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The code I'm trying to compile is named gepol93 which was finished in 1994 with FORTRAN 77.
It's a program designed to compute some chemical properties of a molecule.
I tried to compile the file using the command gfortran GEPOL93.FOR but an error was raised:
EPOL93.FOR:598.72:
IF(LPR)CALL STAT(NCOR)
1
Error: Missing actual argument 'values' in call to 'stat' at (1)
I'm not familar with FORTRAN, and I guess that the code is too old to be compatible with the compiler on my computer.
What I'm certain about is that the code is definitely correct because it has never been modified by other people since 1994 (you can check this with ls -l) and many scholars are using it to conduct research nowadays.
But the question is how to compile the code into a .out file? Can anybody help me?
In the first subroutine PCOUNT, STAT is called.
SUBROUTINE PCOUNT(NATOM,NCOR,KSURF,LPR)
C -------------------------------------------------------------------
C This prints general counters
C -------------------------------------------------------------------
IMPLICIT NONE
LOGICAL FIRST
LOGICAL LPR
INTEGER*2 IUSE
INTEGER*4 I,IUC
INTEGER*4 J
INTEGER*4 MC,MI
INTEGER*4 NATOM,NCOR,NEWS
REAL*4 RE
REAL*4 XE
REAL*4 YE
REAL*4 ZE
CHARACTER*5 KSURF
PARAMETER (MC=100000,MI=6)
COMMON/CSFE/XE(MC),YE(MC),ZE(MC),RE(MC),IUSE(MC)
DIMENSION IUC(MI)
IF(LPR)WRITE(6,'(/A)')' ===> Start Subroutine Pcount '
NEWS=NCOR-NATOM
IF(KSURF.EQ.'ESURF')THEN
WRITE(6,'(3(A,I7/))')
& ' Number of INITIAL coordinates ',NATOM,
& ' Number of NEW coordinates ',NEWS,
& ' Number of TOTAL coordinates ',NCOR
ELSE
WRITE(6,'(A,I7/)')
& ' Number of TOTAL coordinates ',NCOR
END IF
IF(LPR)THEN
DO J=1,MI
IUC(J)=0
END DO
DO I=1,NATOM
J=IUSE(I)
IUC(J)=IUC(J)+1
END DO
FIRST=.TRUE.
DO J=1,MI
IF(IUC(J).NE.0)THEN
IF(FIRST)THEN
FIRST=.FALSE.
WRITE(6,'(A)')
& ' MORE INFORMATION ABOUT INITIAL SET OF COORDINATES'
END IF
WRITE(6,'(A,I2,A,I7)')
& ' Number of coordinates with IUSE ',J,'=',IUC(J)
END IF
END DO
IF(KSURF.EQ.'ESURF')THEN
DO J=1,MI
IUC(J)=0
END DO
DO I=NATOM+1,NCOR
J=IUSE(I)
IUC(J)=IUC(J)+1
END DO
FIRST=.TRUE.
DO J=1,MI
IF(IUC(J).NE.0)THEN
IF(FIRST)THEN
FIRST=.FALSE.
