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Re: stolen uint's

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From:
ToddAndMargo via perl6-users
Date:
January 30, 2020 02:08
Subject:
Re: stolen uint's
Message ID:
161a0f32-805a-c72f-b8ca-3bd028df1652@zoho.com
On 2020-01-29 17:45, Trey Harris wrote:
> 
> 
> On Wed, Jan 29, 2020 at 20:20 ToddAndMargo via perl6-users 
> <perl6-users@perl.org <mailto:perl6-users@perl.org>> wrote:
> 
>     On 2020-01-29 10:28, Trey Harris wrote:
>      > B is not a subset of A. That is the relationship of uint and int—two
>      > distinct types whose values happen to overlap in a way that
>     describes a
>      > subset. Perl isn’t Prolog; a logical relationship between two
>     types is
>      > not a first-class entity of the language.
>      >
>      >
>      >
>      >     I know, I am slicing the baloney thin here, but uint is
>      >     not a static C variable. It can change into an int with
>      >     the position of the moon.
>      >
>      >
>      > I’m STILL waiting for you to show me ONE example of a `uint` turning
>      > into `int`. Not `Int`, via auto-boxing, `int`, via who-knows-what.
>      > Either do that, or stop making the assertion it does that; if you
>     don’t
>      > show a reproducible example, I am going to conclude you are lying
>     if you
>      > persist.
> 
>     $ p6 'my uint8 $u; say $u.^name;'
>     Int
> 
> I’m done. I haven’t killfiled anyone in over twenty years. 
> Congratulations, you’re the first this century.



#!/usr/bin/env perl6

use NativeCall;

constant BYTE     := uint8;
constant BYTES    := CArray[BYTE];

my BYTES  $lpBuffer = CArray[BYTE].new( 0xFF xx 4 );

if    $lpBuffer[0] == -1   { say "$lpBuffer[0] acts like an integer"; }
elsif $lpBuffer[0] == 0xFF { say "$lpBuffer[0] acts like an unsigned 
integer"; }
else                       { say $lpBuffer[0], " unknown"; }


-1 acts like an integer

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