Re: How to copy a referenced data structure?

Check out Clone. Faster than Storable, but not so flexible.

--shmem

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

Comment on Re: How to copy a referenced data structure?

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Re^2: How to copy a referenced data structure? by Anonymous Monk on May 22, 2007 at 06:22 UTC
I am trying to see what the issue is that this question is trying to address. If I make a copy of a reference to a reference, and then undef that reference, the copy to original references data still exists. I know I am missing the point here, I would just like to see how. `#!/usr/bin/perl use Data::Dumper; $ar = [ [ 1,2,3 ] ]; $copy = [ @$ar ]; undef $ar; print Dumper($copy);` [download]	[reply] [d/l]
Re^3: How to copy a referenced data structure? by syphilis (Archbishop) on May 22, 2007 at 06:49 UTC
`$copy = [ @$ar ];` $copy is a reference to an array. The first (and only) element of that array is the array that $ar refers to. When you `undef $ar;` that has no effect on the array that $ar referred to - and hence no effect on $copy. Cheers, Rob	[reply] [d/l] [select]