
      I am liking this F# thing. Thanks! :-)