WRITE(6,'(A)')
& ' MORE INFORMATION ABOUT NEW SET OF COORDINATES'
END IF
WRITE(6,'(A,I2,A,I7)')
& ' Number of coordinates with IUSE ',J,'=',IUC(J)
END IF
END DO
END IF
IF(LPR)CALL STAT(NCOR)
END IF
RETURN
END
SUBROUTINE STAT(NCOR)
C -------------------------------------------------------------
C This prepares some statistics about the set of spheres
C -------------------------------------------------------------
IMPLICIT NONE
INTEGER*2 IUSE
INTEGER*4 CT6,CT4
INTEGER*4 I
INTEGER*4 J
INTEGER*4 MAXI,MC
INTEGER*4 NCOR,NINT
INTEGER*4 TMIN,TMAX
REAL*4 C
REAL*4 R1,R2,RE
REAL*4 VINT
REAL*4 XE
REAL*4 YE
REAL*4 ZE
PARAMETER (MC=100000,MAXI=100)
COMMON/CSFE/XE(MC),YE(MC),ZE(MC),RE(MC),IUSE(MC)
DIMENSION CT4(MAXI),CT6(MAXI)
WRITE(6,'(/A)')' ==> Start subroutine STAT'
VINT=0.1
TMIN=9000
TMAX=0
DO J=1,MAXI
CT4(J)=0
CT6(J)=0
END DO
DO I=1,NCOR
C=RE(I)/VINT
J=INT(C)+1
IF(IUSE(I).EQ.6)THEN
IF(J.LT.MAXI)THEN
CT6(J)=CT6(J)+1
TMAX=MAX(TMAX,J)
TMIN=MIN(TMIN,J)
END IF
ELSE IF(IUSE(I).EQ.4)THEN
IF(J.LT.MAXI)THEN
CT4(J)=CT4(J)+1
TMAX=MAX(TMAX,J)
TMIN=MIN(TMIN,J)
END IF
END IF
END DO
WRITE(6,'(A)')' RADII .GE. and .LT. TYPE 4 TYPE 6'
DO J=TMIN,TMAX
R1=(J-1)*VINT
R2=J*VINT
WRITE(6,'(2F10.5,2I10)')R1,R2,CT4(J),CT6(J)
END DO
RETURN
END
C
Related
I'm writing a subroutine that transform a regular vector into the one with only non-zero elements. Say, vector a=(0,0,1,2,3)' (n by 1). Then the non-zero vector is c=(1,2,3), and the row index is recorded as ic=(0,0,0,1,2,3) where ic(1)=0, ic(i+1)-ic(i) is the number of non-zero elements in i-th row. The vector index jc=(1,1,1) with size 3 as there are 3 non-zero entries. See the sparse matrix wiki for FYI: https://en.wikipedia.org/wiki/Sparse_matrix.
Despite its simplicity, I'm having troubles in running the following code named sparsem.f90
!This subroutine coverts a regular sparse matrix a into a CSR form
MODULE SPARSEM
CONTAINS
SUBROUTINE vsparse(a,c,jc,ic,counta,ierr,myid)
IMPLICIT NONE
REAL(8), INTENT(IN):: a(:)
INTEGER, INTENT(IN):: counta,myid
REAL(8), INTENT(OUT):: c(counta)
INTEGER, INTENT(OUT):: jc(counta),ic(size(a)+1)
INTEGER:: ierr,countaa,i
character(len=90):: filename
ierr=0
jc=0
c=0.0d0
ic=0
PRINT *, 'SIZE OF A IN VSPARSE', size(a),count(a>0.0d0),counta
IF (COUNT(a>0.0d0) /= counta) THEN
ierr=1
PRINT *, 'ERROR: number count of non-zero a(i,j) is not', counta
ELSE
countaa=0
ic(1)=0
DO i=1,size(a)
IF (a(i) > 0.0d0 ) THEN
countaa=countaa+1
c(countaa)=a(i)
ic(i+1)=ic(i)+1
jc(countaa)=1
IF (countaa<100) PRINT *,'checkcheckcheck', a(i), &
countaa,jc(countaa),c(countaa),jc(1:5)
ELSE
ic(i+1)=ic(i)
END IF
END DO
PRINT *, 'JCJCJCJC',jc(1:5)
END IF
IF (myid==7) THEN
WRITE(filename,'("sparsedens_dcheck",I1,".txt")') myid+1
OPEN(UNIT=212101, FILE="/home/wenya/Workspace/Model4/valuef/"//filename,ACTION='write',status='replace')
DO i=1,counta+1
IF (i<=counta) THEN
WRITE(212101,*) c(i),jc(i)
ELSE
WRITE(212101,*) 0.0D0,0
END IF
END DO
CLOSE(212101)
END IF
return
END SUBROUTINE vsparse
END MODULE SPARSEM
So the three print jc codes shall give 1 1 1 1 1.... Yet starting from the second print jc code, the result is 6750960 6750691 6750692 .... The array of jc has size 9,000,000. And I know the first 2250000 element is 0.
