I am writing code in fortran for 2D integral for function func(x,y) with limits on y from y1(x) to y2(x) and limits on x from x1=3 to x2=5.
Basic assumptions are as follows:
Integral[func(x,y),{y=y1 to y2}, {x=x1 to x2}] = Integral[funcx(x),{x = x1 to x2}]. where funcx = Integral[func,{y = y1(x) to y2(x)}]
Please find my question in the middle of code. I write the code as follows,
implicit real*8(a-h,o-z)
external func
external y1
external y2
x1 = 3.d0
x2 = 5.d0
call twodint(func,y1,y2,x1,x2,result)
print*, result
stop
end
!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
subroutine twodint(origfunc,y1,y2,x1,x2,result)
implicit real*8(a-h,o-z)
external funcx
call simpsonintegral(funcx,x1,x2,ss) ! my integral routine funcx = function, x1 and x2 = limits, ss = output
result = ss
return
end
!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
function funcx(x)
implicit real*8(a-h,o-z)
external funcy
external y1
external y2
common/xvalues/xx
xx = x
call simpsonintegral(funcy,y1(x),y2(x),ss)
funcx = ss
return
end
!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
function funcy(y)
implicit real*8(a-h,o-z)
common/xvalues/x
! my question : if i write here funcy = func(x,y), my code works fine. but i want to write here something like funcy = origfunc(x,y), so that it can receive function from subroutine named as twodint. but this does not work. please help...
return
end
!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
function func(x,y)
implicit real*8(a-h,o-z)
func = (x**2)*y*dsin(x*y) ! some function of x and y
return
end
function y1(x)
implicit real*8(a-h,o-z)
y1 = x ! some limit
return
end
function y2(x)
implicit real*8(a-h,o-z)
y2 = x*x ! some limit
return
end
In my opinion, external is mainly included in modern Fortran for legacy reasons. Its not the best way to pass functions to subroutines. Also, implicit type declaration is dangerous ... much better to use implicit none and explicitly type all of your variables ... you'll catch a lot of mistakes that way. So ... here is a sketch of how to pass various functions to a subroutine such as an integrator.
module MyStuff
implicit none
abstract interface
function funcX (x)
real, intent (in) :: x
real :: funcX
end function funcX
end interface
contains
subroutine twodint(origfunc,y1,y2,x1,x2,result)
real :: y1, y2, x1, x2, result, ss
procedure (funcX) :: origfunc
call simpsonintegral (origfunc,x1,x2,ss) ! my integral routine funcx = function, x1 and x2 = limits, ss = output
result = ss
return
end subroutine twodint
real function funcA (x)
real, intent (in) :: x
funcA = x ** 2
end function funcA
real function funcB (x)
real, intent (in) :: x
funcB = x ** 2
end function funcB
end module MyStuff
program test1
use MyStuff
implicit none
real :: x1, x2, y1, y2, result
x1 = 3.d0
x2 = 5.d0
call twodint(funcA,y1,y2,x1,x2,result)
print*, result
call twodint(funcB,y1,y2,x1,x2,result)
print*, result
stop
end program test1
Related
I'm trying to implement Newton's method but I'm getting a confusing error message. In my code you'll see I called external with f1 and f2 which I assumes tells the computer to look for the function but it's treating them as variables based on the error message. I've read the stack overflow posts similar to my issue but none of the solutions seem to work. I've tried with and without the external but the issue still persists. Hoping someone could see what I'm missing.
