On 2020-01-29 17:45, Trey Harris wrote:
On Wed, Jan 29, 2020 at 20:20 ToddAndMargo via perl6-users
<perl6-us...@perl.org <mailto:perl6-us...@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
