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I did a little playing with this, to try to learn something.
%h = (1 => 2, a=>"b", c => 2); %h = reverse (%x = reverse %h); foreach my $n (keys %h) { print "$n $h{$n}\n"; }
Yeah, I know. Anyway, here's what I get:
2 1 2 a b
As if a duplicate element were eliminated by making the key null. But if I do
%h = (1 => 2, a=>"b", c => 2); %x = reverse %h; %h = reverse %x; foreach my $n (keys %h) { print "$n $h{$n}\n"; }
I get
1 2 a b
which is what I'd expect. Now, here's yet another thing: if I create another duplicate, d => 2 (at the end of the list), I still get the same end result; that duplicate is eliminated. In fact, in the first bit of code, I still get three lines of output. If there are duplicates, it doesn't seem to be guaranteed which one is eliminated by the reverse. I just looked at reverse in my Camel book and it doesn't mention anything about duplicates, so I suspect the order can't be relied upon. Typical for a hash.

In reply to Re: whats wrong reverse %x = reverse %h? by melora
in thread whats wrong reverse %x = reverse %h? by borisz

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