To replicate this problem, here is the main program
PROGRAM MAIN
USE SPARSEM
IMPICIT NONE
REAL(8):: dens_last(9000000)
REAL(8), ALLOCATABLE :: dens(:)
INTEGER, ALLOCATABLE :: ic(:),jc(:)
INTEGER:: i
dens_last(1:2250000)=0.0d0
dens_last(2250001:9000000)=1.0d0/6750000.0d0
ncount=count(dens_last>0.0d0)
ALLOCATE(dens(ncount), ic(9000000+1), jc(ncount)_
CALL VSPASEM(dens_last, dens, jc, ic, ncount,ierr)
DEALLOCATE(dens,ic,jc)
END PROGRAM MAIN
I am using gfortran 6.3.0 and openmpi latest version on a UBUNTU 17.04 computer. Although openmpi is not used in this example, it's used in the rest of the program. Any thoughts? Thanks!
There are several threads with similar titles of mine, but I do not believe they are the same. One was very similar fortran pass allocated array to main procedure, but the answer required Fortran 2008. I am after a Fortran 90/95 solution.
Another very good, and quite similar thread is Dynamic array allocation in fortran90. However in this method while they allocate in the subroutine, they don't ever appear to deallocate, which seems odd. My method looks on the surface at least to be the same, yet when I print the array in the main program, only blank spaces are printed. When I print in the subroutine itself, the array prints to screen the correct values, and the correct number of values.
In the following a MAIN program calls a subroutine. This subroutine reads data into an allocatable array, and passes the array back to the main program. I do this by using small subroutines each designed to look for specific terms in the input file. All of these subroutines are in one module file. So there are three files: Main.f90, input_read.f90 and filename.inp.
It seems then that I do not know how to pass an array that is allocatable in program Main.f90 as well as in the called subroutine where it is actually allocated, sized, and then deallocated before being passed to program Main. This perhaps sounds confusing, so here is the code for all three programs. I apologize for the poor formatting when I pasted it. I tried to separate all the rows.
main.f90:
Program main
use input_read ! the module with the subroutines used for reading filename.inp
implicit none
REAL, Allocatable :: epsilstar(:)
INTEGER :: natoms
call Obtain_LJ_Epsilon(epsilstar, natoms)
print*, 'LJ Epsilon : ', epsilstar
END Program main
Next is the module with a subroutine (I removed all but the necessary one for space), input_read.f90:
module input_read
contains
!===============================================================
!===============================================================
Subroutine Obtain_LJ_Epsilon(epsilstar,natoms)
! Reads epsilon and sigma parameters for Lennard-Jones Force-Field and also
! counts the number of types of atoms in the system
!===============================================================
!===============================================================
INTEGER :: error,line_number,natoms_eps,i
CHARACTER(120) :: string, next_line, next_next_line,dummy_char
CHARACTER(8) :: dummy_na,dummy_eps
INTEGER,intent(out) :: natoms
LOGICAL :: Proceed
real, intent(out), allocatable :: epsilstar(:)
error = 0
line_number = 0
Proceed = .true.
open(10,file='filename.inp',status='old')
!=============================================
! Find key word LJ_Epsilon
!=============================================
DO
line_number = line_number + 1
Read(10,'(A120)',iostat=error) string
IF (error .NE. 0) THEN
print*, "Error, stopping read input due to an error reading line"
exit
END IF
IF (string(1:12) == '$ LJ_epsilon') THEN
line_number = line_number + 1
exit
ELSE IF (string(1:3) == 'END' .or. line_number > 2000) THEN
print*, "Hit end of file before reading '$ LJ_epsilon' "
Proceed = .false.
