1. delay loop

Please, could anyone with cpu over 1000mhz. tell me the output of the following?
... 3515 on P166.

--
include get.e
atom tick,ti,z,x tick=0 ti=time()
while time()-ti < .5 do
tick+=1 z=power(56.982,7.3)/tick
end while
function abs(object a)
if a<0 then return-a else return a end if
end function
procedure soft_delay()
for i = 1 to 2 + abs(tick/2400) do
z=power(56.982,7.3)/i end for
end procedure
x=0 ti=time()
while time()-ti < 5 do
soft_delay() soft_delay()
x += 1 end while
printf(1,"%d %s\n",{x/5,":done, any key exits"})
z=wait_key()
--
TIA, Wolf

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2. Re: delay loop

>Please, could anyone with cpu over 1000mhz. tell me the output of the
following?                                                            
>... 3515 on P166.                                                    
>                                                                     
                                                                      
I'm and AMD 1000mhz                                                   
i get 15889                                                           
                                                                      
Dan McGrath

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3. Re: delay loop

----- Original Message -----
From: "Wolf" <wolfritz at KING.IGS.NET>
To: "EUforum" <EUforum at topica.com>
Subject: delay loop


>
> Please, could anyone with cpu over 1000mhz. tell me the output of the
following?
> ... 3515 on P166.
>
> --

I got 14821 on a Pentium III, 550mHz

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4. Re: delay loop

3382, under AMD 800, win XP (exw)

Rubens


At 01:06 17/5/2002, you wrote:
>
>Please, could anyone with cpu over 1000mhz. tell me the output of the 
>following?
>... 3515 on P166.
>
>--
>include get.e
>atom tick,ti,z,x tick=0 ti=time()
>while time()-ti < .5 do
>tick+=1 z=power(56.982,7.3)/tick
>end while
>function abs(object a)
>if a<0 then return-a else return a end if
>end function
>procedure soft_delay()
>for i = 1 to 2 + abs(tick/2400) do
>z=power(56.982,7.3)/i end for
>end procedure
>x=0 ti=time()
>while time()-ti < 5 do
>soft_delay() soft_delay()
>x += 1 end while
>printf(1,"%d %s\n",{x/5,":done, any key exits"})
>z=wait_key()
>--
>TIA, Wolf
>
>
>

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5. Re: delay loop

18500/19700 AMD 800 , DOS, (ex)

At 01:06 17/5/2002, you wrote:
>
>Please, could anyone with cpu over 1000mhz. tell me the output of the 
>following?
>... 3515 on P166.
>
>--
>include get.e
>atom tick,ti,z,x tick=0 ti=time()
>while time()-ti < .5 do
>tick+=1 z=power(56.982,7.3)/tick
>end while
>function abs(object a)
>if a<0 then return-a else return a end if
>end function
>procedure soft_delay()
>for i = 1 to 2 + abs(tick/2400) do
>z=power(56.982,7.3)/i end for
>end procedure
>x=0 ti=time()
>while time()-ti < 5 do
>soft_delay() soft_delay()
>x += 1 end while
>printf(1,"%d %s\n",{x/5,":done, any key exits"})
>z=wait_key()
>--
>TIA, Wolf
>
>
>

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6. Re: delay loop

this seems strange to me...
17950 AMD 1000 Athlon under Windows ME

but then.. same machine booted into Linux Mandrake 8.3
4188!

Can anyone tell me explain this discrepancy?

Ken Rhodes





> At 01:06 17/5/2002, you wrote:
> >
> >Please, could anyone with cpu over 1000mhz. tell me
> the output of the 
> >following?
> >... 3515 on P166.
> >
> >--
> >include get.e
> >atom tick,ti,z,x tick=0 ti=time()
> >while time()-ti < .5 do
> >tick+=1 z=power(56.982,7.3)/tick
> >end while
> >function abs(object a)
> >if a<0 then return-a else return a end if
> >end function
> >procedure soft_delay()
> >for i = 1 to 2 + abs(tick/2400) do
> >z=power(56.982,7.3)/i end for
> >end procedure
> >x=0 ti=time()
> >while time()-ti < 5 do
> >soft_delay() soft_delay()
> >x += 1 end while
> >printf(1,"%d %s\n",{x/5,":done, any key exits"})
> >z=wait_key()
> >--
> >TIA, Wolf
> >
> >
> 
> 
>
> 


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7. Re: delay loop

Hi Wolf,

----------

> > Please, could anyone with cpu over 1000mhz.
> > tell me the output of the following?                                   
                        
> >... 3515 on P166.                                                    
> >                                                                     
>                                                                       
> I'm and AMD 1000mhz                                                   
> i get 15889                                                           
>                                                                       

