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bug in g77
- To: gcc-bugs at gcc dot gnu dot org
- Subject: bug in g77
- From: Steven Castillo <scastill at nmsu dot edu>
- Date: Tue, 13 Feb 2001 00:01:18 -0700 (MST)
Hello:
I am trying to run some fortran codes requiring very large arrays (~ 1
gbyte). When trying to compile a fortran code with a declared single
precision floating point array having a length of more than 134217727
using the command:
g77 test.f
I get the error message:
************************
test.f: In program `MAIN__':
test.f:21:
real testm(134217728)
^
Array `testm' at (^) is too large to handle
************************
where testm is the large array. I am using redhat 7.0 with all redhat
updated rpms applied on a pentium III system with 2 gbytes of memory and
using the stock 2.2.16-22enterprise supplied with redhat 7.0. g77 -v
yields
g77 version 2.96 20000731 (Red Hat Linux 7.0) (from FSF-g77 version 0.5.26
20000731 (Red Hat Linux 7.0))
Driving: g77 -v -c -xf77-version /dev/null -xnone
Reading specs from /usr/lib/gcc-lib/i386-redhat-linux/2.96/specs
gcc version 2.96 20000731 (Red Hat Linux 7.0)
/usr/lib/gcc-lib/i386-redhat-linux/2.96/tradcpp0 -lang-fortran -v
-D__GNUC__=2 -D__GNUC_MINOR__=96 -D__GNUC_PATCHLEVEL__=0 -D__ELF__ -Dunix
-Dlinux -D__ELF__ -D__unix__ -D__linux__ -D__unix -D__linux
-Asystem(posix) -Acpu(i386) -Amachine(i386) -Di386 -D__i386 -D__i386__
-D__tune_i386__ /dev/null /dev/null
GNU traditional CPP version 2.96 20000731 (Red Hat Linux 7.0)
/usr/lib/gcc-lib/i386-redhat-linux/2.96/f771 -fnull-version -quiet
-dumpbase g77-version.f -version -fversion -o /tmp/cctLM9TW.s /dev/null
GNU F77 version 2.96 20000731 (Red Hat Linux
7.0) (i386-redhat-linux) compiled by GNU C version 2.96 20000731 (Red Hat
Linux 7.0).
GNU Fortran Front End version 0.5.26 20000731 (Red Hat Linux 7.0)
as -V -Qy -o /tmp/ccqD1l4I.o /tmp/cctLM9TW.s
GNU assembler version 2.10.90 (i386-redhat-linux) using BFD version
2.10.0.18
ld -m elf_i386 -dynamic-linker /lib/ld-linux.so.2 -o /tmp/ccb360gx
/tmp/ccqD1l4I.o /usr/lib/gcc-lib/i386-redhat-linux/2.96/../../../crt1.o
/usr/lib/gcc-lib/i386-redhat-linux/2.96/../../../crti.o
/usr/lib/gcc-lib/i386-redhat-linux/2.96/crtbegin.o
-L/usr/lib/gcc-lib/i386-redhat-linux/2.96
-L/usr/lib/gcc-lib/i386-redhat-linux/2.96/../../.. -lg2c -lm -lgcc -lc
-lgcc /usr/lib/gcc-lib/i386-redhat-linux/2.96/crtend.o
/usr/lib/gcc-lib/i386-redhat-linux/2.96/../../../crtn.o
/tmp/ccb360gx
__G77_LIBF77_VERSION__: 0.5.26 20000731 (prerelease)
@(#)LIBF77 VERSION 19991115
__G77_LIBI77_VERSION__: 0.5.26 20000731 (prerelease)
@(#) LIBI77 VERSION pjw,dmg-mods 19991115
__G77_LIBU77_VERSION__: 0.5.26 20000731 (prerelease)
@(#) LIBU77 VERSION 19980709
A test code is listed below. thank you in advance for your help.
--
Steve Castillo
Professor and Department Head
Klipsch School of Electrical and Computer Engineering
New Mexico State University
scastill@nmsu.edu
(505)646-3117
C
c PROGRAM CAPEX
c
c Program Capex finds the capacitance
c matrix for a system of wires having
c a geometry specified by AMP specifications.
c The kernal is integrated analytically.
c
c The program first solves for the charge on
c the wires assuming some voltage distribution
c using the method of moments. Point basis
c functions and point matching is used.
