|The stupid question is the question not asked|
It's not as easy as that (I'm playing Ikegamis adovate now ;-). One could make the argument that a hash %h=(0,'a',1,'b',2,'c'); and an array @a=('a','b','c'); are both ordered in the same way. No sequence of equivalent insertions and deletions will change that both will look the same when you print them with
In older languages that would have been the only way to access arrays and print them.
That orderedness of the hash only breaks when you use print @a for which there is no direct substitute on the hash side (but could be built in a way that it comes out ordered). If you access them with a foreach (keys ...) loop, then you might as well use a foreach (sort keys ...) loop.
Of course you get a random list when you call keys(). But what if perl had a built-in function arraykeys() which gave back the numbers of the array in a random ordering. Would arrays then be both ordered and unordered or partially ordered or not ordered at all?. What if you had no keys() and each() functions in perl? Would a hash then be ordered or undefined in its orderedness?
As I said before, I am of the opinion that perl hashes are unordered and arrays are ordered, but proving it is not as simple as one might think.
In reply to Re^5: What makes an array sorted and a hash unsorted?