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### Re^10: What makes an array sorted and a hash unsorted?

by BrowserUk (Patriarch)
 on Jun 05, 2009 at 15:07 UTC ( #768819=note: print w/replies, xml ) Need Help??

the hash must traverse the internal data structures in some order to produce the list, but that sequence is explicitly undefined. That is the essence of being unordered.

I've been watching from the sidelines and for the most part, this seems like a prime example of what killed the Pedants' Conference--they died arguing about whether the ' should be before or after the s--but I've always enjoyed a good freindly debate, so here goes nothing :)

I would say that for all hash implementations, including Perl's current, (and all previous) implementations, they do have an "intrisic ordering". That ordering is a definable and concrete function of 2 factors:

• The hashing algorithm.
• The bucket size at the point of traversal.

And given knowledge of algorithm (use the source), and the current bucket size, (see the second element of scalar %hash), the ordering is knowable; therefore definable; therefore defined by the implementation(s).

However, I agree with the earlier statement--I've lost track of who made it (first)--that there is a significant difference between "order(|ing|ed)" and "sort(|ing|ed)".

And the answer to the OP question is:

Whilst the iteration order of arrays is, (for most (all?) languages), unspecified, it is defined, (by convention), to be the same as the natural ordering of its integer indexes; whereas, the iteration order of hashes, (also unspecified), is defined, (as a matter of practicality), to be the simplest or most efficient, (which are often the same thing), order of traversal of the data-structures underlying the implementation.

As hash implementations differ, the ordering differs with implementation; whereas the natural ordering of integer indexes does not.

QED.

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

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Re^11: What makes an array sorted and a hash unsorted? (nit)
by tye (Sage) on Jun 05, 2009 at 15:29 UTC
And given knowledge of algorithm (use the source), and the current bucket size, (see the second element of scalar %hash), the ordering is knowable; therefore definable; therefore defined by the implementation(s).

Actually, you have overlooked the order of items on the linked list of a bucket that holds more than one entry.

- tye

Woe! A non-pedantic counter-argument. I wasn't expecting that. But then, nobody expects....

And you are, as usual, correct.

I could argue that "knowledge of the algorithm" covers it, but in truth, I hadn't even considered it :)

But then again, I did say:

the iteration order of hashes, (also unspecified), is defined, (as a matter of practicality), to be the simplest or most efficient, (which are often the same thing), order of traversal of the data-structures underlying the implementation.

which absolutlely does cover it. So, halvsies :)

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

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