Yes. I shepherded this change into blead. In the course of doing so I had to fix a surprisingly large amount of code that one way or another had bogus expectations of hash ordering. Some of these were surprisingly subtle. For instance code like this:
my %hash= some_list_of_key_value_pairs();
my @keys= keys %hash;
my %copy= %hash;
my @values= values %copy;
do_something_with($keys[$_],$values[$_]) for 0..$#keys;
This code will work so long as the keys put into %hash all hash into different buckets. However if any collide it will fail. So it is very sensitive to what data is involved and to an uninformed observer would make no sense, especially if test samples happened to work out. With an older perl the same input would always produce the same output so a safe input set would remain safe for ever, on the other hand with hash seed randomization you are pretty much guaranteed to get a collision /sometime/ no matter what input you feed in, which will then make this code fail regularly.
Ive seen all kinds of variants of this. All of them would be broken with old perls with tweaks to the state of the hash, such as pre-sizing the hash buckets to a particular size, or putting the same keys in two hashes with two different bucket array sizes. Things that would fail very very rarely, and would be very hard to debug. All of these kind of bugs will start happening much more often, and therefore become much easier to track down and fix.
with several different versions of Perl. It prints "Different order" with all the versions I tried (5.8.9, 5.10.1, 5.12.5, 5.14.4, 5.16.3, 5.17.10). With all but 5.17.10, the output is the same each time the program is run:
So, if I understand correctly, the code you posted is visibly broken and has been for a very long time.
Well yes, it is broken, that was my point for bringing it up :-). Whether it is visibly broken or not depends on whether you have any collisions. In the case you posted there are two sets of collisions, 4 and 1, and 3 and 0. (You can tell because they reverse each copy in earlier perls.) However if changed it to be 1,2,3,5,7,8 you would not have any collisions and the order would appear the same. Eg try:
In 5.18 the order will be different for every hash. No exceptions. Here is what perlfunc will say in 5.18:
Hash entries are returned in an apparently random order. The actual r
order is specific to a given hash; the exact same series of operations
on two hashes may result in a different order for each hash. Any inser
into the hash may change the order, as will any deletion, with the exc
that the most recent key returned by C<each> or C<keys> may be deleted
without changing the order. So long as a given hash is unmodified you
rely on C<keys>, C<values> and C<each> to repeatedly return the same o
as each other. See L<perlsec/"Algorithmic Complexity Attacks"> for
details on why hash order is randomized. Aside from the guarantees
provided here the exact details of Perl's hash algorithm and the hash
traversal order are subject to change in any release of Perl.