in reply to When does it pay to use the schwartzian transform?
It "pays" when the cost of pre-computing the sort keys for all of your elements is significantly less than re-computing the sort keys each time you make a comparison. It depends on the cost of computing a sort key, and the number of elements.
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Re^2: When does it pay to use the schwartzian transform?
by jfroebe (Parson) on Dec 16, 2005 at 18:59 UTC | |
Re^2: When does it pay to use the schwartzian transform?
by Perl Mouse (Chaplain) on Dec 18, 2005 at 14:14 UTC |
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