exit
ENDIF
ENDDO
!========================================================
! Key word found, now determine number of parameters
! needing to be read
!========================================================
natoms_eps = -1
dummy_eps = 'iii'
do while ((dummy_eps(1:1) .ne. '$') .and. (dummy_eps(1:1) .ne. ' '))
natoms_eps = natoms_eps + 1
read(10,*) dummy_eps
enddo !we now know the number of atoms in the system (# of parameters)
close(10)
Allocate(epsilstar(natoms_eps))
epsilstar = 0.0
!============================================================
! Number of parameters found, now read their values
!============================================================
if(Proceed) then
open(11,file='filename.inp',status='old')
do i = 1,line_number-1
read(11,*) ! note it is not recording anything for this do loop
enddo
do i = 1,natoms_eps
read(11,*) dummy_char
read(dummy_char,*) epsilstar(i) ! convert string read in to real, and store in epsilstar
enddo
close(11)
PRINT*, 'LJ_epsilon: ', epsilstar ! printing to make sure it worked
endif
deallocate(epsilstar)
END Subroutine Obtain_LJ_Epsilon
end module input_read
And finally the input file: filename.inp
# Run_Type
NVT
# Run_Name
Test_Name
# Pressure
1.0
# Temperature
298.15
# Number_Species
# LJ_epsilon
117.1
117.1
117.1
# LJ_sigma
3.251
3.251
3.251
END
And again, I can't figure out how to pass the allocated epsilstar array to the main program. I have tried passing an unallocated array to the subroutine from the main.f90, allocating it inside, passing it back, and deallocating it in the main.f90, but that did not work. I have tried it as the code currently is... the code works (i.e. is bug free) but it does not pass epsilstar from the subroutine where it correctly finds it and creates an array.
It turns out that the mistake I made was in deallocating the array in the subroutine before passing it to the main program. By NOT deallocating, the array was sent back fine. Also, I do not deallocate in the main program either.
I get a bunch of errors "non-numeric character at...." and "unclassifiable argument". I am new to fortran and I don't know what is going on:
program ex
implicit none
include 'support.f90'
integer N
integer info
parameter(N=10)
real*8 D0(0:N,0:N)
integer intep(0:N)
integer j,k
integer np1
np1=N+1
do j=0,N
do k=0,N
D0(j,k)=(j+k)
end do
end do
call dgefa(D0,np1,np1,intep,info)
write(*,*) info
write(*,*) intep
write(*,*) D0
stop
end
Hermite Interpolation woes
I am trying to find the Newton Dividing Differences for the function and derivative values of a given set of x's. I'm running into serious problems with my code working for tiny examples, but failing on bigger one's. As is clearly visible, my answers are very much larger than they original function values.
Does anybody have any idea what I'm doing wrong?
program inter
implicit none
integer ::n,m
integer ::i
real(kind=8),allocatable ::xVals(:),fxVals(:),newtonDivDiff(:),dxVals(:),zxVals(:),zdxVals(:),zfxVals(:)
real(kind=8) ::Px
real(kind=8) ::x
Open(Unit=8,File="data/xVals")
Open(Unit=9,File="data/fxVals")
Open(Unit=10,File="data/dxVals")
n = 4 ! literal number of data pts
m = n*2+1
!after we get the data points allocate the space
allocate(xVals(0:n))
allocate(fxVals(0:n))
allocate(dxVals(0:n))
allocate(newtonDivDiff(0:n))
!allocate the zvalue arrays
allocate(zxVals(0:m))
allocate(zdxVals(0:m))
allocate(zfxVals(0:m))
!since the size is the same we can read in one loop
do i=0,n
Read(8,*) xVals(i)
Read(9,*) fxVals(i)
Read(10,*) dxVals(i)
end do
! contstruct the z illusion
do i=0,m,2
zxVals(i) = xVals(i/2)
zxVals(i+1) = xVals(i/2)
zdxVals(i) = dxVals(i/2)
zdxVals(i+1) = dxVals(i/2)
zfxVals(i) = fxVals(i/2)
zfxVals(i+1) = fxVals(i/2)
end do
!slightly modified business as usual
call getNewtonDivDiff(zxVals,zdxVals,zfxVals,newtonDivDiff,m)
do i=0,n
call evaluatePolynomial(m,newtonDivDiff,xVals(i),Px,zxVals)
print*, xVals(i) ,Px
end do
close(8)
close(9)
close(10)
stop
deallocate(xVals,fxVals,dxVals,newtonDivDiff,zxVals,zdxVals,zfxVals)
end program inter
subroutine getNewtonDivDiff(xVals,dxVals,fxVals,newtonDivDiff,n)
implicit none
integer ::i,k
integer, intent(in) ::n
real(kind=8), allocatable,dimension(:,:) ::table
real(kind=8),intent(in) ::xVals(0:n),dxVals(0:n),fxVals(0:n)
real(kind=8), intent(inout) ::newtonDivDiff(0:n)
allocate(table(0:n,0:n))
table = 0.0d0
