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util.F
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util.F
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c&UTIL
c&ADDARR
subroutine addarr(n,aname,lab,m)
c*********************************************************************
c
c Subroutine to add an array to the vector array. The elements to
c be added are to be in wk(m),...,wk(m+n-1) (passed in common).
c
c**********************************************************************
c
#include 'tslabc'
character aname*15
num=num-n
numarr=numarr+1
xlist(numarr)=aname
label(numarr)=lab
nstart(numarr)=1
if(numarr.gt.1) nstart(numarr)=nend(numarr-1)+1
nend(numarr)=nstart(numarr)+n-1
nn=m - nstart(numarr)
do 50 i=nstart(numarr),nend(numarr)
50 array(i)=wk(nn+i)
c
c
99 continue
return
end
c&CLEAN
subroutine clean(narr)
c********************************************************************
c
c Subroutine to delete the array having index narr. If narr=0, then
c the user is given the opportunity to delete any or all of the
c arrays.
c
c*********************************************************************
c
#include 'tslabc'
character*1 i10
c
c
numr=numarr
do 100 i=numr,1,-1
c
c
if ( narr .eq. 0 ) goto 5
if ( narr .eq. i ) goto 40
goto 100
c
c we know here that narr = 0:
c
5 write(*,10) xlist(i),nend(i)-nstart(i)+1
10 format(' do you want to delete array ',a15,' having ',
1 i5,' elements?(y/n/q)'$)
read(*,20) i10
20 format(a1)
if(i10.eq.'Q'.or.i10.eq.'q') go to 99
if(i10.eq.'N'.or.i10.eq.'n') go to 100
if(i10.eq.'Y'.or.i10.eq.'y') go to 40
write(*,30)
30 format(' improper response')
go to 5
40 continue
c
c shift everything (unless we're deleting the last array) :
c
nn=nend(i)-nstart(i)+1
numarr=numarr-1
num=num+nn
if(narr.ne.0.and.i.gt.numarr) go to 99
if(i.gt.numarr) go to 100
do 50 j=i,numarr
xlist(j)=xlist(j+1)
50 label(j)=label(j+1)
do 55 j=i,numarr
ii=1
if(j.gt.1) ii=nend(j-1)+1
nstart(j)=ii
nend(j)=ii+nend(j+1)-nstart(j+1)
55 continue
c do 60 j=i,numarr
c ns=nstart(j)-1
c n=nend(j)-nstart(j)+1
c do 65 k=1,n
c 65 array(ns+k)=array(ns+k+nn)
c 60 continue
do 60 k = nstart(i),nend(numarr)
60 array(k) = array(k+nn)
if(narr.ne.0) go to 99
100 continue
c
c
99 continue
return
end
c&FINDCH
subroutine findch(l,n1,n2,char,nocc,ncol)
c********************************************************
c
c subroutine to find the first occurrence ncol of the
c single character char in l(n1),...,l(n2), and the
c number of occurrences nocc.
c
c*********************************************************
c
character*1 char,l(n2)
c
nocc=0
ncol=0
do 10 i=n1,n2
if(l(i).ne.char) go to 10
nocc=nocc+1
if(nocc.eq.1) ncol=i
10 continue
return
end
c&FORMNM
subroutine formnm(l,n1,n2,name)
c**********************************************************
c
c subroutine to form the character*20 variable name from
c l(n1),...,l(n2).
c
c***********************************************************
c
character*1 l(n2)
character*15 name
if(n2-n1+1.gt.15.or.n2-n1+1.lt.1) go to 5
write(name,2) (l(i),i=n1,n2)
2 format(20a1)
go to 10
5 name='***************'
10 continue
return
end
c&LASTNB
subroutine lastnb(l,n,m)
c********************************************************************
c
c Subroutine to find the number (m) of the last nonblank character in t
c the character*1 array l(1),...,l(n).
c
c*********************************************************************
c
character*1 l(n),apst
apst=char(39)
c
c Let m=position of last nonblank character.
c If last non blank character is apostrophe, rturn m. Otherwise
c continue on looking for a ;. If there isn't one, return m. If
c there is, and it's in column one, return m. If there is and it's
c not in column one, blank out from ; to m and return m=position
c to left of ;.
c
do 10 i=n,1,-1
if(l(i).ne.' ') go to 20
10 continue
20 m=i
if(l(m).eq.apst) go to 99
do 30 i=m,1,-1
if(l(i).eq.';') go to 40
30 continue
go to 99
40 if(i.eq.1.or.i.eq.2) go to 99
do 50 j=i,m
50 l(j)=' '
m=i-1
99 continue
return
end
c<OUP
subroutine ltoup(l,n)
c********************************************************************
c
c Subroutine to convert all characters having ASCII codes between 97
c and 122 to the corresponding character having code 32 less.
c
c*********************************************************************
c
character*1 l(n)
do 10 i=1,n
ii=ichar(l(i))
10 if(97.le.ii.and.122.ge.ii) l(i)=char(ii-32)
return
end
c&PARSE
subroutine parse(l,pname,args,nargs,vname,ierr)
c*******************************************************************
c
c subroutine to find the procedure name (pname), arguments
c (args), and number of arguments (nargs) for a type 1 entry line
c l(1),...,l(72) . The error flag ierr is
c 0 if no error is encountered. If the line is type 3 then the
c variable name is also returned in vname.
