in reply to
Re: Benchmarking for loops?
in thread
Benchmarking for loops?
Do you mean to say that sub "a2" is identical to "b", and "b2 is the same code as "a"? (I just wasn't sure what you meant.)
Comment on
Re^2: Benchmarking for loops?
Replies are listed 'Best First'.
Re^3: Benchmarking for loops?
by
keszler
(Priest)
on Jul 13, 2004 at 02:40 UTC
Yes, exactly. Cut&paste, rename a -> b2, b -> a2.
[reply]
In Section
Seekers of Perl Wisdom