do i=0,n
table(i,0) = fxVals(i)
end do
do k=1,n
do i = k,n
if( k .eq. 1 .and. mod(i,2) .eq. 1) then
table(i,k) = dxVals(i)
else
table(i,k) = (table(i,k-1) - table(i-1,k-1))/(xVals(i) - xVals(i-k))
end if
end do
end do
do i=0,n
newtonDivDiff(i) = table(i,i)
!print*, newtonDivDiff(i)
end do
deallocate(table)
end subroutine getNewtonDivDiff
subroutine evaluatePolynomial(n,newtonDivDiff,x,Px,xVals)
implicit none
integer,intent(in) ::n
real(kind=8),intent(in) ::newtonDivDiff(0:n),xVals(0:n)
real(kind=8),intent(in) ::x
real(kind=8), intent(out) ::Px
integer ::i
Px = newtonDivDiff(n)
do i=n,1,-1
Px = Px * (x- xVals(i-1)) + newtonDivDiff(i-1)
end do
end subroutine evaluatePolynomial
Values
x f(x) f'(x)
1.16, 1.2337, 2.6643
1.32, 1.6879, 2.9989
1.48, 2.1814, 3.1464
1.64, 2.6832, 3.0862
1.8, 3.1553, 2.7697
Output
1.1599999999999999 62.040113431002474
1.3200000000000001 180.40121445431600
1.4800000000000000 212.36319446149312
1.6399999999999999 228.61845650513027
1.8000000000000000 245.11610836104515
You are accessing array newtonDivDiff out of bounds.
You are first allocating it as 0:n (main program's n) then you are passing to subroutine getNewtonDivDiff as 0:n (the subroutine's n) but you pass m (m=n*2+1) to the argument n. That means you tell the subroutine that the array has bounds 0:m which is 0:9, but it has only bounds 0:4.
It is quite difficult to debug the program as it stands, I had to use valgrind. If you move your subroutines to a module and change the dummy arguments to assumed shape arrays (:,:) then the bound checking in gfortran (-fcheck=all) will catch the error.
Other notes:
kind=8 is ugly, 8 can mean different things for different compilers. If you want 64bit variables, you can use kind=real64 (real64 comes from module iso_fortran_env in Fortran 2008) or use selected_real_kind() (Fortran 90 kind parameter)
You do not have to deallocate your local arrays in the subroutines, they are deallocated automatically.
Your deallocate statement in the main program is after the stop statement, it will never be executed. I would just delete the stop, there is no reason to have it.
I am trying to pass an array of strings from C to a Fortran subroutine as well as from Fortran to that same Fortran subroutine. I have managed to pass single strings (i.e. 1D character arrays) successfully from both C and Fortran. However, I'm having trouble with arrays of strings. I am using ISO C binding on the Fortran side, and ideally I'd like this to be as seamless as possible on the calling side.
I have read some related questions and answers. Some, (i.e. this and this) are simply "Use ISO C" without further details, which doesn't help much. This answer was very helpful (similar answer to a different question), but only works for single strings, where it seems that the c_null_char is recognized in the single Fortran string. I can't figure out what to do for the array case without having two separate routines.
What I currently have is a C routine which I want to pass the array of strings (string) from:
#include <iostream>
extern "C" void print_hi_array(char input_string[][255]);
using namespace std;
int main() {
char string[3][255] = {"asdf","ghji","zxcv"};
print_hi_array(string);
return 0;
}
And, a similar Fortran routine:
program main
implicit none
call print_hi_array( (/"asdf", "ghji", "zxcv"/) )
end program
Thus far, this is what I have for the receiving end:
subroutine print_hi_array(input_string) bind(C)
use iso_c_binding, only: C_CHAR, c_null_char
implicit none
character (kind=c_char, len=1), dimension (3,255), intent (in) :: input_string
character (len=255), dimension (3) :: regular_string
character (len=255) :: dummy_string
integer :: i,j,k
write (*,*) input_string
do j = 1 , 3
dummy_string(:) = c_null_char
k = 1
do i = 1 + (j-1)*255, j*255,1
if (input_string(i) .ne. c_null_char) then
write (*,*) "i ",i,j, input_string(i)
dummy_string(k:k) = input_string(i)
endif
k = k +1
enddo
regular_string(j) = dummy_string
enddo
write (*,*) regular_string
end subroutine print_hi_array
This works for the C function; I get this output:
asdfghjizxcv
j= 1
i 1 1 a
i 2 1 s
i 3 1 d
i 4 1 f
j= 2
i 256 2 g
i 257 2 h
i 258 2 j
i 259 2 i
j= 3
i 511 3 z
i 512 3 x
i 513 3 c
i 514 3 v
asdf ghji zxcv
However, when it's done through Fortran I get nonsense out:
asdfghjizxcv#O,B�#(P,B�]B]6(P,B�# .......
