3 ms·

The language is happy to accommodate any style you like best, including oneliners. sub work($n) { say $n } my @numbers = 1..100; # rubyish @nu

by bmn__ 8y ago

The language is happy to accommodate any style you like best, including oneliners. sub work($n) { say $n } my @numbers = 1..100; # rubyish @numbers.grep(* > 10 && * %% 2).map(&work); # lol turbo haskal map &work <== grep * %% 2 <== grep * > 10 <== @numbers; # same, but in proper reading direction @numbers ==> grep * > 10 ==> grep * %% 2 ==> map &work; The traditional function composition operator exists, see https://docs.perl6.org/routine/%E2%88%98 https://docs.perl6.org/routine/%E2%88%98

b2gills 8y ago
The following doesn't do what you think it does * > 10 && * %% 2 That is two WhateverCode objects which each take one argument. Since both are definite it gives you the second one. my &a = * > 10; my &b = * %% 2; my &c = &a && &b; &c === &b; # True If it wasn't split up by the `&&`, it would be a code object that took two arguments. # using multiplication (×) as a boolean and # (it always cooperates in the WhateverCode lambda syntax) my &c = (* > 10) × (* %% 2); say so c(10,2); # True say (1..20).grep(&c); # ((11 12) (13 14) (15 16) (17 18) (19 20)) Note that `grep` is written in terms of `map` (1..20).map({ ($^a,$^b) if ($^a > 10) × ($^b %% 2) }) # ((11 12) (13 14) (15 16) (17 18) (19 20)) If you need to refer to an argument more than once, you (generally) can't do it with the WhateverCode lambda syntax. -> $n { $n > 10 && $n %% 2 } { $^n > 10 && $^n %% 2 } sub ($n){ $n > 10 && $n %% 2 } I say generally because array indexing will give you the number of elements for all the arguments you ask for. @a[ (* × ⅓) .. (* × ⅔) ] @a[ (@a.elems × ⅓) .. (@a.elems × ⅔) ]
- raiph 8y ago
  I noticed that error too but decided I didn't have the energy to provide a response worthy of the situation. I'm so glad I didn't try. As always you've provided a helpful answer and I learned something new. I hadn't twigged that `[...]` context would treat each Whatever as the same value -- though in retrospect I can see of course it would given the Perlish principle of doing something very useful rather than doing something useless (generating an error). I noticed what looked like a mistake in your post: my &c = (* > 10) × (* %% 2); say so c(10,2); # True I thought "surely that returns False" and when I tried, it did.
  - b2gills 8y ago
    Right it was late, and I was writing the code in different ways when writing here and writing in the REPL.

by bmn__ 8y ago

b2gills 8y ago

The following doesn't do what you think it does * > 10 && * %% 2 That is two WhateverCode objects which each take one argument. Since both are definite it gives you the second one. my &a = * > 10; my &b = * %% 2; my &c = &a && &b; &c === &b; # True If it wasn't split up by the `&&`, it would be a code object that took two arguments. # using multiplication (×) as a boolean and # (it always cooperates in the WhateverCode lambda syntax) my &c = (* > 10) × (* %% 2); say so c(10,2); # True say (1..20).grep(&c); # ((11 12) (13 14) (15 16) (17 18) (19 20)) Note that `grep` is written in terms of `map` (1..20).map({ ($^a,$^b) if ($^a > 10) × ($^b %% 2) }) # ((11 12) (13 14) (15 16) (17 18) (19 20)) If you need to refer to an argument more than once, you (generally) can't do it with the WhateverCode lambda syntax. -> $n { $n > 10 && $n %% 2 } { $^n > 10 && $^n %% 2 } sub ($n){ $n > 10 && $n %% 2 } I say generally because array indexing will give you the number of elements for all the arguments you ask for. @a[ (* × ⅓) .. (* × ⅔) ] @a[ (@a.elems × ⅓) .. (@a.elems × ⅔) ]

raiph 8y ago
I noticed that error too but decided I didn't have the energy to provide a response worthy of the situation. I'm so glad I didn't try. As always you've provided a helpful answer and I learned something new. I hadn't twigged that `[...]` context would treat each Whatever as the same value -- though in retrospect I can see of course it would given the Perlish principle of doing something very useful rather than doing something useless (generating an error). I noticed what looked like a mistake in your post: my &c = (* > 10) × (* %% 2); say so c(10,2); # True I thought "surely that returns False" and when I tried, it did.
- b2gills 8y ago
  Right it was late, and I was writing the code in different ways when writing here and writing in the REPL.