Re: Benchmarking for loops?

In Benchmark.pm: Does subroutine testing order bias results? it was suggested that the first test is consistently faster, even if the tests are identical.

I added a 2nd "timethese" to your code, with the tests reversed:

Benchmark: timing 30 iterations of a, b...
         a: 26 wallclock secs (26.64 usr +  0.00 sys = 26.64 CPU) @  1
+.13/s (n=30)
         b: 32 wallclock secs (31.64 usr +  0.00 sys = 31.64 CPU) @  0
+.95/s (n=30)
Benchmark: timing 30 iterations of a2, b2...
        a2: 20 wallclock secs (20.10 usr +  0.00 sys = 20.10 CPU) @  1
+.49/s (n=30)
        b2: 28 wallclock secs (27.25 usr +  0.00 sys = 27.25 CPU) @  1
+.10/s (n=30)
[download]

It appears that at least some of the difference is due to the order of the tests.

Comment on Re: Benchmarking for loops? Download Code

Replies are listed 'Best First'.
Re^2: Benchmarking for loops? by graff (Chancellor) on Jul 13, 2004 at 01:37 UTC
Do you mean to say that sub "a2" is identical to "b", and "b2 is the same code as "a"? (I just wasn't sure what you meant.)	[reply]
Re^3: Benchmarking for loops? by keszler (Priest) on Jul 13, 2004 at 02:40 UTC
Yes, exactly. Cut&paste, rename a -> b2, b -> a2.	[reply]