Re: Re: Why "Bizzare copy" ?

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Re*3: Why "Bizzare copy" ? by Tomte (Priest) on Mar 30, 2004 at 10:34 UTC
`$#{@1}` [download] 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 `$1` beeing a default-variable and the "multi-type" nature of perl variables. I'd love to be corrected/educated on this rather blunt assumption I wasn't able to backup with my first dive into Devel::Peek and the like. regards, tomte Hlade's Law: If you have a difficult task, give it to a lazy person -- they will find an easier way to do it.	[reply] [d/l]
Re: Re: Re: Why "Bizzare copy" ? by bunnyman (Hermit) on Mar 30, 2004 at 21:33 UTC
Actually, it does die, but only in the debugger. % perl -d -e '$#{@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): $#{@1} DB<1> c Bizarre copy of ARRAY in leave at -e line 1. Debugged program terminated. [download] Without the debugger, it works fine. This error message is created in sv.c, in the function Perl_sv_setsv. This function copies a scalar value from one location to another. "Bizarre copy of ARRAY in leave" means that an opcode called "leave" tried to copy an array to a scalar, should be impossible. "leave" is the opcode for leaving a block of code. Why this happened, I cannot figure out.	[reply] [d/l]
Re: Re: Re: Why "Bizzare copy" ? by Anonymous Monk on Mar 30, 2004 at 11:04 UTC
correction: in that case `$#{@{"1"}}` [download] would die in the same way. This however, does work: `@a=@{//};$#a` [download] Why?	[reply] [d/l] [select]
Re: Re: Re: Re: Why "Bizzare copy" ? by CountZero (Bishop) on Mar 30, 2004 at 19:22 UTC
`$#{@{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]