c
c Programmer : Steven P. Castillo
c Last Mod. : 2/28/96
c Language : Fortran 77
c Libraries : linpack
c
c
real x(15000),y(15000),z(11585,11585),g(15000),pi,wk(15000)
real testm(134217728)
real c(40,40), char(15000)
real mo,im,mj,cap(40,40),l(40,40),nmax
integer count1,count2,test,ipvt(15000)
character*12 ctime, answer
data ctime /'the time is'/
c
write(*,*)'enter the number of conductors (even #)'
read(*,*)n
write(*,*)'enter the spacing'
read(*,*)s
write(*,*)'enter the conductor thickness'
read(*,*)t
write(*,*)'enter the number of basis functions per'
write(*,*)'side of conductor '
read(*,*)nb
write(*,*)'Do you want the matrix output? (yes/no)'
read(*,1000)answer
1000 format(a)
c
open(unit=1,file='result',status='unknown')
open(unit=2,file='matrix',status='unknown')
open(unit=3,file='solution',status='unknown')
rewind(1)
rewind(2)
rewind(3)
c
c
c write(*,*)'the start time is'
c call gettim(ihr,imin,isec,i100th)
c write(*,'(1x,a,i2.2,1h:,i2.2,1h:,i2.2,1h.,i2.2)')
c >ctime,ihr,imin,isec,i100th
s=s*.0254
t=t*.0254
pi=3.141592654
uo=4.*pi*1.e-7
eo=8.85e-12
tpi=2.*pi
fpi=4.*pi
n2=n*2
n4=n*4
nt=n4*nb
i5=4
tb=t/nb
c
c Fill x,y coordinate vectors.
c
call geom(x,y,t,s,n,n2,n4,nb)
c
c
c
count1=1
count2=1
do 15 i=1,nt
if(i.ne.1)then
test=count2
16 test=test-2
if(test.eq.0)then
ep1=y(i)-tb/2.
ep2=y(i)+tb/2.
g1=ep2*alog(ep2**2+x(i)**2)-2.*ep2+
+ 2.*x(i)*atan2(ep2,x(i))
g2=ep1*alog(ep1**2+x(i)**2)-2.*ep1+
+ 2.*x(i)*atan2(ep1,x(i))
g(i)=1./fpi/eo*(g1-g2)
go to 17
endif
if(test.lt.0)then
ep1=x(i)-tb/2.
ep2=x(i)+tb/2.
g1=ep2*alog(ep2**2+y(i)**2)-2.*ep2+
+ 2.*y(i)*atan2(ep2,y(i))
g2=ep1*alog(ep1**2+y(i)**2)-2.*ep1+
+ 2.*y(i)*atan2(ep1,y(i))
g(i)=1./fpi/eo*(g1-g2)
go to 17
endif
go to 16
17 continue
else
g(i)=tb/tpi/eo*(alog(tb/2.)-1.)
endif
count1=count1+1
if(count1.gt.nb)then
count2=count2+1
count1=1
endif
15 continue
c
c Fill impedence matrix.
c
do 30 i=1,nt
z(1,i)=tb/eo
30 continue
c
do 40 j=2,nt
count1=1
count2=1
do 45 i=1,nt
if(i.eq.j)then
c self term
z(j,i)=tb/tpi/eo*(1-alog(tb/2.))+g(i)
go to 46
c non-self term
else
20 test=count2
47 test=test-2
if(test.eq.0)then
ep1=y(j)-y(i)+tb/2.
ep2=y(j)-y(i)-tb/2.
arg=abs(x(j)-x(i))
g1=ep2*alog(ep2**2+(x(j)-x(i))**2)-2.*ep2+
+ 2.*abs(x(j)-x(i))*atan2(ep2,arg)
g2=ep1*alog(ep1**2+(x(j)-x(i))**2)-2.*ep1+
+ 2.*abs(x(j)-x(i))*atan2(ep1,arg)
z(j,i)=g(i)+1./fpi/eo*(g1-g2)
go to 46
endif
if(test.lt.0)then
ep1=x(j)-x(i)+tb/2.
ep2=x(j)-x(i)-tb/2.
arg=abs(y(j)-y(i))
g1=ep2*alog(ep2**2+(y(j)-y(i))**2)-2.*ep2+
+ 2.*abs(y(j)-y(i))*atan2(ep2,arg)
g2=ep1*alog(ep1**2+(y(j)-y(i))**2)-2.*ep1+
+ 2.*abs(y(j)-y(i))*atan2(ep1,arg)
z(j,i)=g(i)+1./fpi/eo*(g1-g2)
go to 46
endif
go to 47
endif
46 count1=count1+1
if(count1.gt.nb)then
count2=count2+1
count1=1
endif
45 continue
40 continue
c
c Solve the system
c
c if(answer.eq.'yes')then
c do 100 i=1,nt
c do 105 j=1,nt
c write(2,*)i,j,z(i,j)
c105 continue
c write(2,*)i,char(i)
c100 continue
c endif
c call sgefa(z,15000,nt,ipvt,info)
job=1
c call sgedi(z,15000,nt,ipvt,det,wk,job)
c call sgesl(z,15000,nt,ipvt,char,0)
c do 106 i=1,nt
c write(3,*)i,char(i)
c 106 continue
stop
c
c Fill the generalized capacitance matrix for
c the system.
