Re: Revolution, Python and bytecode optimization - A teeny math benchmark

[Alex Tweedly]

At 08:33 07/12/2004 -0800, Gordon wrote:
Teeny math benchmark:
10,000,000 iterations of a loop, calculating the
square root of the loop index on each pass.
Revolution 2.5
  repeat with n = 1 to niter
    put sqrt(n) into tmp
  end repeat
Time = 7.133 seconds (elapsed)
Python 2.3
while n <= niter:
   s = math.sqrt(n)
   n += 1
Time = 18.1 seconds (elapsed)

<big hand-waving disclaimer>
I know I know ... the author of this little benchmark
is fully aware that this is just one tiny aspect of
each langauge and that it doesn't prove anything and
you shouldn't sell the farm based upon this etc. etc.
... He was just curious and wanted to make the point
about bytecode optimization.
</big hand-waving disclaimer>
I'll get around to a longer reply on the underlying issues ..... but a 
quick comment on your benchmark. I read your disclaimer - but even allowing 
for that, I don't understand why you used such different forms of loops to 
compare.

If you're doing the Python version
while n <= niters:
    n += 1
why not do the Rev as
repeat while n <= niters
   add 1 to n
end repeat
(For rev, the "repeat while" version is 60% slower than "repeat for".)
Or, otoh, why not compare
repeat for n=1 to niters
with
for n in range(niters):
(the "for" version is about 25% faster in Python than the "while" version)
And of course, I still can't figure out why a bytecode optimizer wouldn't 
achieve what a good compiler optimizer would - i.e. eliminate the entire 
loop, setting the terminal values for n and tmp.

-- Alex.
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