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André,
On 1/19/2011 6:52 AM, André Warnier wrote:
> #include <stdio.h>
>
> int main(int argc, char *argv[]) {
>
> int thousand = 1000;
> int sixty = 60;
> int twentyfour = 24;
> int hundred = 100;
> int three = 3;
>
> int cacheTime1 = (1000 * 60 * 60 * 24 * 100 * 3);
> long cacheTime2 = ((long)1000 * 60 * 60 * 24 * 100 * 3);
> long long cacheTime3 = ((long long)1000 * 60 * 60 * 24 * 100 * 3);
> long long cacheTime4 = (1000 * 60 * 60 * 24 * 100 * (long long)3);
> long long cacheTime5 = (thousand * sixty * sixty * twentyfour *
> hundred * three);
>
> printf("%d\n%ld\n%lld\n%lld\n%lld\n", cacheTime1, cacheTime2, cacheTime3,
> cacheTime4,cacheTime5);
>
> return 0;
> }
>
> aw@arthur:~/tests$ cc -ansi -pedantic -Wall -o longtime2 longtime2.c &&
> ./longtime2
> longtime2.c: In function 'main':
> longtime2.c:11: warning: integer overflow in expression
> longtime2.c:12: warning: integer overflow in expression
> longtime2.c:13: warning: ISO C90 does not support 'long long'
> longtime2.c:13: warning: ISO C90 does not support 'long long'
> longtime2.c:14: warning: ISO C90 does not support 'long long'
> longtime2.c:14: warning: integer overflow in expression <=== ****
Line 14 looks like "long long cacheTime4..." not the one you're looking
for. Yours is on line 15 where no warning was indicated.
> Maybe the Java compiler cannot afford to do that, because it is used in
> "just-in-time" compilations ?
No, Java has a well-defined conversion system regardless of the
underlying platform, so the compiler should be able to check. It just
doesn't apparently.
>> I couldn't make head or tail out of the timeit() function's return
>> value, so I just used 'time' from the command line to see what really
>> happened.
>>
> Neither could I, so I rewrote your test program somewhat (and decreased
> the loop by a factor 10, to be practical):
(Just get a bigger machine :)
> #!/usr/bin/perl
>
> use Benchmark;
>
> my $loops = 10000000;
> NORMAL:{
> my $left = 10;
> my $right = 3;
> my $res = $left / $right;
> print "\'NORMAL\' result : $res\n";
> my $t0 = Benchmark->new;
> for (my $i=0; $i<$loops; $i++) {
> my $res = $left / $right;
> }
> my $t1 = Benchmark->new;
> my $td = timediff($t1, $t0);
> print "\'NORMAL\' times :",timestr($td),"\n";
> }
>
> INTEGER:{
> use integer;
So, does "use integer" set the arithmetic evaluation strategy? I've
never see that before (though I'm not a seasoned Perl programmer).
> Note that for the BIGFLOAT and BIGINT sections, you need to be *really*
> patient.
> I interrupted at after maybe 10 minutes..
What, you didn't run it yourself? :( I ran mine for you :)
> Funny thing is, the INTEGER section seems actually slower than the
> standard one.
I'll have to run your test and see what happens.
- -chris
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