Re: Why is this code so much slower than the same algorithm in C?

At first glance your two algorithms look different.

Re-writing as

#!/usr/bin/perl

sub test {
    my ( $i, $j );

    for ( $i = 20; ; $i += 20 ) {
        for ( $j = 1; $j < 20; $j++ ) {
            last if ( $i % $j );
        }

        if ( $j == 20 ) {
            printf( "Number: %d\n", $i );
            last;
        }
    }

    return( 0 );
}

test();
[download]

should more closely match your original C code.

But even after re-writing to match your original algorithm this takes 20 odd seconds on my machine. When I add use integer; it cuts it down to 13 seconds.

Comment on Re: Why is this code so much slower than the same algorithm in C? Select or Download Code

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Re^2: Why is this code so much slower than the same algorithm in C? by wanna_code_perl (Friar) on Dec 09, 2008 at 04:04 UTC
Thanks for the comments. Here are a few more results, based on your suggestions and comments: Yes, I acknowledged in the original post that the C algorithm is actually a bit less efficient. Putting the missing comparison into the Perl version instead of the loop label, as you have done, actually had no measurable effect on runtime. Moving `my ($i, j)` outside the loop had no measurable effect on runtime. Also not too surprising. Running your version exactly as listed took on average 48.8 +/- 0.1 sec—again no measurable change from my previous results. True, I could use more rigorous methods to more accurately measure the effect of the above changes, but at best we'd be looking at a fraction of a percentage difference. Nowhere near the 3000+ % difference in the C version. Adding `use integer;` to your version caused it to run in 46.8 sec, which is a marginal improvement. (My test machine has an FPU). Not forgetting my original question of why this runs so slow, your optimization ideas do help shed a little light on a few of the factors that may be influencing performance. It's perhaps important to underscore that my question is not, "what can I do to speed up this random little demonstration program", but instead, "why is Perl so much slower than C on some arbitrary, computationally expensive algorithm". Perhaps more to the point, "what are the factors influencing the performance of tight loops in Perl". (Or where could I read more about that topic!) Hope that makes a bit more sense. Thanks again for your insights!	[reply] [d/l] [select]
Re^3: Why is this code so much slower than the same algorithm in C? (ops) by tye (Sage) on Dec 09, 2008 at 04:32 UTC
Because the C code and the Perl code end up doing roughly equivalent things except that the C code runs through a few dozen machine-language instructions while the Perl code runs through a few dozen "opnodes". Machine-language instructions are the fastest unit of execution on a computer. While the slowness of opnode dispatch (each requires several dozen or even hundreds of machine-language instructions) is one of the motivations for some of the design changes in Perl 6. - tye	[reply]


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