|Perl: the Markov chain saw|
Re^2: What makes an array sorted and a hash unsorted?by herveus (Parson)
|on Jun 04, 2009 at 17:25 UTC||Need Help??|
In practice perls data structures are as shifty as the language itself and you can implement hashes in arrays as well as arrays in hashes. So the sentence "arrays are ordered and hashes are not" (I hope we all concur by now that 'sorted' was the wrong word) is merely a short form for "arrays provide built-in infrastructure to manipulate ordered values in an efficient manner, while hashes do not"
Hashes and arrays are reasonably well defined fundamental data types in Perl. I'd say "clearly defined", but this thread casts doubt on that assertion. Only at a conceptual level can you really assert that equivalency in the second sentence. Arrays and hashes are very strong data types in Perl. You cannot cast one into the other. You can use certain operations on each to produce instances of the other (conflating lists with arrays here). Those operators are not type casters, however.
In your code example, you first put a value in to the array at index 4. This causes indexes 0 through 3 to spring into existence, but, naturally, they have no defined value. When you then shift off the first value to print it, you get index 0 (the "first" value) and, being undefined, triggers the warning that you asked to have happen. It is unclear what that was meant to demonstrate.
Later you speak of push/pop and how they are not defined for hashes. Of course. Push and pop (and shift and unshift) are all curried versions of splice, which is the general purpose function for mucking about with arrays. Since a hash isn't (and can't be coerced into) an array, splice can't do anything with one. Likewise, keys() and values() and each() are not defined for arrays. The two data types are very distinct. I'm not sure what you are trying to say there. It makes no sense.