Sorry, I sent my answer just for you.
So, the problem is you didn't call the same var you had declared.
my $foo = * **2;
Then you call
foo(2).say
Missing the $
Try:
$foo(2).say
or
say $foo(2)
About the
my @a = * **2;
Your suggestion works
@a[0](2)
or
@a[0].(2)
But I would appreciate an explanation about why `$a[0](0)` didn't work.
On Tue, Feb 11, 2020 at 9:45 PM Andy Bach <andy_b...@wiwb.uscourts.gov>
wrote:
> > I think it should be like this:
>
> > my $foo = * **2;
> { ... }
> > say $foo(4)
> 16
>
> That's what the doc says, but that's not what my install version says. I
> do get
> > my $foo = * **2;
> { ... }
>
> but say foo get the "unknown sub" error
>
> > But I have another point::
>
> > my @a = * **2;
> > @a(2)
> Invocant of method 'CALL-ME' must be a type object of type 'List', not an
> object instance of type 'Array'. Did you forget a 'multi'?
> in block <unit> at <unknown file> line 1
> Yeah, I'd be surprised if that worked
>
> > $a[0](2)
> ===SORRY!=== Error while compiling:
> Variable '$a' is not declared. Did you mean '@a'?
> ------> <BOL>⏏$a[0](2)
>
> raku doesn't swap sigils anymore, so it should be
> @a[0](2)
>
> maybe, pass the param, to the first bucket in @a which is holding a sub,
> so run it - works here
> > my @a = * **2;
> [{ ... }]
> > say @a[0](4);
> 16
>
> as does ".()"
> > say @a[0].(5);
> 25
> ------------------------------
> *From:* Aureliano Guedes <guedes.aureli...@gmail.com>
> *Sent:* Tuesday, February 11, 2020 6:36 PM
> *To:* Andy Bach <andy_b...@wiwb.uscourts.gov>
> *Subject:* Re: variable as subroutine?
>
> I think it should be like this:
>
> > my $foo = * **2;
> { ... }
> > say $foo(4)
> 16
>
> But I have another point::
>
> > my @a = * **2;
> [{ ... }]
> > @a(2)
> Invocant of method 'CALL-ME' must be a type object of type 'List', not an
> object instance of type 'Array'. Did you forget a 'multi'?
> in block <unit> at <unknown file> line 1
>
> > $a[0](2)
> ===SORRY!=== Error while compiling:
> Variable '$a' is not declared. Did you mean '@a'?
> ------> <BOL>⏏$a[0](2)
>
>
>
> On Tue, Feb 11, 2020 at 8:43 PM Andy Bach <andy_b...@wiwb.uscourts.gov>
> wrote:
>
> >The * * * call generates a WhateverCode block. This is expecting 2
> arguments.
>
> -> $x { $x * $x } is taking one argument.
>
> > The best documentation would probably be :
> https://docs.raku.org/type/Whatever
>
> so, from:
> Multiple * in one expression generate closures with as many arguments:
>
> my $c = * + *; # same as -> $x, $y { $x + $y }
> Using * in complex expressions will also generate closures:
>
> my $c = 4 * * + 5; # same as -> $x { 4 * $x + 5 }
>
> The * * * the parser says "one whatever, one math op (*) and one more
> whatever"
> my $foo = $x, $y { $x + $y };
>
> so,
> my $foo = * **2;
> should do $x * $x? Though I see
>
> > my $foo = * **2;
> { ... }
> say foo(4);
> ===SORRY!=== Error while compiling:
> Undeclared routine:
> foo used at line 1
>
> but '&' works
> > my &foo = * **2;
> { ... }
> > foo(4);
> 16
> > my &c = * **2;
> { ... }
> > say c(4);
> 16
>
>
>
>
>
> ------------------------------
> *From:* Simon Proctor <simon.proc...@gmail.com>
> *Sent:* Tuesday, February 11, 2020 9:27 AM
> *To:* Andy Bach <andy_b...@wiwb.uscourts.gov>
> *Cc:* perl6-users <perl6-us...@perl.org>
> *Subject:* Re: variable as subroutine?
