Re: bitwise NOT

[Tom Browder]

guys, interesting thread, but it's "complement"

-Tom

Re: bitwise NOT

[Todd Chester via perl6-users]

On Tue, Jan 14, 2020 at 7:45 AM Paul Procacci > 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
mailto:perl6-users@perl.org>> wrote:

 >> On Mon, Jan 13, 2020 at 11:30 PM ToddAndMargo via perl6-users
 >> mailto:perl6-users@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

Re: bitwise NOT

[Gerard ONeill]

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..

On Tue, Jan 14, 2020 at 7:45 AM Paul Procacci  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-users@perl.org> wrote:
>
>> >> On Mon, Jan 13, 2020 at 11:30 PM ToddAndMargo via perl6-users
>> >> mailto:perl6-users@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?
>
>
>>
>
> --
> __
>
> :(){ :|:& };:
>

Re: bitwise NOT

[Paul Procacci]

 >> 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-users@perl.org> wrote:

> >> On Mon, Jan 13, 2020 at 11:30 PM ToddAndMargo via perl6-users
> >> mailto:perl6-users@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?
>


-- 
__

:(){ :|:& };:

Re: bitwise NOT

[ToddAndMargo via perl6-users]

On Mon, Jan 13, 2020 at 11:30 PM ToddAndMargo via perl6-users 
mailto:perl6-users@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?

Re: bitwise NOT

[Paul Procacci]

 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)

On Mon, Jan 13, 2020 at 11:30 PM ToddAndMargo via perl6-users <
perl6-users@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
>


-- 
__

:(){ :|:& };: