On Tue, Jan 14, 2020 at 7:45 AM Paul Procacci <pproca...@gmail.com
<mailto:pproca...@gmail.com>> wrote:
>> 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 <mailto: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>
<mailto: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?
On 2020-01-14 05:14, Gerard ONeill wrote:
A negative number (-A5) is the twos compliment of the positive number.
A ones compliment is all the bits flipped. A twos compliment is a ones
compliment plus one. So a ones compliment of (A5) is (-A5 - 1), which
is -A6.
So presumably, the twos compliment operator is (-). And I suppose for
consistency, +-A5 gives you -A5, which makes +- the twos compliment
bitwise operator..
It is a pain to keep the variables from being "Coerced to Int",
but this pretty much shows it
$ p6 'my byte $x = 0xA5; say $x.base(16), " original";
my byte $y = +^$x; say $y.base(16), " ones compliment";
my byte $z = $x + $y + 1;
say $z.base(16), " original + 2s compliment";'
A5 original
5A ones compliment
0 original + 2s compliment
