1. Timeing Question
- Posted by Euman <euman at bellsouth.net> Feb 08, 2001
- 510 views
------=_NextPart_000_000B_01C09182.98B441A0 charset="iso-8859-1" If There are 64K (65,536) ticks in one hour Why is it I cant seem to get this to work 18.204 ticks per second Or, is this too OLD SCHOOL? I had been useing the equivalent Euphoria Routine: tick_rate(100) atom t t =3D time() t =3D time()-t ? t for some reason I thought about this earlier today and thought I'd ask....... Euman ------=_NextPart_000_000B_01C09182.98B441A0 charset="iso-8859-1" <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=3DContent-Type content=3D"text/html; = charset=3Diso-8859-1"> <META content=3D"MSHTML 5.50.4611.1300" name=3DGENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=3D#ffffff> <DIV><FONT face=3DArial size=3D2>If There are 64K (65,536) ticks in one=20 hour</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>Why is it I cant seem to get this to=20 work</FONT></DIV> <DIV><FONT face=3DArial size=3D2>18.204 ticks per second</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>Or, is this too OLD = SCHOOL?</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>I had been useing the equivalent = </FONT><FONT=20 face=3DArial size=3D2>Euphoria Routine:</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>tick_rate(100)<BR>atom t<BR>t =3D = time()</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>t =3D time()-t</FONT></DIV> <DIV><FONT face=3DArial size=3D2>? t</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>for some reason I thought about this = earlier today=20 and thought</FONT></DIV> <DIV><FONT face=3DArial size=3D2>I'd ask.......</FONT></DIV> <DIV><FONT face=3DArial size=3D2></FONT> </DIV> <DIV><FONT face=3DArial size=3D2>Euman</FONT></DIV> ------=_NextPart_000_000B_01C09182.98B441A0--
2. Re: Timeing Question
- Posted by Colin Taylor <ctaylor at racsa.co.cr> Feb 08, 2001
- 471 views
There are 65,536 ticks in one hour until you use tick_rate(100), then you have 360,000 ticks in one hour. I tried this: include machine.e atom t t = time() while time() = t do -- wait next tick end while ? 1/(time()-t) tick_rate(100) t = time() while time() = t do end while ? 1/(time()-t) For some reason I don't ever get 18.2 ticks per second in the first instance; I get either 16.66 or 20. In the second instance I always get a result close to 100. - Colin Taylor Euman wrote: If There are 64K (65,536) ticks in one hour Why is it I cant seem to get this to work 18.204 ticks per second
3. Re: Timeing Question
- Posted by Euman <euman at bellsouth.net> Feb 08, 2001
- 462 views
> There are 65,536 ticks in one hour until you use tick_rate(100), then you > have 360,000 ticks in one hour. > > I tried this: > > include machine.e > atom t > > t = time() > while time() = t do -- wait next tick > end while > ? 1/(time()-t) > > tick_rate(100) > > t = time() > while time() = t do > end while > ? 1/(time()-t) > > For some reason I don't ever get 18.2 ticks per second in the first > instance; I get either 16.66 or 20. In the second instance I always get a > result close to 100. > > - Colin Taylor > > Euman wrote: > > If There are 64K (65,536) ticks in one hour > > Why is it I cant seem to get this to work > 18.204 ticks per second > > > This would explain the difference in timeing of loops (profiling) If you run a program to test for speed occasionally the times are different for the same algorytm. Maybe Robert hasnt been told about this either. or maybe he knows but hasnt done anything to correct it. Robert? Euman
4. Re: Timeing Question
- Posted by Kat <gertie at PELL.NET> Feb 08, 2001
- 479 views
On 8 Feb 2001, at 11:00, Euman wrote: > > > > There are 65,536 ticks in one hour until you use tick_rate(100), then you > > have 360,000 ticks in one hour. > > > > I tried this: > > > > include machine.e > > atom t > > > > t = time() > > while time() = t do -- wait next tick > > end while > > ? 1/(time()-t) > > > > tick_rate(100) > > > > t = time() > > while time() = t do > > end while > > ? 1/(time()-t) > > > > For some reason I don't ever get 18.2 ticks per second in the first > > instance; I get either 16.66 or 20. In the second instance I always get a > > result close to 100. > > > > - Colin Taylor > > > > Euman wrote: > > > > If There are 64K (65,536) ticks in one hour > > > > Why is it I cant seem to get this to work > > 18.204 ticks per second > > > > > > > > This would explain the difference in timeing of loops > (profiling) > > If you run a program to test for speed > occasionally the times are different for the same algorytm. > > Maybe Robert hasnt been told about this either. > or maybe he knows but hasnt done anything to correct it. The OS determines how many ticks to use on each program. If one app has a higher priority than your's, it will get available ticks before your app will. And it may interrupt your program to do OS stuff, like swap disc space. Kat
