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I'm currently reading the book Higher Order Perl and I'm on the section where it covers memoization. The book says a single multiplication won't be improved on by memoization because of the overhead involved and because multiplication is already fast. I'm surprised that your lookup with ord() and substr() is faster than a single multiplication. The only difference I see is that my function is using a float constant and the ord() and substr() don't need to do float calculations.

#!/usr/bin/env perl use warnings; use strict; use Benchmark 'cmpthese'; my $lookup = join '', map{ chr( $_ / 255 * 100 ) } 0 .. 255; my $const = 100 / 255; cmpthese(-2, { lookup => sub { my @output = map { ord( substr $lookup, $_, 1 )} 0 .. 255; }, calc => sub { my @output = map {$_ / 255 * 100} 0 .. 255; }, calc2 => sub { my @output = map {$_ * $const} 0 .. 255; }, }); my @output1 = map { ord( substr $lookup, $_, 1 )} 0 .. 5; my @output2 = map {$_ * $const} 0 .. 5; print "@output1\n"; print "@output2\n"; __END__ # Results on my machine (v5.22.1 built for MSWin32-x64-multi-thread): Rate calc calc2 lookup calc 12009/s -- -24% -31% calc2 15753/s 31% -- -9% lookup 17376/s 45% 10% -- 0 0 0 1 1 1 0 0.392156862745098 0.784313725490196 1.17647058823529 1.5686274509803 +9 1.96078431372549

In reply to Re^2: Normalizing a range of numbers to a percentage by Lotus1
in thread Normalizing a range of numbers to a percentage by stevieb

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