Re: my $a outside sort block incompatibility

I assume you want to declare $a beforehand but not use its value in sort. If that is the case, you could declare a local sub before $a:

   local *_sort = sub { return sort { $a <=> $b } @_ };
   my $a  = 'whatever';
   print join(' ', &_sort(@arr2)), "\n";
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But obviously it would be easier to just use a different variable name...

Update: I just read up on $a and $b in perlvar:

Special package variables when using sort(), see sort. Because of this specialness $a and $b don't need to be declared (using use vars, or our()) even when using the strict 'vars' pragma. Don't lexicalize them with my $a or my $b if you want to be able to use them in the sort() comparison block or function.

Comment on Re: my $a outside sort block incompatibility Download Code

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Re^2: my $a outside sort block incompatibility by Anonymous Monk on Feb 21, 2009 at 01:43 UTC
D'oh! Reading your comment reminded me of the solution--using "our"--that I had found earlier but misremembered as using "local". Thanks much. `# Works. my $a = 'whatever'; # just happens to be in scope of below sort block @arr2 = sort { our $a; $a <=> $b } @arr1; print join(' ', @arr2), "\n";` [download]	[reply] [d/l]