--- Jonathan Scott Duff <[EMAIL PROTECTED]> wrote:
> Maybe we've gone over this before but, if so, I don't remember ...
>
> On Tue, Oct 29, 2002 at 05:16:48PM -0800, Michael Lazzaro wrote:
> > hyperoperators:
> >
> > [op] - as prefix to any unary/binary operator, "vectorizes" the
>
> > operator
>
> > . - method call on current topic
>
> What would [.]method() mean?
@a = @b[.]method(); # Sets each @a[x] to @b[x].method()
Thinking out loud ...
@a = @b[.method()]; # Sets @a to the value stored in @b indexed by
# current context.method()
>
> > < > <= >= == != <=> - comparision
> > lt gt le ge eq ne cmp
>
> What do these do?
>
> if $a [<] @b { ... } # if $a < all(*@b)
Credible.
> if @a [<] @b { ... } # if $a[0] < all(*@b) &&
> # $a[1] < all(*@b) &&
> # $a[2] < all(*@b)
My inclination here is that $a[0] < $b[0] && $a[1] < $b[1] ...
> $a [&] @b # all($a,*@b) ???
Questionable. Perhaps an array s.t. @result[x] = $a & @b[x] ?
On the other hand, I like the idea of being able to distribute an
operator using this syntax...
@a = @b op $c
versus
?? = @b [op] $c
What's the difference? One maybe produces @b[x] op $c while the other
produces -> { $result = $c; for @b -> $b { $c op= $b; } }
But then it's not legal any more in array context unless @b is a LoL.
Erk!
> @a [&] @b # all(*@a,*@b) ???
Array s.t. @result[x] = @a[x] & @b[x]
> @a = ($b [,] @c); # @a = ($b, *@c); ???
LoL: @a[x] = ($b, @c[x])
> I'm not even sure how to hyper these two. I guess if I had an array
> of "range objects" I could hyper ;
Hypersemi is probably just the same as hypercomma, since the
circumlocutions needed to use it are such that you'll probably specify
the construction clearly.
But if not:
@a = @b [;] @c --> @a[x] = ((@b[x]) ; (@c[x]))
> Would this write to several filehandles?
>
> print @file_handles [:] "fooey!\n";
>
> > .. - range
>
> And this is the one that made me start thinking about hypering the
> others
>
> @a = @b[..]@c # @a = ($b[0]..$c[0], $b[1]..$c[1], ...) ???
> @a = $b[..]@c # @a = ($b..$c[0], $b..$c[1], ...) ???
> @a = @b[..]$c # @a = ($b[0]..$c, $b[1]..$c, ...) ???
>
You, too? Range was the first thing I started trying to hyper.
> I know that this stuff probably seems obvious to everyone, but I'd
> rather have it explicit just in case :-)
if $x == any(@b[..]@c)
print "In one of the ranges...\n";
or
if $x ~~ any(@b[..]@c)
print "In one of the ranges...\n";
Does .. create an implicit flexpr in numeric context, or does it
require smartmatch?
=Austin
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