in reply to Assign a reference to a non reference

References are "normal" variables:

sub foo { return ({},[]); } my ($baz, $bar); ($baz,$bar) = &foo(); print ref $baz," ",ref $bar,"\n";

\%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;

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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];

    But if I can eliminate the need for @tmp, i'd be happy

      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();

      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());

      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);

      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";

      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'