>
> The * * * call generates a WhateverCode block. This is expecting 2
> arguments.
>
> -> $x { $x * $x } is taking one argument.
>
> The best documentation would probably be :
> https://docs.raku.org/type/Whatever
>
> Hope that helps.
>
> (For giggles earlier I made this dumb example of functional programming)
>
>
> my &ident = {$_};
> my &sq = {$_ * $_};
> sub trinar( &test, &true, &false, *@values ) { @values.map( -> $v {
> &test($v) ?? &true($v) !! &false($v) } ) };
> trinar( *.is-prime, &sq,&ident, ^30 ).say
>
> Enjoy. ;)
>
> On Tue, 11 Feb 2020 at 15:22, Andy Bach <andy_b...@wiwb.uscourts.gov>
> wrote:
>
> I have a few less related questions
> >> those are 3 ways to write the same sub:
>
> sub foo ($x) { $x * $x }
> my &foo = -> $x { $x * $x }
> my &foo = * * *;
>
> > A Note on Marc's comment:
> my &foo = * * *
> is not the same as:
> my &foo = -> $x { $x * $x }
> it is the same as:
> my &foo = -> $x, $y { $x * $y }
>
> Okay, "* * *" - how does that work? How is it different than
> -> $x { $x * $x }
> ? It needs two params?
>
> I followed the callable link but that left me with more questions:
>
> method CALL-ME
> method CALL-ME(Callable:D $self: |arguments)
> This method is required for postfix:«( )» and postfix:«.( )». It's what
> makes an object actually call-able and needs to be overloaded to let a
> given object act like a routine. If the object needs to be stored in a
> &-sigiled container, is has to implement Callable.
>
> class A does Callable {
> submethod CALL-ME(|c){ 'called' }
> }
> my &a = A;
> say a(); # OUTPUT: «called»
>
> That second "postfix" operator, means
> say a.(); # also outputs "called"
>
> but what is the "pipe c" signature doing for the submethod?
> ------------------------------
> *From:* Simon Proctor <simon.proc...@gmail.com>
> *Sent:* Tuesday, February 11, 2020 3:17 AM
> *To:* ToddAndMargo <toddandma...@zoho.com>
> *Cc:* perl6-users <perl6-us...@perl.org>
> *Subject:* Re: variable as subroutine?
>
> If you can store a subroutine in a variable then you can pass said
> subroutine to another one as an argument.
>
> This leads us into the joys of functional programming.
>
> And you may have used it already and not even realised.
>
> The .map and .grep methods (and .reduce and bunch of others) all expect a
> callable code block (that might be a subroutine) as a function.
>
> This :
>
> my @a = (1..10).map( * ** 2 )
>
> and this :
>
> my &sq = sub ($v) { $v ** 2 };
> my @a = (1..10).map( &sq );
>
> are doing the same thing. Except the second one has the &sq function
> available for other things.
>
> (A Note on Marc's comment * * * is not the same as -> $x { $x * $x } it is
> the same as -> $x, $y { $x * $y } )
>
> You can then start doing things like storing functions as values in hashes
> and doing all *kinds* of fun stuff.
>
> Welcome to the tip of the iceberg.
>
> Simon
>
>
> On Tue, 11 Feb 2020 at 03:21, ToddAndMargo via perl6-users <
> perl6-us...@perl.org> wrote:
>
> Hi All,
>
> Is Larry using his magic powder again?
>
> Can I declare a subroutine as a variable?
>
> my $abc = my sub (UInt $u, Str $s, Int $I) {
>
> How would I use it?
>
> And why would do such a thing?
>
> -T
>
>
>
> --
> Simon Proctor
> Cognoscite aliquid novum cotidie
>
> http://www.khanate.co.uk/
>
>
>
> --
> Simon Proctor
> Cognoscite aliquid novum cotidie
>
> http://www.khanate.co.uk/
>
>
>
> --
> Aureliano Guedes
> skype: aureliano.guedes
> contato: (11) 94292-6110
> whatsapp +5511942926110
>
--
Aureliano Guedes
skype: aureliano.guedes
contact: (11) 94292-6110
WhatsApp +5511942926110
