Consider the following snippet: 

#!/usr/bin/perl

use Data::Dumper;

my %h;
{
   local $h{key} = 'blah';
   # even "local $h{key};" will do
   # inititialization is only for clarity
}

print Dumper \%h;

[download]

In perl 5.6.1 the output is: 

$VAR1 = {
          'key' => undef
        };

[download]

while in 5.8.0 it is: 

$VAR1 = {};

[download]

Digging through the documentation I found the following statement, in perldoc perlsub (in both 5.6.1 and 5.8.0 distributions):

The behavior of local() on non-existent members of composite types is subject to change in future.

Question 1: Can you provide me with some insight into the history of this development? Also, what lies ahead? The 5.8 behaviour looks more DWIMish IMHO - but the doc says that the behaviour may change. In which ways?

Question 2: While experimenting I wrote the following code to emulate 5.8 behaviour on 5.6: 

# ...
my %h;
# ...
my @local_keys = qw/foo bar baz/;
my @inexistent = grep {! exists $h{$_}} @local_keys;
{
   local @h{@local_keys};
   # ...
}
delete @h{@inexistent};
# ...

[download]

You may notice the use of a hash slice as an argument to local. It came to me naturally (and it works). The question is: is this supported? My worries are based on my (possibly incorrect) belief that local and such are "special forms" and should be used only in ways sanctioned by the documentation (perldoc -f local is minimal and ambiguous). Also, why it works?

In reply to Perl 5.6 vs. 5.8: 'local' behaviour difference by calin

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