NetWallah has asked for the wisdom of the Perl Monks concerning the following question:

Bretheren and Sisteren .. I'm trying to understand the "for <ZIP syntax> -> (param)" syntax in perl6, using pugs Version: 6.2.13, on Windows. I'm getting confusing results - I cannot tell if this is a pugs issue, or a perl6 syntax problem, documentation problem, or (most likely) my brain malfunction. Given: 
my @names = ('Jack', 'Emma', 'Robert');
my @ages = (20, 19, 35);

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I'm trying to print them in NAME, AGE pairs.
Here is my code with results from pugs: 
pugs> for zip(@names; @ages) -> [ $name, $age ] {say "  $name is $age 
+years old"}
Internal error while running expression:
***
    Unexpected "["
    expecting subroutine parameters, trait or block
    at <interactive> line 1, column 27

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This (to me) seems like the documentation on the pugs website is wrong. So I try something different - remove the brackets: 
pugs> for zip(@names; @ages) ->  $name, $age  {say "  $name is $age ol
+d"}
  Jack 20 is Emma 19 old
  Robert 35 is  old
undef

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OK - that does run, but seems to mix up the 2 entities into a single variable, probably a array-ref. 
pugs> for zip(@names; @ages) ->  $name, $age  {say  "   $name.[0] is $
+name[1] old; while  $age.[0] may be $age
[1]"}
   Jack is 20 old; while  Emma may be 19
   Robert is 35 old; while   may be
undef

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OK - making SOME progress above - at least I extracted all the data. Still, it did not put info into the parameters in a manner I would like. 
pugs> for zip(@names; @ages) -> @_  {say  "   @_[0] is @_[1] years old
+;"}
   Jack is 20 years old;
   Emma is 19 years old;
   Robert is 35 years old;

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OK !! That worked perfectly. But - I would like to use NAMED variables, not @_. None of the attempts below do what I want: 
pugs> for zip(@names; @ages) -> ($n,$a)  {say  "   $n is $a old; "}
   Jack 20 is Emma 19 old;
   Robert 35 is  old;
undef
pugs> for @names Z @ages -> ($n,$a)  {say  "   $n is $a old; "}
   Jack 20 is Emma 19 old;
   Robert 35 is  old;
undef
pugs> for @names Z @ages -> ($n;$a)  {say  "   $n is $a old; "}
   Jack 20 is Emma 19 old;
   Robert 35 is  old;
undef
pugs> for @names Z @ages -> $n,$a  {say  "   $n is $a old; "}
   Jack 20 is Emma 19 old;
   Robert 35 is  old;
undef
pugs> for @names Z @ages -> $n;$a  {say  "   $n is $a old; "}
   Jack 20 is Emma 19 old;
   Robert 35 is  old;
undef

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Could a fellow monk point me in the right direction - where can I RTFM on syntax related to this, and how should I do this ?

     "As you get older three things happen. The first is your memory goes, and I can't remember the other two... " - Sir Norman Wisdom

Replies are listed 'Best First'.
Re: perl6/pugs zip + for + subroutine params
by moritz (Cardinal) on Oct 09, 2007 at 06:23 UTC
    Pugs doesn't implement slice and list contexts correctly at the moment.

    The idea is that if there is no explicit slice context (=nested array context), the lists are automatically flattened.

    So 1, 2, 3 Z  <a b c> should become 1, 'a', 2, 'b', 3, 'c' (unless in slice context). When you use this list in a for loop, you can supply a "pointy block" with two paramters:

    for @list -> $a, $b { say "$a: $b" }

    There's no way you can ever use [$a, $b] as a formal parameter since it's not an lvalue (we don't have pattern matching in sub calls like haskell has... at least not yet ;-)

    You can read more about the syntax at the Official Perl 6 Documentation.

    Perl 6 in German

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Re: perl6/pugs zip + for + subroutine params
by perlfan (Parson) on Oct 09, 2007 at 01:39 UTC
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