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subroutine where
c extrait l'arbre d'appel de l'instruction courante
c cette routine est issue de la fin du sous programme error
c Copyright INRIA
c Revised and corrected by Francois VOGEL, July/August 2004
c (bugs 908, 922 and 911 are fixed by this version)
include '../stack.h'
integer iadr,sadr
integer p,lpts(6),lcts,r,vol,rios
logical first
iadr(l)=l+l-1
sadr(l)=(l/2)+1
if (rhs .ge. 0) then
call error(39)
return
endif
if (lhs .ne. 2) then
call error(41)
return
endif
c preserve datas
call icopy(6,lpt,1,lpts,1)
lcts=lct(8)
p=pt
rios=rio
c initialize
top=top+1
il=iadr(lstk(top))
first=.true.
il0=il
nn=0
ll=0
c depilement de l'environnement
p=p+1
1001 p=p-1
if(p.eq.0) goto 1010
r=rstk(p)
goto(1002,1002,1004) r-500
goto 1001
c
c on depile une macro
1002 k=lpt(1)-(13+nsiz)
lpt(1)=lin(k+1)
lpt(2)=lin(k+2)
lpt(6)=k
c
c recherche du nom de la macro correspondant a ce niveau
lk=sadr(lin(k+6))
if(lk.le.lstk(top+1)) then
km=0
else
km=lin(k+5)-1
endif
1003 km=km+1
if(km.gt.isiz)goto 1013
if(lstk(km).ne.lk) goto 1003
c
1013 continue
ilk=lin(k+6)
if(istk(ilk).ne.10) then
if(first) then
first=.false.
nlc=0
else
call whatln(lpt(1),lpt(2),lpt(6),nlc,l1,ifin)
endif
err=sadr(il+2+nlgh)-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
istk(il)=lct(8)-nlc
il=il+1
if (km.le.isiz) then
call namstr(idstk(1,km),istk(il+1),istk(il),1)
ll=ll+istk(il)
il=il+1+istk(il)
else
istk(il)=0
il=il+1
endif
nn=nn+1
else
istk(il)=0
istk(il+1)=7
call cvstr(7,istk(il+2),'execstr',0)
ll=ll+7
il=il+9
nn=nn+1
endif
lct(8)=lin(k+12+nsiz)
c
goto 1001
c
c on depile un exec ou une pause
1004 if(rio.ne.rte) then
c exec
k=lpt(1)-(13+nsiz)
lpt(1)=lin(k+1)
lpt(2)=lin(k+4)
lpt(6)=k
c
if(first) then
first=.false.
nlc=0
endif
err=sadr(il+2+nlgh)-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
istk(il)=lct(8)-nlc
istk(il+1)=4
call cvstr(4,istk(il+2),'exec',0)
ll=ll+4
il=il+6
nn=nn+1
lct(8)=lin(k+12+nsiz)
c
1005 p=p-1
if(rstk(p).ne.902) goto 1005
rio=pstk(p)
goto 1001
else
c pause
k = lpt(1) - (13+nsiz)
lpt(1) = lin(k+1)
lpt(2) = lin(k+2)
lpt(3) = lin(k+3)
lpt(4) = lin(k+4)
lpt(6) = k
if(first) first=.false.
err=sadr(il+2+nlgh)-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
istk(il)=0
istk(il+1)=5
call cvstr(5,istk(il+2),'pause',0)
ll=ll+5
il=il+7
nn=nn+1
lct(8)=lin(k+12+nsiz)
goto 1001
endif
c restaure datas
1010 call icopy(6,lpts,1,lpt,1)
lct(8)=lcts
rio=rios
c create return variables
ill=il0
ll=sadr(ill+4)
ilm=iadr(ll+nn)
if(nn.eq.0) then
err=sadr(ill+8)-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
istk(ill)=1
istk(ill+1)=0
istk(ill+2)=0
istk(ill+3)=0
lstk(top+1)=ll
top=top+1
istk(ilm)=1
istk(ilm+1)=0
istk(ilm+2)=0
istk(ilm+3)=0
lstk(top+1)=sadr(ilm+4)
return
endif
c compute total size of names strings
vol=0
il1=il0
do 05 i=1,nn
n1=istk(il1+1)
vol=vol+n1
il1=il1+2+n1
05 continue
c check memory
ilw=ilm+4+nn+1+vol
err=sadr(ilw+il-il0)-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
call icopy(il-il0,istk(il0),-1,istk(ilw),-1)
istk(ill)=1
istk(ill+1)=nn
istk(ill+2)=1
istk(ill+3)=0
c
istk(ilm)=10
istk(ilm+1)=nn
istk(ilm+2)=1
istk(ilm+3)=0
istk(ilm+4)=1
c
il=ilw
ln=1
do 10 i=1,nn
stk(ll-1+i)=istk(il)
n=istk(il+1)
istk(ilm+4+i)=ln+n
il=il+2
if(n.gt.0) then
call icopy(n,istk(il),1,istk(ilm+4+nn+ln),1)
il=il+n
endif
ln=ln+n
10 continue
lstk(top+1)=ll+nn
top=top+1
lstk(top+1)=sadr(ilm+4+nn+ln)
return
end
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