c
c*******************************************************************
c
character*1 l(72),comma,equal
character*15 pname,args(21),vname
dimension nleft(21),nright(21)
data comma,equal/',','='/
ierr=0
c
c find left and right parentheses :
c
call findch(l,1,72,'(',noc1,ncol1)
call findch(l,1,72,')',noc2,ncol2)
c
c find where arguments are :
c
nargs=1
nleft(1)=ncol1+1
do 20 i=ncol1,ncol2
if(l(i).ne.comma) go to 20
nright(nargs)=i-1
nargs=nargs+1
nleft(nargs)=i+1
20 continue
nright(nargs)=ncol2-1
c
c form vector of arguments :
c
do 30 i=1,nargs
nrmnl=nright(i)-nleft(i)+1
if(nrmnl.gt.15.or.nrmnl.lt.1) then
ierr=1
go to 99
endif
30 call formnm(l,nleft(i),nright(i),args(i))
c
c find procedure name :
c
nn=ncol1-1
40 if(nn.eq.0) go to 50
if(l(nn).eq.equal) go to 50
nn=nn-1
go to 40
50 continue
nrmnl=ncol1-1-nn
if(nrmnl.lt.1.or.nrmnl.gt.15) then
ierr=1
go to 99
endif
call formnm(l,nn+1,ncol1-1,pname)
c
c see if type 3 :
c
if(nn.eq.0) go to 99
if(nn-1.lt.1.or.nn-1.gt.15) then
ierr=1
go to 99
endif
call formnm(l,1,nn-1,vname)
99 return
end
c&RMBLNK
subroutine rmblnk(l,n,lnb)
c*********************************************************************
c
c Subroutine to remove any embedded blanks in the character*1 array
c l(1),...,l(n). The output integer lnb is the number of the last
c nonblank element in the new version of the array. The array is
c assumed to have at least one nonblank character.
c
c Blanks are not removed from between apostrophes.
c
c*********************************************************************
c
character*1 l(n),blnk,l1(72)*1,apost*1
data blnk,apost/' ',''''/
call lastnb(l,n,lnb)
j=0
iap=0
do 10 i=1,n
if(l(i).eq.apost) iap=mod(iap+1,2)
if(l(i).eq.blnk.and.iap.eq.0) go to 10
j=j+1
l1(j)=l(i)
10 continue
call movct(l,lnb,blnk)
lnb=j
call movxy(l,l1,lnb)
c
return
end
c&TYPCHK
subroutine typchk(l,lnb,itype)
c*******************************************************************
c
c Subroutine to determine the command type of the command line made
c up of the character*1 array l(1),...,l(lnb).
c It is assumed that there are no blanks in l and that lnb>1.
c
c The output integer itype is 1,2,3,4,or 5 according to :
c
c TYPE 1 : x...x(x....x) i.e. no equal sign, exactly one (,
c with l(lnb)=) and location of (
c not the first element.
c
c TYPE 2 : x...x=x...x i.e. exactly 1 equal sign not at
c beginning or end and no (,),<,>
c
c TYPE 3 : x...x=x...x(x...x) i.e. one equal sign not at begin
c or end, exactly one ( not next to
c + and not at end, and exactly one
c which is at end. Also no <.
c
c TYPE 4 : x...x=<x...> i.e. exactly one = not at begin or
c end, exactly one < next to = and
c exactly one > at the end.
c
c TYPE 5 : x......x i.e. no =,(,),<,>
c
c The integer itype is 0 if l fits none of these types.
c
c**********************************************************************
c
character*1 l(lnb),ll
itype=0
c
c check for type 5 :
c
if(lnb.gt.15) go to 11
do 1 i=1,lnb
ll=l(i)
if(ll.eq.'='.or.ll.eq.'('.or.ll.eq.')') go to 11
1 if(ll.eq.'<'.or.ll.eq.'>') go to 11
itype=5
go to 99
c
c check for '=' (if more than one or at beginning or end then illegal
c if exactly one and not at beginning or end, could be legal type2-4,
c if none, could be legal type 1)
c
11 call findch(l,1,lnb,'=',nocc,neq)
if(nocc.gt.1.or.neq.eq.1.or.neq.eq.lnb) go to 99
if(nocc.eq.1) go to 10
c
c check if legal type 1 :
c
call findch(l,1,lnb,'(',nocc1,nc1)
call findch(l,1,lnb,')',nocc2,nc2)
if(nocc1*nocc2.ne.1.or.lnb.ne.nc2.or.nc1.eq.1) go to 99
itype=1
go to 99
c
c check whether legal type 2,3,4 (first look for <)
c
10 continue
call findch(l,1,lnb,'<',nocc1,nc1)
if(nocc1.eq.0) go to 20
c
c could be type 4 :
c
if(nocc1.gt.1.or.nc1.ne.neq+1) go to 99
call findch(l,1,lnb,'>',nocc2,nc2)
if(nocc2.ne.1.or.nc2.ne.lnb) go to 99
itype=4
go to 99
c
c could be types 2 or 3 :
c
20 continue
c
c type 2 if no >,(,), type 3 if no >, exactly one ( more than one
c element to the right of = and exactly one ) which is at the end.
c
call findch(l,1,lnb,'>',nocc2,nc2)
if(nocc2.ne.0) go to 99
call findch(l,1,lnb,'(',nocc3,nc3)
call findch(l,1,lnb,')',nocc4,nc4)
if(nocc3+nocc4.ne.0) go to 30
itype=2
go to 99
30 if(nocc3*nocc4.ne.1.or.nc3.le.neq+1.or.nc4.ne.lnb) go to 99
itype=3
c
c
99 continue
return
end
c&WAITK
subroutine waitk(n)
c*****************************************************************
c
c
c*****************************************************************
c
integer*4 status,getc
character ch
status=getc(ch)
n=ichar(ch)
return
end