5. Re: Timeing Question
- Posted by Mike Swayze <mswayze at TRUSWOOD.COM> Feb 08, 2001
- 475 views
is this the reason for RS232 communications(timing critical) needing to be done in DOS? Swayze mswayze at truswood.com kswayze at bellsouth.net ----- Original Message ----- From: "Kat" <gertie at PELL.NET> To: <EUforum at topica.com> Sent: Thursday, February 08, 2001 3:05 PM Subject: Re: Timeing Question .......................... > > The OS determines how many ticks to use on each program. If one app has a higher > priority than your's, it will get available ticks before your app will. And it may interrupt > your program to do OS stuff, like swap disc space. > > Kat
6. Re: Timeing Question
- Posted by Mike The Spike <mtsreborn at yahoo.com> Feb 08, 2001
- 481 views
Euman is still alive? Oh, that's right! I don't get any messages from him since he is banned from my mailbox. Mike The Spike --- Kat <gertie at PELL.NET> wrote: > On 8 Feb 2001, at 11:00, Euman wrote: > > > > > > > > There are 65,536 ticks in one hour until you use > tick_rate(100), then you > > > have 360,000 ticks in one hour. > > > > > > I tried this: > > > > > > include machine.e > > > atom t > > > > > > t = time() > > > while time() = t do -- wait next tick > > > end while > > > ? 1/(time()-t) > > > > > > tick_rate(100) > > > > > > t = time() > > > while time() = t do > > > end while > > > ? 1/(time()-t) > > > > > > For some reason I don't ever get 18.2 ticks per > second in the first > > > instance; I get either 16.66 or 20. In the > second instance I always get a > > > result close to 100. > > > > > > - Colin Taylor > > > > > > Euman wrote: > > > > > > If There are 64K (65,536) ticks in one hour > > > > > > Why is it I cant seem to get this to work > > > 18.204 ticks per second > > > > > > > > > > > > > This would explain the difference in timeing of > loops > > (profiling) > > > > If you run a program to test for speed > > occasionally the times are different for the same > algorytm. > > > > Maybe Robert hasnt been told about this either. > > or maybe he knows but hasnt done anything to > correct it. > > The OS determines how many ticks to use on each > program. If one app has a higher > priority than your's, it will get available ticks > before your app will. And it may interrupt > your program to do OS stuff, like swap disc space. > > Kat > >
7. Re: Timeing Question
- Posted by Kat <gertie at PELL.NET> Feb 08, 2001
- 474 views
On 8 Feb 2001, at 12:09, Mike Swayze wrote: > is this the reason for RS232 communications(timing critical) needing to be > done in DOS? That and the windoze habit of "virtualizing" the ports by putting a software layer between the hardware and the software, so the app doesn't see the real port or interrupt. Kat > Swayze > mswayze at truswood.com > kswayze at bellsouth.net > > ----- Original Message ----- > From: "Kat" <gertie at PELL.NET> > To: <EUforum at topica.com> > Sent: Thursday, February 08, 2001 3:05 PM > Subject: Re: Timeing Question > > > ........................... > > > > The OS determines how many ticks to use on each program. If one app has a > higher > > priority than your's, it will get available ticks before your app will. > And it may interrupt > > your program to do OS stuff, like swap disc space. > > > > Kat > > >
8. Re: Timeing Question
- Posted by Mike Swayze <mswayze at TRUSWOOD.COM> Feb 09, 2001
- 469 views
are there any work arounds for this 'virtual' layer, or hooks? Swayze mswayze at truswood.com kswayze at bellsouth.net ----- Original Message ----- From: "Kat" <gertie at PELL.NET> To: "EUforum" <EUforum at topica.com> Sent: Thursday, February 08, 2001 4:08 PM Subject: Re: Timeing Question > On 8 Feb 2001, at 12:09, Mike Swayze wrote: > > > is this the reason for RS232 communications(timing critical) needing to be > > done in DOS? > > That and the windoze habit of "virtualizing" the ports by putting a software layer > between the hardware and the software, so the app doesn't see the real port or > interrupt. > > Kat > > > Swayze > > mswayze at truswood.com > > kswayze at bellsouth.net > > > > ----- Original Message ----- > > From: "Kat" <gertie at PELL.NET> > > To: <EUforum at topica.com> > > Sent: Thursday, February 08, 2001 3:05 PM > > Subject: Re: Timeing Question > > > > > > ........................... > > > > > > The OS determines how many ticks to use on each program. If one app has a > > higher > > > priority than your's, it will get available ticks before your app will. > > And it may interrupt > > > your program to do OS stuff, like swap disc space. > > > > > > Kat > > > > > > > > >