Re: progress bar in fortran

In article <d46a483c-164d-4cd8-8f4e-04bb0ca26566@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
rudra <bnrj.rudra@xxxxxxxxx> wrote:

is there any way to generate something like that using fortran? where
the progress percentage will update in the same line?

Probably non-standard, but writing char(13) writes a carriage-return
which moves the cursor back to the start of the line.

Really crappy somewhat non-standard F77 code follows: works under g77 on
linux and IRIX and IFC on Linux. It uses an external C function
hires_time() to get a double precision time value (secs since the epoch
in our case): I'm sure you can roll your own version, and there may
well be an F90/F95 equivalent.

Not thread safe. Use at own risk. Warranty void if protective sticker
removed. You should really be using something more modern than F77
anyway :).

c print a 'progress bar' to unit 0.
c Call with ndone=0 before each bar starts
c Call with ndone=ntotal to finish
c
subroutine progress(string,ndone,ntotal)
implicit none
character*(*) string
character*255 prog,oldprog
double precision oldtime,hires_time,tl
integer ndone,ntotal,i
save oldprog,oldtime

if (ndone.eq.0) oldtime=hires_time()
tl=hires_time()-oldtime
if (tl.lt.0) tl=0
if (ndone.gt.0) tl=(1.0*ntotal/ndone)*tl-tl

c When finished, print the total time taken rather
c than just 00:00!
if (ndone.eq.ntotal) tl=hires_time()-oldtime

write(prog,'(a25,1x,''['')') string
do i=1,40
prog(27+i:27+i)=' '
enddo
write(prog(43:51),'(f7.1,''%'')') 100.0*ndone/ntotal
do i=1,40
if ((1.0*ndone/ntotal).gt.(1.0*i/40)) then
if (prog(27+i:27+i).eq.' ') prog(27+i:27+i)='#'
endif
enddo
prog(67:67)=']'
write(prog(70:72),'(i2.2,'':'')')int(tl/3600)
write(prog(73:75),'(i2.2,'':'')')int((tl-int(tl/3600)*3600)/60)
write(prog(76:77),'(i2.2)')int((tl-int(tl/60)*60))
if (prog.ne.oldprog) write(0,'(a,a,$)') prog(1:77),char(13)
oldprog=prog
if (ndone.eq.ntotal) write(0,*)
return
end


--
Mark Mackey http://www.swallowtail.org/
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