c
nmax=nb*4+1
do 60 i=1,n
do 65 j=1,n
c(i,j)=0.
65 continue
60 continue
count2=1
count3=1.
do 70 i=1,nt
count1=1
count=1.
do 75 j=1,nt
c if(j.ne.1)then
c(count2,count1)=c(count2,count1)+z(i,j)*tb
c endif
count=count+1
if(count.eq.nmax)then
count=1.
count1=count1+1
endif
75 continue
count3=count3+1.
if(count3.eq.nmax)then
count3=1.
count2=count2+1
endif
70 continue
c
c
c
do 80 i=1,n
do 85 j=1,n
write(1,*)'c(',i,',',j,')=',c(j,i)
85 continue
80 continue
c
c Compute the transmission line capacitance
c matrix from the generalized capacitance
c matrix with reference to the first
c conductor.
c
mo=0.
do 200 i=1,n
do 205 j=1,n
mo=mo+c(i,j)
205 continue
200 continue
c
do 210 i=2,n
do 215 j=2,n
im=0.
do 220 i2=1,n
if(i2.ne.i)then
im=im+c(i2,j)
endif
220 continue
c
mj=0.
do 225 i2=1,n
if(i2.ne.j)then
mj=mj+c(i,i2)
endif
225 continue
c
cap(i-1,j-1)=(c(i,j)*(mo-im-mj-c(i,j))-im*mj)/mo
write(1,*)'cap(',i,',',j,')=',cap(i-1,j-1)
c
215 continue
210 continue
c
call sgefa(cap,40,n-1,ipvt,info)
job=1
call sgedi(cap,40,n-1,ipvt,det,wk,job)
c call linv1f(cap,n-1,10,l,0,wk,ier)
do 230 i=1,n-1
do 235 j=1,n-1
cap(i,j)=cap(i,j)*eo*uo
write(1,*)'l(',i,',',j,')=',cap(i,j)
235 continue
230 continue
c
c write(*,*)'the finish time is'
c call gettim(ihr,imin,isec,i100th)
c write(*,'(1x,a,i2.2,1h:,i2.2,1h:,i2.2,1h.,i2.2)')
c >ctime,ihr,imin,isec,i100th
c
stop
end
c
c SUBROUTINE GEOM
c
c Subroutine Geom fills the x and y vectors
c for the AMP conector problem.
c
c Last Mod: 2/5/85
c
subroutine geom(x,y,t,s,n,n2,n4,nb)
c
real x(1),y(1)
c
no2=n/2
count1=1.
count2=1.
x1=0.
y1=0.
i1=1
c
do 10 i=1,n4
x(i1)=x1
y(i1)=y1
c
if(count1.eq.1)then
do 15 j=1,nb-1
x1=x1+t/(nb+1)
i1=i1+1
x(i1)=x1
y(i1)=y1
15 continue
x1=x1+t/(nb+1)
y1=y1+t/(nb+1)
endif
c
if(count1.eq.2)then
do 20 j=1,nb-1
y1=y1+t/(nb+1)
i1=i1+1
x(i1)=x1
y(i1)=y1
20 continue
x1=x1-t/(nb+1)
y1=y1+t/(nb+1)
endif
c
if(count1.eq.3)then
do 25 j=1,nb-1
x1=x1-t/(nb+1)
i1=i1+1
x(i1)=x1
y(i1)=y1
25 continue
x1=x1-t/(nb+1)
y1=y1-t/(nb+1)
endif
c
if(count1.eq.4)then
do 30 j=1,nb-1
y1=y1-t/(nb+1)
i1=i1+1
x(i1)=x1
y(i1)=y1
30 continue
x1=x1+s+t/(nb+1)
y1=y1-t/(nb+1)
count1=0.
endif
count1=count1+1
i1=i1+1
c
if(count2.eq.n2)then
x1=x1-no2*s
y1=y1+s
count2=0.
endif
count2=count2+1
c
10 continue
c
return
end