      On 386-25MHz I have:

    - 398 under Windows 95
    - 453 on plain DOS-32

    Where is a progress ???
           I ask !!!
There are no over-1000-machines here,
      it seems to be !?!?!?
 There is Euphoria here, I am sure!

smile(-:smile(-:smile(-:smile(-:smile(-:smile(-:

Regards,
Igor Kachan
kinz at peterlink.ru

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8. Re: delay loop

> this seems strange to me...
> 17950 AMD 1000 Athlon under Windows ME
> but then.. same machine booted into Linux Mandrake 8.3
> 4188!
Thank's everyone, but I'm still 'stupid'.
My first loop just checks the cpu speed, and the second was supposed to
compensate, so it was a *lot* closer to the same
on any machine. Guess I got it wrong in every way   :(

Perhaps changing:
> > >for i = 1 to 2 + abs(tick/2400) do
to:
for i = 1 to (1 + (power(tick,1.2)/22000)) do

... might get a little closer ???
486/90  =  2360
P166  =  3330

The real number's not critical, as long as it doesn't go over 80,000 on an AMD
10,000   blink
Wolf

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9. Re: delay loop

Hello Ken:

----------
> Îò: Ken Rhodes <ken_rhodes30436 at yahoo.com>
> Êîìó: EUforum <EUforum at topica.com>
> Òåìà: Re: delay loop
> Äàòà: ïÿòíèöà, Ìàé 17, 2002 22:17
> 
> this seems strange to me...
> 17950 AMD 1000 Athlon under Windows ME
> 
> but then.. same machine booted into Linux Mandrake 8.3
> 4188!
> 
> Can anyone tell me explain this discrepancy?
> 
> Ken Rhodes

I think, Linux has more clever multitasking than 
Windows ME has.
Then Mandrake 8.3 is optimized just for Athlon,
it seems to be.
I have disks with RH 7.2, it is optimized for Athlon.
What is this optimisation? Partly it may be just a kind
of the soft delays (like Wolf's procedure) for the old software to make it
more similar (for user) to itself 
on the old CPUs.

To have more clear picture with the real *performance*
in this case, you can try to run 3...4 separate processes with Wolf's
soft_delay() on your Athlon under Linux.
Just some thoughts, I do not know exactly.

New CPUs have extra power for most of the usual tasks
supported with the old software, I think.

Old CPUs are slow now, yes, but good and the best old 
software may to look 'extra' fast now, on new CPUs.

Regards,
Igor Kachan
kinz at peterlink.ru

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10. Re: delay loop

Hello Wolf,

I timed it on a 1.7Ghz P4
 3225 under WinMe,
23329 in a WinMe Dosbox

antoine tammer
a.tammer at hetnet.nl

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11. Re: delay loop

>> > >for i = 1 to 2 + abs(tick/2400) do
>to:
>for i = 1 to (1 + (power(tick,1.2)/22000)) do
 >
>.... might get a little closer ???
>486/90  =  2360
>P166  =  3330
>

On my P100 Win98 laptop I get:

 ewx                    =      4795
 ex (DOS box)    =      3619
 ex (plain DOS)  =      4069

On a old IBM 386DX/25 running DOS I get 157. sad


   



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12. Re: delay loop

11637 400Mhz Celeron w/29 process runing.
Alvin 

=====
See my homepage at
http://ka9qlq.tripod.com/home/
and where I live at
http://ka9qlq.tripod.com/CCC/

This is a 256 Megabyte 400 Mhz 13 Gig. plug and pray system.

__________________________________________________
Do You Yahoo!?
LAUNCH - Your Yahoo! Music Experience
http://launch.yahoo.com

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13. Re: delay loop

Pentium4 1.8Ghz
873

Wolf wrote:

>
>Please, could anyone with cpu over 1000mhz. tell me the output of the
>following?
>... 3515 on P166.
>
>--
>include get.e
>atom tick,ti,z,x tick=0 ti=time()
>while time()-ti < .5 do
>tick+=1 z=power(56.982,7.3)/tick
>end while
>function abs(object a)
>if a<0 then return-a else return a end if
>end function
>procedure soft_delay()
>for i = 1 to 2 + abs(tick/2400) do
>z=power(56.982,7.3)/i end for
>end procedure
>x=0 ti=time()
>while time()-ti < 5 do
>soft_delay() soft_delay()
>x += 1 end while
>printf(1,"%d %s\n",{x/5,":done, any key exits"})
>z=wait_key()
>--
>TIA, Wolf
>
>
>
>

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14. Re: delay loop

Sorry I forgot to add Win 98 OSR1 DOS box
Alvin 

=====
See my homepage at
http://ka9qlq.tripod.com/home/
and where I live at
http://ka9qlq.tripod.com/CCC/

This is a 256 Megabyte 400 Mhz 13 Gig. plug and pray system.

