> watch about a minute of this Damian Conway video--where he shows the new Raku > (Perl6) sigil table:
https://youtu.be/Nq2HkAYbG5o?t=568 Watch the whole thing - D.C. is one of the most entertaining instructors I've ever had the pleasure to list to[1], plus he's smarter than a sitting on a bushel of thumbtacks! Okay, and the accent. a [1] I really wish there were somewhere a copy of his presentation at the Chicago YAPC long ago where he presented a clip on why to go to YAPC in Australia the next year - "of the 10 most deadly snakes in the world, 11 of them are native to Australia". Before that YAPC, he had a meet at a bar in Chicago with the PM group there where he "explained" his obfu Conway (no relation) "game of life"/quine script https://everything2.com/title/SelfGOL ________________________________ From: William Michels <w...@caa.columbia.edu> Sent: Wednesday, February 12, 2020 1:27 PM To: Aureliano Guedes <guedes.aureli...@gmail.com> Cc: Andy Bach <andy_b...@wiwb.uscourts.gov>; perl6-users <perl6-users@perl.org> Subject: Re: variable as subroutine? On Wed, Feb 12, 2020 at 8:12 AM Aureliano Guedes <guedes.aureli...@gmail.com<mailto:guedes.aureli...@gmail.com>> wrote: On Wed, Feb 12, 2020 at 1:09 PM Andy Bach <andy_b...@wiwb.uscourts.gov<mailto:andy_b...@wiwb.uscourts.gov>> wrote: > 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 $ D'oh! Thanks. > About the my @a = * **2; > Your suggestion works @a[0](2) > or @a[0].(2) > But I would appreciate an explanation about why `$a[0](0)` didn't work. Same reason as mine didn't work "$a" of "$a[0]" is *not* the same variable as @a - raku doesn't swap sigils, so arrays always use @ even when they're being dereferenced (?) to a single element - unlike Perl5 Now I see. I din't know that. Thanks. I must study better Raku. Hi Aureliano, watch about a minute of this Damian Conway video--where he shows the new Raku (Perl6) sigil table: https://youtu.be/Nq2HkAYbG5o?t=568 HTH, Bill. ________________________________ From: Aureliano Guedes <guedes.aureli...@gmail.com<mailto:guedes.aureli...@gmail.com>> Sent: Tuesday, February 11, 2020 7:00 PM To: Andy Bach <andy_b...@wiwb.uscourts.gov<mailto:andy_b...@wiwb.uscourts.gov>>; perl6-users <perl6-users@perl.org<mailto:perl6-users@perl.org>> Subject: Re: variable as subroutine? 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<mailto: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<mailto:guedes.aureli...@gmail.com>> Sent: Tuesday, February 11, 2020 6:36 PM To: Andy Bach <andy_b...@wiwb.uscourts.gov<mailto: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<mailto: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<mailto:simon.proc...@gmail.com>> Sent: Tuesday, February 11, 2020 9:27 AM To: Andy Bach <andy_b...@wiwb.uscourts.gov<mailto:andy_b...@wiwb.uscourts.gov>> Cc: perl6-users <perl6-users@perl.org<mailto:perl6-users@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<mailto: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 $self: |arguments) This method is required for postfix:«( )» and postfix:«.( )». It's what makes an object actually call-able and needs to be overloaded to let a given object act like a routine. If the object needs to be stored in a &-sigiled container, is has to implement Callable. class A does Callable { submethod CALL-ME(|c){ 'called' } } my &a = A; say a(); # OUTPUT: «called» That second "postfix" operator, means say a.(); # also outputs "called" but what is the "pipe c" signature doing for the submethod? ________________________________ From: Simon Proctor <simon.proc...@gmail.com<mailto:simon.proc...@gmail.com>> Sent: Tuesday, February 11, 2020 3:17 AM To: ToddAndMargo <toddandma...@zoho.com<mailto:toddandma...@zoho.com>> Cc: perl6-users <perl6-users@perl.org<mailto:perl6-users@perl.org>> Subject: Re: variable as subroutine? If you can store a subroutine in a variable then you can pass said subroutine to another one as an argument. This leads us into the joys of functional programming. And you may have used it already and not even realised. The .map and .grep methods (and .reduce and bunch of others) all expect a callable code block (that might be a subroutine) as a function. This : my @a = (1..10).map( * ** 2 ) and this : my &sq = sub ($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-users@perl.org<mailto:perl6-users@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 -- Aureliano Guedes skype: aureliano.guedes contato: (11) 94292-6110 whatsapp +5511942926110
