Abigail

Well, in prototyped version (\%) i will get a reference. However nothing would stop me from doing same sub with ($) and using that to pass a reference (in this case hash_ref). Now that would not be nice to the interface of that sub (eq. is actually a hash_ref instead of scalar), but would it then be faster?

What i'm hoping here is that by using prototypes (and \%) perl is not transfering a hash as whole but a reference to hash, which it seems from my limited understanding.

With:

sub foo (\%) { ... }
foo(%$hashref);
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perl doesn't have to copy or expand the hash, but it will generate a new reference to it.

You could test this yourself using top(1) or whatever the quivalent is on Windows if you're on that platform. There's also a module that will tell you how much memory your data structures are using, but I forget the name of it.

Simple tests of my own show that when I create a large hash, the program uses about 71M. When I pass it as a hash ref (sub foo { ...}  foo(\%hash);), the program still uses about 71M. When I pass it as a hash (sub foo { ...}  foo(%hash);), the program uses about 117M. When I pass it as a hash to a prototyped sub ((sub foo (\%) { ...}  foo(%hash);), the program uses about 117M. Of course, I could have screwed up so test it yourself! :-)

Here's the program I used while I had top(1) running in another window. I just ran it 3 times with two of the three alternatives commented out each time and while it was sleeping I looked at the top window.

#!/usr/bin/perl

print "creating hash\n";
$k = "aaaaaaaa";
for (0..1000000) {  $hash{$k++} = $_ };
print "sleeping\n";
sleep (10);
print "copying\n";
foo(\%hash);
#bar(%hash);
#baz(%hash);
print "done";
exit;

sub foo { print "sleeping in foo\n"; sleep(10); }
sub bar (\%) { print "sleeping in bar\n"; sleep(10); }
sub baz (%) { print "sleeping in baz\n"; sleep(10); }
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duff