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Id: 1429
Status: resolved
Priority: 0/
Queue: perl5

Owner: Nobody
Requestors:
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Operating System: (no value)
PatchStatus: (no value)
Severity: low
Type: docs
Perl Version: (no value)
Fixed In: (no value)



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Thanks Jarkko for pointing it out. - Barrie --- /var/perls/perl5.005_61/lib/Benchmark.pm Tue Jul 20 13:17:59 1999 +++ lib/Benchmark.pm Fri Sep 17 06:47:03 1999 @@ -273,7 +273,9 @@ sub debug { $debug = ($_[1] != 0); } -sub clearcache { delete $cache{$_[0]}; } +# The cache needs two branches: 's' for strings and 'c' for code. The +# emtpy loop is different in these two cases. +sub clearcache { delete $cache{"$_[0]c"}; delete $cache{"$_[0]s"}; } sub clearallcache { %cache = (); } sub enablecache { $cache = 1; } sub disablecache { $cache = 0; } @@ -362,11 +364,18 @@ croak "runloop unable to compile '$c': $@\ncode: $subcode\n" if $@; print STDERR "runloop $n '$subcode'\n" if $debug; - $t0 = Benchmark->new(0); + # Wait for the user timer to tick. This makes the error range more like -0.01, +0. If + # we don't wait, then it's more like -0.01, +0.01. This may not seem important, but it + # significantly reduces the chances of getting too low initial $n in the initial, 'find + # the minimum' loop in &runfor. This, in turn, can reduce the number of calls to + # &runloop a lot, and thus reduce additive errors. + my $tbase = Benchmark->new(0)->[1]; + do { + $t0 = Benchmark->new(0); + } while ( $t0->[1] == $tbase ) ; &$subref; $t1 = Benchmark->new($n); $td = &timediff($t1, $t0); - timedebug("runloop:",$td); $td; } @@ -377,12 +386,12 @@ my($wn, $wc, $wd); printf STDERR "timeit $n $code\n" if $debug; - - if ($cache && exists $cache{$n}) { - $wn = $cache{$n}; + my $cache_key = $n . ( ref( $code ) ? 'c' : 's' ) ; + if ($cache && exists $cache{$cache_key} ) { + $wn = $cache{$cache_key}; } else { - $wn = &runloop($n, ''); - $cache{$n} = $wn; + $wn = &runloop($n, ref( $code ) ? sub { undef } : '' ); + $cache{$cache_key} = $wn; } $wc = &runloop($n, $code); @@ -414,24 +423,23 @@ my ($n, $td, $tc, $ntot, $rtot, $utot, $stot, $cutot, $cstot ); - # First find the minimum $n that gives a non-zero timing. + # First find the minimum $n that gives a significant timing. my $nmin; - for ($n = 1, $tc = 0; $tc <= 0; $n *= 2 ) { + for ($n = 1, $tc = 0; ; $n *= 2 ) { $td = timeit($n, $code); $tc = $td->[1] + $td->[2]; + last if $tc > 0.1 ; } $nmin = $n; my $ttot = 0; my $tpra = 0.05 * $tmax; # Target/time practice. - # Double $n until we have think we have practiced enough. - for ( $n = 1; $ttot < $tpra; $n *= 2 ) { + for ( ; $ttot < $tpra; $n *= 2 ) { $td = timeit($n, $code); - $tc = $td->cpu_p; $ntot += $n; $rtot += $td->[0]; $utot += $td->[1]; @@ -446,7 +454,7 @@ # Then iterate towards the $tmax. while ( $ttot < $tmax ) { $r = $tmax / $ttot - 1; # Linear approximation. - $n = int( $r * $n ); + $n = int( $r * $ntot ); $n = $nmin if $n < $nmin; $td = timeit($n, $code); $ntot += $n;


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