>> What is the syntax for a twos complement anyway? I'm not sure I understand the question. Two's compliment is +^ ... the routine you've been using.
On Tue, Jan 14, 2020 at 12:33 AM ToddAndMargo via perl6-users < perl6-us...@perl.org> wrote: > >> On Mon, Jan 13, 2020 at 11:30 PM ToddAndMargo via perl6-users > >> <perl6-us...@perl.org <mailto:perl6-us...@perl.org>> wrote: > >> > >> Hi All, > >> > >> This works, > >> > >> $ p6 'my uint8 $c = 0xA5; my uint8 $d = +^$c; say > $d.base(16);' > >> 5A > >> > >> But this does not: > >> > >> $ p6 'my uint8 $c = 0xA5; say (+^$c).base(16);' > >> -A6 > >> > >> 1) who turned it into an negative integer? > >> > >> 2) how do I turn it back? > >> > >> Many thanks, > >> -T > > On 2020-01-13 21:18, Paul Procacci wrote: > > If you read the signature for +^, you'll notice it returns an Int. > > > > In your first working example, you're taking a uint8 with binary value > > 10100101, zero extending it to 64 bits via +^, applying a two's > > compliment, and then assigning bits [0:7] to another uint8 which at that > > point contains the binary value of 01011010 (or hex value 0x5A). > > In your second example that isn't working, you're taking uint8 with > > binary value 10100101, zero extending it to 64 bits via +^, applying a > > two's compliment, and then displaying this *Int* (64 bits) as hex.[1] > > To turn it back you need to mask off bits [8:63] with: say ((+^$e) +& > > 0x0FF).base(16);" [2] > > > > [1] I'd show you the 64 bit value but it's a bunch of 1's followed by > > the value -0xA6. > > [2] Note, since the type has been promoted to an Int there' no going > > back to uint8 without an explicit assignment (afaik) > > > > That explains it. Thank you. > > I used uint8 to keep the ones to a mild torrent! > > If I am remembering correctly, 0xA5 going to 0x5A is > a ones compliment. > > What is the syntax for a twos complement anyway? > -- __________________ :(){ :|:& };:
