ccsupos%prism@GATECH.EDU ("SCHREIBER, O. A.") (09/21/89)
I would like to time a FORTRAN program.
Would anybody tell me how to do it on an
IRIS4D120GTX and IRIS4D50GT.
I have tried call time(t) without luck.
Thanks in advance.
Olivier Schreiber (404)894 6147, Office of Computing Services
Georgia Institute of Technology, Atlanta Georgia, 30332
uucp: ...!{allegra,amd,hplabs,seismo,ut-ngp}!gatech!prism!ccsupos
ARPA: ccsupos@prism.gatech.educalvin@dinkum.wpd.sgi.com (Calvin H. Vu) (09/22/89)
In article <8909211423.AA06605@prism.gatech.edu>, ccsupos%prism@GATECH.EDU ("SCHREIBER, O. A.") writes: > I would like to time a FORTRAN program. > Would anybody tell me how to do it on an > IRIS4D120GTX and IRIS4D50GT. > I have tried call time(t) without luck. > Thanks in advance. > If you don't need better resolution than seconds, you can use time(). The way to use it is: integer t1, t2, time external time t1 = time() ......................... do something here t2 = time() - t1 ! t2 should be the elapsed time The reason for the EXTERNAL statement is that there are two versions of time(): an intrinsic VMS-compatible version, and a library version. Using time() without the EXTERNAL statement will default to the intrinsic version and give you an error message if you don't provide an argument. This is documented in the 3.1 release notes. If you want microsecond resolution you will have to use C and call gettimeofday(). You can look up in the manual to find out how to call a C subroutine from Fortran. > Olivier Schreiber (404)894 6147, Office of Computing Services > Georgia Institute of Technology, Atlanta Georgia, 30332 > uucp: ...!{allegra,amd,hplabs,seismo,ut-ngp}!gatech!prism!ccsupos > ARPA: ccsupos@prism.gatech.edu Hope it helps, Calvin Vu Silicon Graphics Computer Systems calvin@sgi.sgi.com
glennrp@BRL.MIL (Glenn Randers-Pehrson, WMB) (09/22/89)
Here's what I use for timing Fortran programs. The Fortran program
contains
CALL CPUSEC(TIME)
to get the elapsed user+system time, in seconds.
I wouldn't mind having a version that properly reports total cpu time
for multiprocessor jobs on the Power Series.
----cpusec.f-----
subroutine cpusec (time)
c
c Glenn Randers-Pehrson, US Army Ballistic Research Laboratory
c <glennrp@brl.mil> 11 Jan 89
c
c Silicon Graphics IRIS version, works on 4D's and 3xxx's
c Returns elapsed system+user cpu time in seconds, accurate to
c 1/60th second. If more than 4295 seconds elapse between calls
c to cpusec, n*4295 seconds will not be recorded.
c 4295 (actually 4294.967296) is (2**32)/1000000.
c With mp jobs on Power series 4D's, may only report elapsed
c time for one of the cpu's.
c
real lasttime,thistime iris
save lasttime, s4295 iris
data lasttime /0.0/, s4295 /0.0/ iris
call ccpusec(now) iris
thistime=now/1 000 000. iris
if(thistime.lt.lasttime) s4295 = s4295 + 4294.967296 iris
lasttime=thistime iris
time = s4295 + thistime iris
c
c Cray version
C call ccpusec(time) cray
c
C Alliant version
C dimension ait(2) alliant
C xxx=etime(ait) alliant
C time = ait(1)+ait(2) alliant
c
return
end
----ccpusec.c-----
#ifdef sgi
/*
* get cpu + system time for f77 program on IRIS
* Glenn Randers-Pehrson, BRL, September 87
*/
#ifdef mips
fortran ccpusec_(thistime)
#else
fortran ccpusec(thistime)
#endif
/* returns unsigned integer, counting from 0 to (2**32 - 1)
* in 16666 microsecond steps, representing elapsed system+user
* time. If more than 4295 seconds elapse between calls to clockf,
* n*4295 seconds will not be recorded. */
long *thistime;
{
long clock();
*thistime = clock();
}
#endif
#ifdef cray
/*
* get cpu + system time for CFT & CFT77 program
* Gary Kuehl, BRL, 11 Jan 89
*/
#include <sys/types.h>
#include <sys/times.h>
#ifdef CRAY2
#define CYCLE_TIME 4.1E-9
#else
#define CYCLE_TIME 8.5E-9
#endif
int CCPUSEC(sec)
float *sec;
{
struct tms *t;
times(t);
*sec=(float)(t->tms_utime+t->tms_stime)*CYCLE_TIME;
return;
}
#endif
------
Glenn Randers-Pehrson
US Army Ballistic Laboratory
Aberdeen Proving Ground, MD 21005-5066
<glennrp@brl.mil>jmb@patton.sgi.com (Jim Barton) (09/25/89)
In article <8909220905.aa22118@TBD.BRL.MIL>, glennrp@BRL.MIL (Glenn Randers-Pehrson, WMB) writes: > > > Here's what I use for timing Fortran programs. The Fortran program > contains > CALL CPUSEC(TIME) > to get the elapsed user+system time, in seconds. > > I wouldn't mind having a version that properly reports total cpu time > for multiprocessor jobs on the Power Series. ... > Glenn Randers-Pehrson > US Army Ballistic Laboratory > Aberdeen Proving Ground, MD 21005-5066 > <glennrp@brl.mil> If you are willing to do a little work, you can get the sum of all CPU time used by calling the 'times()' system call yourself from the the initial process of the share group (i.e., the parent). Then sum in the child user+sys times to the current process times. You can see how this might work by timing a parallel command: $ time command [args] ... You'll notice that the real time is usually less than the user+sys time used, if the threads were computing for a significant amount of time. I suspect that the CPUSEC call in Fortran just calls 'times()' and returns the current process user+sys. -- Jim Barton Silicon Graphics Computer Systems "UNIX: Live Free Or Die!" jmb@sgi.sgi.com, sgi!jmb@decwrl.dec.com, ...{decwrl,sun}!sgi!jmb