It seems there is no c_null_char in this approach.
So, how can I write a Fortran subroutine to take in arrays of strings from both C and Fortran?
Fortran uses spaces to fill the rest of the string if it is declared longer than its stored text. It is not zero delimited, the declared length is stored in a hidden variable. It does not contain c null char and therefore you are reading some garbage (buffer overflow). What Fortran should print when tlit prints a string with \000 is undefined by the standard and depends on the implementation.
In particular, you are also passing a character(4) array with dimension 3 to a subroutine that expects much more data (255 chars, though I am not shure about the index order). Only pointers are passed so I think it can not be checked.
It is possible to define the length of the strings in the array constructor this way:
[character(255) :: "a","ab","abc"]
I see actually two ways to do that. Either, you write a loop in C and pass the strings one by one to Fortran, as you already did it before. Alternatively, if you want to pass the entire array and you want to handle the Fortran and the C arrays with the same routine, you will have to make an appropriate copy of your C-string array. Below a working, but not too much tested example:
extern "C" void print_array_c(int nstring, char input_string[][255]);
using namespace std;
int main() {
char string[3][255] = {"asdf","ghji","zxcv"};
print_array_c(3, string);
return 0;
}
Please note that I also pass the number of the strings, so that the example can handle arrays with various sizes. (The length of the strings is, however, assumed to be 255 characters.) On the Fortran size, one would need a routine to convert it Fortran strings. One possible visualization could be:
module arrayprint_module
use, intrinsic :: iso_c_binding
implicit none
integer, parameter :: STRLEN = 255
contains
!> The printing routine, works with Fortran character arrays only.
subroutine print_array(strings)
character(len=STRLEN), intent(in) :: strings(:)
integer :: ii
do ii = 1, size(strings)
write(*,*) ii, strings(ii)
end do
end subroutine print_array
!> Converts C string array to Fortran string array and invokes print_array.
subroutine print_array_c(nstring, cptr) bind(C)
integer(c_int), value :: nstring
type(c_ptr), intent(in), value :: cptr
character(kind=c_char), pointer :: fptr(:,:)
character(STRLEN), allocatable :: fstrings(:)
integer :: ii, lenstr
call c_f_pointer(cptr, fptr, [ STRLEN, nstring ])
allocate(fstrings(nstring))
do ii = 1, nstring
lenstr = cstrlen(fptr(:,ii))
fstrings(ii) = transfer(fptr(1:lenstr,ii), fstrings(ii))
end do
call print_array(fstrings)
end subroutine print_array_c
!> Calculates the length of a C string.
function cstrlen(carray) result(res)
character(kind=c_char), intent(in) :: carray(:)
integer :: res
integer :: ii
do ii = 1, size(carray)
if (carray(ii) == c_null_char) then
res = ii - 1
return
end if
end do
res = ii
end function cstrlen
end module arrayprint_module
Please note, that the array you pass from C must be contigous for this to work and I assumed that the character(kind=c_char) is compatible with the fortran character type, which usually it should be.
One approach which I've come up with is to modify the calling Fortran routine to also use ISO C binding:
program main
use iso_c_binding, only: C_CHAR
implicit none
character (kind=c_char, len=255), dimension (3) :: input_string
input_string = (/ "asdf", "ghji", "zxcv" /)
call print_hi_array(input_string)
end program