Re: Re: Assign a reference to a non reference

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

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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]

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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]