"Was thinking about using a qw(continue 0 next 1 redo 2 last 3), as an initializer but the randomness of hashes had me wondering why that worked -- but they'll still be kept as pairs in any randomized hash..."
Yes, because a list is always ordered. Here's a very common idiom in parameter parsing:
sub blah { my %args = @_; ... }
It simply takes the elements in order, assigning the first as the key and the second as the value until all elements have been assigned to the hash.
You can call the sub in numerous ways, and get the same result:
blah(qw(continue 0 next 1 redo 2 last 3)); blah('continue', 0, 'next', 1, 'redo', 2, 'last', 3); blah(continue => 0, next => 1, redo => 2, last => 3); my %h = (continue => 0, next => 1, redo => 2, last => 3); blah(%h);
In reply to Re^3: Curious: are anon-hashes in random order?
by stevieb
in thread Curious: are anon-hashes in random order?
by perl-diddler
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