Re^14: Why does each() always re-evaluate its argument? ("for

Of course this can only be used if it isn't magical or tied and so on

Does it really matter if it's faster if it doesn't work?

Comment on Re^14: Why does each() always re-evaluate its argument? ("for_list" )

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Re^15: Why does each() always re-evaluate its argument? ("for_list" ) by NERDVANA (Priest) on Dec 11, 2023 at 22:20 UTC
Yes, because you separate the common case (non-magical) from the rare case with an `if` block, and then use the fast implementation most of the time. I didn't make up that example above, I copied it out of perl's own hv.c, where other various routines have been able to be optimized like that. I also understand that there is probably a reason why they didn't optimize this case, but back to my original statement, it surprises me that this code path is using the normal iterator. I would think dumping the elements of a hash would be a frequent enough operation to be worth optimizing.	[reply] [d/l]
Re^16: Why does each() always re-evaluate its argument? ("for_list" ) by ikegami (Patriarch) on Dec 12, 2023 at 05:16 UTC
Discounting the magic and the code for `HV_ITERNEXT_WANTPLACEHOLDERS`, that still leaves 20 lines of code.	[reply] [d/l]