Re: inlined DEBUG constant versus $DEBUG

stefp,

Let's look at the numbers:

#!/usr/local/bin/perl
use Benchmark;

use constant DEBUG_C => (1);
sub DEBUG_S() { 1 }
$DEBUG_V = 1;


timethese( 4000000, {
        'constant'   => sub { my $x = 1 if DEBUG_C; my $y=2; },
        'subroutine' => sub { my $x = 1 if DEBUG_S; my $y=2; },
        'variable'   => sub { my $x = 1 if $DEBUG_V; my $y=2; },
  }
);
[download]

produces the following (for DEBUG set to 1 and 0)

DEBUG 0, Run 1
Benchmark: timing 4000000 iterations of constant, subroutine, variable...
  constant:  3 wallclock secs ( 2.00 usr +  0.01 sys =  2.01 CPU) @ 1990049.75/s (n=4000000)
subroutine:  3 wallclock secs ( 1.99 usr +  0.00 sys =  1.99 CPU) @ 2010050.25/s (n=4000000)
  variable:  5 wallclock secs ( 3.93 usr +  0.00 sys =  3.93 CPU) @ 1017811.70/s (n=4000000)

DEBUG 0, Run 2
Benchmark: timing 4000000 iterations of constant, subroutine, variable...
  constant:  1 wallclock secs ( 1.99 usr +  0.00 sys =  1.99 CPU) @ 2010050.25/s (n=4000000)
subroutine:  1 wallclock secs ( 2.13 usr +  0.00 sys =  2.13 CPU) @ 1877934.27/s (n=4000000)
  variable:  3 wallclock secs ( 3.85 usr +  0.00 sys =  3.85 CPU) @ 1038961.04/s (n=4000000)

DEBUG 0, Run 3
Benchmark: timing 4000000 iterations of constant, subroutine, variable...
  constant:  1 wallclock secs ( 2.07 usr +  0.00 sys =  2.07 CPU) @ 1932367.15/s (n=4000000)
subroutine:  1 wallclock secs ( 2.19 usr +  0.00 sys =  2.19 CPU) @ 1826484.02/s (n=4000000)
  variable:  3 wallclock secs ( 3.78 usr +  0.00 sys =  3.78 CPU) @ 1058201.06/s (n=4000000)

DEBUG 1, Run 1
Benchmark: timing 4000000 iterations of constant, subroutine, variable...
  constant:  8 wallclock secs ( 7.47 usr +  0.00 sys =  7.47 CPU) @ 535475.23/s (n=4000000)
subroutine:  8 wallclock secs ( 7.49 usr +  0.00 sys =  7.49 CPU) @ 534045.39/s (n=4000000)
  variable:  9 wallclock secs ( 9.13 usr +  0.00 sys =  9.13 CPU) @ 438116.10/s (n=4000000)

DEBUG 1, Run 2
Benchmark: timing 4000000 iterations of constant, subroutine, variable...
  constant:  8 wallclock secs ( 7.53 usr +  0.00 sys =  7.53 CPU) @ 531208.50/s (n=4000000)
subroutine:  7 wallclock secs ( 7.41 usr +  0.00 sys =  7.41 CPU) @ 539811.07/s (n=4000000)
  variable:  8 wallclock secs ( 9.27 usr +  0.00 sys =  9.27 CPU) @ 431499.46/s (n=4000000)

DEBUG 2, Run 3
Benchmark: timing 4000000 iterations of constant, subroutine, variable...
  constant:  8 wallclock secs ( 7.38 usr +  0.00 sys =  7.38 CPU) @ 542005.42/s (n=4000000)
subroutine:  8 wallclock secs ( 7.23 usr +  0.00 sys =  7.23 CPU) @ 553250.35/s (n=4000000)
  variable:  9 wallclock secs ( 9.06 usr +  0.00 sys =  9.06 CPU) @ 441501.10/s (n=4000000)

Looking at those numbers - sure $DEBUG is a bit slower but I'd say the savings are a wash - especially if there's some network or database code also in the script. The amount of time you save from not using $DEBUG is really down in the noise. The key here is to choose one method and be consistent with it's application across all scripts/modules in your project(s).

-derby

Comment on Re: inlined DEBUG constant versus $DEBUG Download Code

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Re: Re: inlined DEBUG constant versus $DEBUG by stefp (Vicar) on Sep 19, 2001 at 18:23 UTC
I never claimed significant speed improvement. What I had in mind is debugging thru the perl debugger. In that case the print statement becomes noise. I don't want anything to print, I even don't want to bother to go thru the statement that contains the print. Oddly, with `use constant DEBUG => 0` I indeed step thru the statement that contains the print: `print "whatever\n" if DEBUG`. Worse I also step thru one line of constant.pm! With `sub DEBUG() { 0 }` I still step thru the statement that contains the print. When debugging one goes from one nextstate opcode to the next. We see that consecutive nextstate statements are not fusionned as I expected them to be: Both oneliners give me the same tree: perl -e ' sub DEBUG() { 0 } ; use O qw( Concise -exec); print "toto" if DEBUG; print "toto" if DEBUG' `perl -e ' use constant DEBUG =>0; use O qw( Concise -exec); print "tot +o" if DEBUG; print "toto" if DEBUG' 1 <0> enter 2 <;> nextstate(main 1 -e:1) v 3 <;> nextstate(main 1 -e:1) v 4 <@> leave[t1] vKP/REFC` [download] I don't understand how I get thru one line of constant.pm when debugging using constants!!! I tested using perl 5.6.1 -- stefp	[reply] [d/l] [select]
Re: Re: Re: inlined DEBUG constant versus $DEBUG by derby (Abbot) on Sep 19, 2001 at 19:14 UTC
stefp, That's why I tend to favor 'n' over 's'. However, when I debug code that contains `sub DEBUG () { 0 }` [download] I still step through the DEBUG subroutine. Which is what I would expect while debugging. Try: perl -de 'use constant DEBUG =>0; use O qw( Concise -exec); print "toto" if DEBUG; print "toto" if DEBUG' perl -de 'sub DEBUG() { 0 }; use O qw( Concise -exec); print "toto" if DEBUG; print "toto" if DEBUG' -derby	[reply] [d/l]