__________________________________________________
Do You Yahoo!?
LAUNCH - Your Yahoo! Music Experience
http://launch.yahoo.com

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15. Re: delay loop

Ken Rhodes wrote:

> but then.. same machine booted into Linux Mandrake 8.3 
> 4188!

Do you mean 8.2, or did I miss a release?

-- David Cuny

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16. Re: delay loop

OOPS! it is in fact Mandrake 8.2 :)

Ken Rhodes



--- David Cuny <dcuny at LANSET.COM> wrote:
> 
> Ken Rhodes wrote:
> 
> > but then.. same machine booted into Linux Mandrake
> 8.3 
> > 4188!
> 
> Do you mean 8.2, or did I miss a release?
> 
> -- David Cuny
> 
>
> 
> 
>
>

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17. Re: delay loop

EX: 4783
EXW: 1755
With PEU recompiled with BCC I get 858993459. But this seems to be a bug :)

Winchip 200 Mhz

    Martin

> Please, could anyone with cpu over 1000mhz. tell me the output of the
> following?
> ... 3515 on P166.
> 
> --
> include get.e
> atom tick,ti,z,x tick=0 ti=time()
> while time()-ti < .5 do
> tick+=1 z=power(56.982,7.3)/tick
> end while
> function abs(object a)
> if a<0 then return-a else return a end if
> end function
> procedure soft_delay()
> for i = 1 to 2 + abs(tick/2400) do
> z=power(56.982,7.3)/i end for
> end procedure
> x=0 ti=time()
> while time()-ti < 5 do
> soft_delay() soft_delay()
> x += 1 end while
> printf(1,"%d %s\n",{x/5,":done, any key exits"})
> z=wait_key()
> --
> TIA, Wolf

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18. Re: delay loop

On 17 May 2002, at 0:06, Wolf wrote:

> 
> Please, could anyone with cpu over 1000mhz. tell me the output of the
> following? ... 3515 on P166.
> 
> --
> TIA, Wolf
> 

Wolf:

No GHz here though I got some strange values:

1. Intel Pentium MMX 200MHz running Windows98SE

   loop.ex
   run 1 = 3604
       2 = 3878
       3 = 3501 (3552.5 avg.)

   loop.exw
   run 1 = 2423
       2 = 3084
       3 = 3899 (3135.3 avg.)

2. Intel 486 DX2 66MHz running WindowsNT4/sp6

   loop.ex
   run 1 = 569
       2 = 532
       3 = 531 (544.0 avg.)

   loop.exw
   run 1 = 2383
       2 = 3091
       3 = 2383 (2619.0 avg.)

Looks that ex.exe (or loop.ex) don't like 486s, at least as much as 
exw.exe (or loop.exw) do.

I also run sieve.ex on 486 and than sieve.exw with these following 
results:

3. Intel 486 DX2 66MHz running WindowsNT4/sp6

   sieve.ex
   run 1 = 637.9 sieves
       2 = 640.0 sieves
       3 = 641.2 sieves (639.7 sieves avg.)

   loop.exw
   run 1 = 650.0 sieves
       2 = 651.2 sieves
       3 = 651.2 sieves (650.8 sieves avg.)

Pretty much the same, huh? Any clues?

Be good,

-- Euler

--

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19. Re: delay loop

> No GHz here though I got some strange values:
I got similar strange results between 486/90 and P166, all in DOS box, however
my 486 still has EU 2.2 !
P166        .exw > 3521      .ex > 6898
486/90     .exw > 1956      .ex > 1102
.. in pure DOS this was still only 1746.

.btw, in case anyone's interested, I solved my problem with:
       for i = 1 to (2 + (floor(power(tick,1.1))/20000)) do

.. which gave counts of 6690 on a P166, and only 7230 on a Duron950, using .exw
.. not much difference, close enuf  :)

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20. Re: delay loop

"Wolf" <wolf- at KING.IGS.NET> wrote:

> Please, could anyone with cpu over 1000mhz. tell me the output of the
> following?
>  ... 3515 on P166.

I'm still with my 400 MHz AMD, and I'm also somewhat late, but this
are my results:

WinNT, under cmd.exe, with ex.exe:

796 :done, any key exits
806 :done, any key exits
809 :done, any key exits
799 :done, any key exits

WinNT, under cmd.exe with exw.exe:

3226 :done, any key exits
3268 :done, any key exits
3219 :done, any key exits
3167 :done, any key exits

MS-DOS 6.22: 

11168 :done, any key exits
10325 :done, any key exits
10327 :done, any key exits
11278 :done, any key exits

Have a nice day, Rolf

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