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subroutine indxg(il,siz,ilr,mi,mx,lw,iopt1)
c!Purpose
c Converts a scilab index variable to a vector of indices
c!Calling sequence
c subroutine indxg(il,siz,ilr,mi,lw,iopt)
c il : beginning of a a scilab variable structure.
c siz : integer, matrix size, used for implicits index descriptions
c ilr : adress of first elment of resulting vector of indices in
c istk
c mi : size of resulting vector of indices
c mx : maximum value of resulting vector of indices
c lw : pointer to free space in stk (modified by execution)
c iopt1 : flag with decimal form n+10*i
c if n==0 null indices are accepted
c else null indices are rejected
c if i==0
c implicit indices ":" gives a vector istk(ilr)=1:siz, mi=siz,mx=siz
c else
c implicit indice ":" gives mi=-1,mx=siz
c!
c Copyright INRIA
include '../stack.h'
integer siz,iopt1,iopt
double precision e1,v(3)
integer iadr,sadr
c
iadr(l)=l+l-1
sadr(l)=(l/2)+1
c
c
impl=iopt1/10
iopt=iopt1-10*impl
c
if(istk(il).lt.0) il=istk(il+1)
if(istk(il).eq.1.or.istk(il).eq.8) then
c Index is a vector of scalars
m=istk(il+1)
n=istk(il+2)
if(m.ge.1) then
c . general case
l=sadr(il+4)
ilr=iadr(lw)
lw=sadr(ilr+m*n)
err=lw-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
if (istk(il).eq.1) then
if(istk(il+3).ne.0) then
call error(21)
return
endif
call entier(m*n,stk(l),istk(ilr))
else
call tpconv(istk(il+3),4,m*n,istk(il+4),1,istk(ilr),1)
endif
mi=m*n
mx=0
do 05 i=0,m*n-1
if(iopt.eq.1.and.istk(ilr+i).le.0) then
call error(21)
return
else
mx=max(mx,istk(ilr+i))
endif
05 continue
elseif(m.eq.0) then
c . index is []
ilr=il
mi=0
mx=0
elseif(m.eq.-1) then
c . index is :
ilr=iadr(lw)
if(impl.eq.0) then
if(siz.gt.0) then
lw=sadr(ilr+siz)
err=lw-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
do 10 i=1,siz
istk(ilr-1+i)=i
10 continue
endif
mi=siz
else
mi=-1
endif
mx=siz
endif
elseif (istk(il).eq.2) then
c . Index is a vector of polynomial
m=istk(il+1)
n=istk(il+2)
if(istk(il+3).ne.0) then
call error(21)
return
endif
mi=m*n
l=sadr(il+9+mi)
lr=lw
ilr=iadr(lr)
lw=lr+mi
err=lw-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
c . evaluate it for siz
e1=siz
call ddmpev(stk(l),istk(il+8),1,e1,stk(lr),1,1,mi)
call entier(mi,stk(lr),istk(ilr))
lw=sadr(ilr+mi)
mx=0
do 15 i=0,mi-1
if(istk(ilr+i).le.0) then
call error(21)
return
else
mx=max(mx,istk(ilr+i))
endif
15 continue
elseif (istk(il).eq.129) then
c . Index is an implicit polynomial vector (beg:step:end)
e1=siz
l=sadr(il+12)
call ddmpev(stk(l),istk(il+8),1,e1,v,1,1,3)
ideb=v(1)
ipas=v(2)
ifin=v(3)
c sign used to avoid integer overflow
if(ipas.eq.0.or.(ifin-ideb)*sign(1,ipas).lt.0) then
mi=0
mx=0
else
if(ipas.lt.0.and.ifin.le.0.or.ipas.gt.0.and.ideb.le.0) then
call error(21)
return
endif
mi=int((abs(ifin-ideb)+1)/abs(ipas))
ilr=iadr(lw)
lw=sadr(ilr+mi+1)
err=lw-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
k=0
do 20 i=ideb,ifin,ipas
istk(ilr+k)=i
k=k+1
20 continue
mi=k
if(ipas.gt.0) then
mx=istk(ilr-1+mi)
else
mx=istk(ilr)
endif
endif
elseif (istk(il).eq.4) then
c . index is a boolean vector
m=istk(il+1)
n=istk(il+2)
c if(m*n.ne.siz) then
c call error(21)
c return
c endif
ilr=iadr(lw)
lw=sadr(ilr+m*n)
err=lw-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
mi=0
do 30 i=1,m*n
if(istk(il+2+i).eq.1) then
istk(ilr+mi)=i
mi=mi+1
endif
30 continue
if(mi.eq.0) then
mx=0
else
mx=istk(ilr-1+mi)
endif
lw=sadr(ilr+mi)
elseif (istk(il).eq.6) then
c . index is a boolean vector
m=istk(il+1)
n=istk(il+2)
nel=istk(il+4)
ir=il+5
ic=ir+m
ilr=iadr(lw)
ilw=ilr+nel
mx=nel
mi=0
if (nel.gt.0) then
lw=sadr(ilw+nel)
err=lw-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
do 35 i=0,m-1
if(istk(ir).gt.0) then
do 31 kk=0,istk(ir)-1
istk(ilr+mi)=1+i+(istk(ic+kk)-1)*m
mi=mi+1
31 continue
ic=ic+istk(ir)
endif
ir=ir+1
35 continue
call isort1(istk(ilr),nel,istk(ilw),1)
endif
lw=sadr(ilr+nel)
else
call error(21)
return
endif
return
end
subroutine indxgc(il,siz,ilr,mi,mx,lw)
c!Purpose
c Converts a scilab index variable to the complementary vector of indices
c!Calling sequence
c subroutine indxg(il,siz,ilr,mi,lw)
c il : beginning of a scilab variable structure.
c siz : integer, matrix size, used for implicits index descriptions
c ilr : adress of first elment of resulting vector of indices in
c istk
c mi : size of resulting vector of indices
c mx : maximum value of resulting vector of indices
c lw : pointer to free space in stk (modified by execution)
c!
* modification by Bruno so as to use a faster algorithm (7 May 2002)
implicit none
include '../stack.h'
integer il, siz, ilr, mi, mx, lw
integer i, k, ilc
integer l, iadr,sadr
c
iadr(l)=l+l-1
sadr(l)=(l/2)+1
call indxg(il,siz,ilr,mi,mx,lw,1)
if(err.gt.0) return
ilc=iadr(lw)
lw=sadr(ilc+siz)
err=lw-lstk(bot)
if(err.gt.0) then
call error(17)
return
endif
if(mi.eq.0) then
do i=1,siz
istk(ilc+i-1)=i
enddo
mx=istk(ilc+siz-1)
mi=siz
else
* computes complement (part of the code modified by Bruno)
*
* given
* 1/ a "vector" w of mi indices stored from istk(ilr)
* so that w=[istk(ilr), ....., istk(ilr+mi-1)] is this vector
* 2/ the "vector" v of indices v=[1,2,..., siz]
*
* computes the vector v minus w
* this new vector is stored from istk(ilc) and its final number of
* components will be stored in mi
do i = 0, siz-1
istk(ilc+i) = 1
end do
do i = 0, mi-1
k = istk(ilr+i)
if (k .le. siz) istk(ilc+k-1) = 0
end do
k = 0
do i = 1, siz
if (istk(ilc+i-1) .eq. 1) then
istk(ilc+k) = i
k = k+1
end if
end do
mi = k
mx=istk(ilc-1+k)
endif
ilr=ilc
lw=sadr(ilr+mi)
return
end
|
