Re: Re^6: Counting keys with defined or undefined elements in a hash (behaviour of values())

> I think that just throws the created scalar away, but doesn't avoid its creation.

Well that's lame.

> ... and requires a function call for every pair in the hash, rather than a single one for all of them.

Right; I hadn't considered that.

> It does use less memory though, yes, that's what I meant when I said it's desirable when dealing with large hashes.

Ah, good, then I'm not completely insane. :-) I had switched from values to each on an ETL project last year, and I was remembering that I'd done it for speed reasons, but now that I think about it, it may very well have been for memory purposes instead.

Thanx for the clarifications.

Comment on Re: Re^6: Counting keys with defined or undefined elements in a hash (behaviour of values())

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Re^8: Counting keys with defined or undefined elements in a hash (behaviour of values()) by Aristotle (Chancellor) on Jun 07, 2003 at 10:17 UTC
Glad to help. `:)` As for it throwing the created scalar away - well, each gets called in list context, but it can't know what list its return value is being assigned to. Makeshifts last the longest.	[reply]
Re: Re^8: Counting keys with defined or undefined elements in a hash (behaviour of values()) by Oberon (Monk) on Jun 13, 2003 at 18:52 UTC
> ... but it can't know what list its return value is being assigned to. No, I suppose not. Well, not without a change to the language anyway ... I wonder if Perl 6 would be able to tell. Then again, perhaps such a detailed level of introspection isn't generally useful.	[reply]