- Speed test. Clarification

From: Marco Ciot <"Marco>
Date: Sun, 5 Nov 2006 12:56:28 +0100

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Content-type: text/plain

Dear Rex

You must be really having a fast computer.
When I tried your code using global index variables it took 84s.
Changing to local variables except for the big arrays accelerated the speed
by a factor 10!
This is because appearently BB allocates register variables when available
for local variables, BUT NOT for global ones.

You should be careful with the use of global variables as they are meant to
represent the state of a module and not much more.

Unfortunately the BB Compiler does not do many optimisations. It very much
unfolds constant expressions and as I discovered, now, uses registers as
local variables. But for example it doesn't reuse results of identical
operations even within short distance, i.e. your i-j.

I by the way have started looking into the code of the compiler and have
added some small syntactical features.
I had even considered making it a personal project of integrating some
advanced intermediate program representation like Static Single Assignment
in order to stepwise add more optimisations.
But the thing is, that as soon as memory becomes the bottle neck there is
not use in compiling better.
And if the CPU only has to wait for Memory, than things are still ok.,
mostly it waits for the IO anyway!

That is why I'm working on the framework allowing to improve development
speed, which is much more crucial than code performance, which anyway by 90%
depends on the programmer and not on the comiler.

Marco (mc)


-----Original Message-----
From: blackbox{([at]})nowhere.xy
Couture
Sent: Friday, November 03, 2006 22:05
To: BlackBox Mailing List
Subject: [BlackBox] - Speed test. Clarification


        That ought to teach me not to post partial code. The complete
programs are posted below (I added timers).

        I made the arrays big, (1) so I didn't have to bother with timers in
three compilers, and (2) to make sure they wouldn't fit in the cache. For
comparing compilers, I wanted to avoid erratic results due to the cache.

        I like Wojtek's program, though. Caching makes almost a threefold
improvement in speed on my computer, so yes, memory is the bottleneck.

        I played a little more with unrolling the loop. Unrolling 5 times
gave an 18% improvement with an array length of 100, but it was 2% worse
with a length of 1,000,000. Your mileage may vary.

        There is one other inconvenient truth, however. With the FORTRAN
program, with an array size of around 15,000 or lower (number of arithmetic
operations constant), so everything fit in the cache, the operation was
astonishingly fast -- 2 1/2 seconds, compared to 10 seconds at best for BB.
But again, that's an unfair comparison, because with range checking, FORTRAN
takes 10 minutes 30 seconds!

Rex Couture

=======
MODULE TestTime;
IMPORT StdLog, Services;

VAR
        a, b: ARRAY 1000000 OF REAL;
        i, j: INTEGER;
        sum: REAL;
        t0: LONGINT;

PROCEDURE Do*;
BEGIN
t0:= Services.Ticks();
sum:= 0;
FOR j:= 0 TO 199 DO
        FOR i:= 0 TO 999999 DO
                a[i]:= i-j;
                b[i]:= i-j;
                END;
        FOR i:= 0 TO 999999 DO
                sum:= sum +a[i]*b[i];
                END;
        END;
StdLog.Int((Services.Ticks() - t0));
StdLog.Real( sum); StdLog.Ln
END Do;

========
MODULE TestTime2; (* Unrolls loop *)

IMPORT StdLog, Services;

VAR
        a, b: ARRAY 1000000 OF REAL;
        i, j: INTEGER;
        sum: REAL;
        t0: LONGINT;

PROCEDURE Do*;
BEGIN
t0:= Services.Ticks();
sum:= 0;
FOR j:= 0 TO 199 DO
        FOR i:= 0 TO 999999 BY 5 DO
                a[i]:= i-j;
                b[i]:= i-j;
                a[i+1]:= i-j;
                b[i+1]:= i-j;
                a[i+2]:= i-j;
                b[i+2]:= i-j;
                a[i+3]:= i-j;
                b[i+3]:= i-j;
                a[i+4]:= i-j;
                b[i+4]:= i-j;
                END;
        FOR i:= 0 TO 999999 BY 5 DO
                sum:= sum +a[i]*b[i];
                sum:= sum +a[i+1]*b[i+1];
                sum:= sum +a[i+2]*b[i+2];
                sum:= sum +a[i+3]*b[i+3];
                sum:= sum +a[i+4]*b[i+4];
                END;
        END;
StdLog.Int((Services.Ticks() - t0));
StdLog.Real( sum); StdLog.Ln
END Do;

END TestTime2.

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--- BlackBox
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Received on Sun Nov 05 2006 - 12:56:28 UTC

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