Assign a reference to a non reference

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

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Re: Assign a reference to a non reference by fglock (Vicar) on Oct 03, 2002 at 11:36 UTC
References are "normal" variables: `sub foo { return ({},[]); } my ($baz, $bar); ($baz,$bar) = &foo(); print ref $baz," ",ref $bar,"\n";` [download] `\%hash` is a reference, but you can't do `\%hash = $ref` because `\%hash` is a constant. assign references into named non-reference vars `%hash = %$baz; @array = @$bar;` [download]	[reply] [d/l] [select]
Re: Re: Assign a reference to a non reference by Anonymous Monk on Oct 03, 2002 at 13:30 UTC
hrm, I figured \%hash couldn't be an lvalue .. :( I was trying to find an "elegant" way to fit this into someone else's code. I'm currently doing it like this: `my @tmp = &foo(); %baz = %$tmp[0]; @bar = @$tmp[1];` [download] But if I can eliminate the need for @tmp, i'd be happy	[reply] [d/l]
Re:x3 Assign a reference to a non reference by grinder (Bishop) on Oct 03, 2002 at 14:02 UTC
You just can't do it. Look at it from the other end, you're trying to return two lists. To assign them to two variables without using a temporary you must do a listwise assignment of the returned values. This means your code is going to look something like `my (%baz, @bar) = foo();` [download] But that doesn't work with Perl, the two lists are flattened, and the first variable is in a winner-take-all situation. @bar gets nothing. At first I thought it was simply a matter of coercing on the fly `my (%baz, @bar) = sub { %{$_[0]}, @{$_[1]} }->(foo());` [download] But that still doesn't work. To understand, have a look at what adding the following does: `use Data::Dumper; sub foo { return ({J=>'A', P=>'H', }, [4.019,5.8]); } # ... original listy assignment here ... print Dumper(\%baz);` [download] There is a way around the problem of course, and that is to use references henceforth: `my ($baz, $bar) = foo(); print "$baz->{J} $bar->[1]\n";` [download] This will do what you want, at the expense of having to dereference via `->`. This may or may not be a problem to retrofit into your project. print@_{sort keys %_},$/if%_=split//,'= & *a?b:e\f/h^h!j+n,o@o;r$s-t%t#u'	[reply] [d/l] [select]
Re: Assign a reference to a non reference by merlyn (Sage) on Oct 03, 2002 at 14:36 UTC
Ugliness abounds, but this should do it: `(%baz = %{$_->[0]}), (@bar = @{$_->[1]}) for [&foo()];` [download] I'm using $_ as a holder for an array ref to the return value. Hey, it's one line. No, I can't think of a quick idiom, unless you permit global variables: `our (%baz, @bar); (baz, bar) = &foo();` [download] But mucking with the symbol table is generally bad form as well. -- Randal L. Schwartz, Perl hacker	[reply] [d/l] [select]
Re: Assign a reference to a non reference by antifun (Sexton) on Oct 03, 2002 at 14:08 UTC
You're kind of stuck here. Perl won't let you assign to the non-lvaluable thing you tried in your original example. You'll have a lot of trouble doing something like `(@a, %h) = anything()` [download] because `@a` will always get everything that is returned. (An artifact of the way perl handles nested list contexts.) So that leaves you with something like `($href, $aref) = (@{&foo})` [download] which is basically what you came up with on your own. Once you've got the references, you can dereference them one at a time, as you are doing, but good luck trying to assign (copy) them all to "real" arrays/hashes at once because of that second point above. Anyone who knows any better than I do care to comment? --- "I hate it when I think myself into a corner." Matt Mitchell	[reply] [d/l] [select]