Re: How expensive is reverse?

I'm not sure how they are implemented or what the (O) is (I suspect it's O(n)). Here's what a quick benchmark says from my Dual Celeron 400:


#!/usr/bin/perl
use Benchmark;

@one   = ('a' x 100);
@two   = ('a' x 1000);
@three = ('a' x 10000);

timethese( 100000, {
                     'rev 100'    => sub { reverse @one   },
                     'rev 1000'   => sub { reverse @two   },
                     'rev 10000'  => sub { reverse @three },
                   }
         );
[download]

Gives me this:


Benchmark: timing 100000 iterations of rev 100, rev 1000, rev 10000...
   rev 100:  1 wallclock secs ( 0.24 usr +  0.00 sys =  0.24 CPU) @ 41
+6666.67/s (n=100000)
            (warning: too few iterations for a reliable count)
  rev 1000:  1 wallclock secs ( 0.87 usr +  0.00 sys =  0.87 CPU) @ 11
+4942.53/s (n=100000)
 rev 10000:  8 wallclock secs ( 7.65 usr +  0.00 sys =  7.65 CPU) @ 13
+071.90/s (n=100000)
[download]

At a 100,000 iterations of a 1,000 element list, it took 0.87 seconds of CPU time. Not too expensive I'd say.

Updated CRAP! nardo is quite right. I hate it when I make silly mistakes.

/\/\averick
perl -l -e "eval pack('h*','072796e6470272f2c5f2c5166756279636b672');"

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Re: Re: How expensive is reverse?
by nardo (Friar) on Sep 01, 2001 at 22:08 UTC

@one   = (('a') x 100);
@two   = (('a') x 1000);
@three = (('a') x 10000);
[download]

Benchmark: timing 10000 iterations of rev 100, rev 1000, rev 10000...
   rev 100:  0 wallclock secs ( 0.54 usr + -0.01 sys =  0.53 CPU) @ 18
+867.92/s (n=10000)
  rev 1000:  4 wallclock secs ( 5.52 usr + -0.01 sys =  5.51 CPU) @ 18
+14.88/s (n=10000)
 rev 10000: 59 wallclock secs (58.33 usr + -0.00 sys = 58.33 CPU) @ 17
+1.44/s (n=10000)
[download]

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