in reply to Why "Bizzare copy" ?

// is an empty regexp.

@{ ... } dereferences a reference into an array. In this case, you're trying to coerce the return value of $_ =~ m// into being an array. That actually attempts to create a symbolic reference, creating a variable named @1.

$#{ ... } tries to find the highest index of @1.

If you turn on use strict; and use warnings; you'll get all sorts of clues.


Dave

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Re: Re: Why "Bizzare copy" ?
by Anonymous Monk on Mar 30, 2004 at 08:49 UTC
    In that case 
    $#{@1}

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    would die in the same way. It does not.
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[d/l]

      $#{@1}

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      would die in the same way. It does not.

      No, it wouldn't, as it doesn't use a symbolic reference...

      If you take the time to dump or peek at @1, you'll see that it exists as empty array...as do all @NUM I tested.

      My guess was, that @1 exists as a side effect of $1 beeing a default-variable and the "multi-type" nature of perl variables. I'd love to be corrected/educated on this rather blunt assumption I wasn't able to backup with my first dive into Devel::Peek and the like.

      regards,
      tomte

      Hlade's Law:

      If you have a difficult task, give it to a lazy person --
      they will find an easier way to do it.

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[d/l]

      Actually, it does die, but only in the debugger. 

      % perl -d -e '$#{@1}'
Default die handler restored.

Loading DB routines from perl5db.pl version 1.07
Editor support available.

Enter h or `h h' for help, or `man perldebug' for more help.

main::(-e:1):    $#{@1}
  DB<1> c
Bizarre copy of ARRAY in leave at -e line 1.
Debugged program terminated.

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      Without the debugger, it works fine.

      This error message is created in sv.c, in the function Perl_sv_setsv. This function copies a scalar value from one location to another. "Bizarre copy of ARRAY in leave" means that an opcode called "leave" tried to copy an array to a scalar, should be impossible. "leave" is the opcode for leaving a block of code. Why this happened, I cannot figure out.

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[d/l]
      correction: in that case 
      $#{@{"1"}}

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      would die in the same way. This however, does work: 
      @a=@{//};$#a

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      Why?
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[d/l]
[select]
        $#{@{1}}, $#{@{'1'}} and $#{@{"1"}} all die in the same way.

        CountZero

        "If you have four groups working on a compiler, you'll get a 4-pass compiler." - Conway's Law

[reply]
[d/l]
[select]

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