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

Hello. Some monks and I were discussing perl 6, and BooK said, why not just %elbow = 1..10, 11..20;
So I test it out, like i do everything, and i do: perl -e"%a=( 1..10, 11..20 );print keys %a" and I get no output.

I try all my activeperl installs, nada. I try all my indigoperl installs, same thing (btw this be 5.6.0 and 5.6.1 builds 618 and up)

Meanwhile, BooK says, yes it does, and spits Dumper in the cb: perl -MData::Dumper -e'print { 1..10, 11..20 } and sure enough, Dumper dumps a hash like: 

$VAR1 = {
          '13' => 14,
          '7' => 8,
          '15' => 16,
          '9' => 10,
          '1' => 2,
          '17' => 18,
          '3' => 4,
          '19' => 20,
          '11' => 12,
          '5' => 6
        };

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So I ssh over to jcwrens machine and sure enough i get some keys (111313515717919).

So now I says to me, this must be a bug, and to confirm, i try: perl -MData::Dumper -e "%a= 1 .. 10, 2 .. 11 ;print Dumper \%a" which yields nada, and then perl -MData::Dumper -e "$a={ 1 .. 10, 11.. 20};print Dumper $a" which spits our (as expected) 

$VAR1 = {
          '13' => 14,
          '7' => 8,
          '15' => 16,
          '9' => 10,
          '1' => 2,
          '17' => 18,
          '3' => 4,
          '19' => 20,
          '11' => 12,
          '5' => 6
        };

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So, I think this is a win32 bug. Can anyone confirm/deny? Anyone have any idea where it is, sourcecode wise? (like foo.c or something)

 
___crazyinsomniac_______________________________________
Disclaimer: Don't blame. It came from inside the void
perl -e "$q=$_;map({chr unpack qq;H*;,$_}split(q;;,q*H*));print;$q/$q;"

Replies are listed 'Best First'.
Re: perl -e hash assignment and range operator bug
by japhy (Canon) on Aug 26, 2001 at 20:29 UTC
    Windows uses % as its environment variable token. You must double it to get a single one.

    The code %a = (1,2,3,4); print %a is reduced to a because the commandline sees a = (1,2,3,4); print as the name of an environment variable.

    _____________________________________________________
    Jeff[japhy]Pinyan: Perl, regex, and perl hacker.
    s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

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        No, the problem is you're using Windows.

        _____________________________________________________
        Jeff[japhy]Pinyan: Perl, regex, and perl hacker.
        s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

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Re: perl -e hash assignment and range operator bug
by Rudif (Hermit) on Aug 27, 2001 at 01:31 UTC
    crazyinsomniac said

    >> So I test it out, like i do everything, and i do:  perl -e"%a=( 1..10, 11..20 );print keys %a" and I get no output.

    Well, when I test in on my one and only Perl (5.6.1 build 628 on Win2k), I get this 

    H:\>perl -e"%a=( 1..10, 11..20 );print keys %a"
137159117319115

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    or in a more readable version 
    H:\>perl -e"%a=( 1..10, 11..20 );print qq{@{[keys %a]}}"
13 7 15 9 1 17 3 19 11 5

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    both of which look good to me.

    crazyinsomniac continued

    >> So now I says to me, this must be a bug, and to confirm, i try: 

    perl -MData::Dumper -e "%a= 1 .. 10, 2 .. 11 ;print Dumper \%a"

    [download]
    which yields nada, and then ...

    Me again, when I test it, I get this - as expected. 

    H:\>perl -MData::Dumper -e "%a= 1 .. 10, 2 .. 11 ;print Dumper \%a"
$VAR1 = {
          '7' => 8,
          '9' => 10,
          '1' => 2,
          '3' => 4,
          '5' => 6
        };

    [download]
    Seems to work nicely. The % character that japhy mentioned is tucked inside the double quotes and is seen by Perl, not the cmd shell.

    So where is the problem? Did I miss something here?

    I am all for bashing the cmd shell (pun intended) when justified, but here I cannot confirm the problem.

    Can you help me out on this, crazyinsomniac ?

    Rudif

    Update Could somemonk please set me straight on the operator precedence in the second example above?

    I think that the assignment binds 1..10 to %a, whereas 2..11 has no effect other than being the return value from the comma operator, which is thrown away. Is this correct?

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        Thank you for clarifying - I forgot about that possibility.

        Rudif

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