implicit none
contains
subroutine solve(f1,f2,x0,n, EPSILON)
implicit none
real(kind = 2), external:: f1, f2
real (kind = 2), intent(in):: x0, EPSILON
real (kind = 2):: x
integer, intent(in):: n
integer:: iteration
x = x0
do while (abs(f1(x))>EPSILON)
iteration = iteration + 1
print*, iteration, x, f1(x)
x = x - (f1(x)/f2(x))
if (iteration >= n) then
print*, "No Convergence"
stop
end if
end do
print*, iteration, x
end subroutine solve
end module newton
Program Lab10
use newton
implicit none
integer, parameter :: n = 1000 ! maximum iteration
real(kind = 2), parameter :: EPSILON = 1.d-3
real(kind = 2):: x0, x
x0 = 3.0d0
call solve(f(x),fp(x),x0,n, EPSILON)
contains
real (kind = 2) function f(x) ! this is f(x)
implicit none
real (kind = 2), intent(in)::x
f = x**2.0d0-1.0d0
end function f
real (kind = 2) function fp(x) ! This is f'(x)
implicit none
real (kind = 2), intent(in)::x
fp = 2.0d0*x
end function fp
end program Lab10```
You seem to be passing function results to your subroutine and not the functions themselves. Remove (x) when calling solve() and the problem will be resolved. But more importantly, this code is a prime example of how to not program in Fortran. The attribute external is deprecated and you better provide an explicit interface. In addition, what is the meaning of kind = 2. Gfortran does not even comprehend it. Even if it comprehends the kind, it is not portable. Here is a correct portable modern implementation of the code,
module newton
use iso_fortran_env, only: RK => real64
implicit none
abstract interface
pure function f_proc(x) result(result)
import RK
real(RK), intent(in) :: x
real(RK) :: result
end function f_proc
end interface
contains
subroutine solve(f1,f2,x0,n, EPSILON)
procedure(f_proc) :: f1, f2
real(RK), intent(in) :: x0, EPSILON
integer, intent(in) :: n
real(RK) :: x
integer :: iteration
x = x0
do while (abs(f1(x))>EPSILON)
iteration = iteration + 1
print*, iteration, x, f1(x)
x = x - (f1(x)/f2(x))
if (iteration >= n) then
print*, "No Convergence"
stop
end if
end do
print*, iteration, x
end subroutine solve
end module newton
Program Lab10
use newton
integer, parameter :: n = 1000 ! maximum iteration
real(RK), parameter :: EPSILON = 1.e-3_RK
real(RK) :: x0, x
x0 = 3._RK
call solve(f,fp,x0,n, EPSILON)
contains
pure function f(x) result(result) ! this is f(x)
real (RK), intent(in) :: x
real (RK) :: result
result = x**2 - 1._RK
end function f
pure function fp(x) result(result) ! This is f'(x)
real (RK), intent(in) :: x
real (RK) :: result
result = 2 * x
end function fp
end program Lab10
If you expect to pass nonpure functions to the subroutine solve(), then remove the pure attribute. Note the use of real64 to declare 64-bit (double precision) real kind. Also notice how I have used _RK suffix to assign 64-bit precision to real constants. Also, notice I changed the exponents from real to integer as it is multiplication is more efficient than exponentiation computationally. I hope this answer serves more than merely the solution to Lab10.
program prob_1
implicit real*8(a-h, o-z)
f(x) = x**2-cos(x)
df(x) = 2*x+sin(x)
x0 = 0, x1 = 1
do i = 1, 3
if (f((x0+x1)/2) < 0)
x0 = (x0+x1)/2
else
x1 = (x0+x1)/2
end do
print *,"x = ", x
end program
I'm just starting to use Fortran 90.
Now I'm using Code::blocks but I don't know exactly which line the error exists on.
I guess the problem is f((x0+x1)/2) < 0 but don't know actually what is the real error.
what's problem is here?
Be advised that statement functions, the function definitions the OP uses, are obsolescent.
B.3.4 Statement functions
Statement functions are subject to a number of non intuitive restrictions and are a potential source of error because their syntax is easily confused with that of an assignment statement.
The internal function is a more generalized form of the statement function and completely supersedes it.
source: F2018 Standard
Also the notation REAL*8 or anything of that form has never been part of any Fortran standard (see here):
I would suggest to rewrite the code as:
program prob_1
implicit none
double precision :: x1,x0
integer :: i
x0 = 0; x1 = 1
do i = 1, 3
if (f((x0+x1)/2.0D0) < 0) then
x0 = (x0+x1)/2.0D0
else
x1 = (x0+x1)/2.0D0
endif
end do
print *,"x = ", (x0+x1)/2.0D0
contains
function f(x)
double precision, intent(in) :: x
double precision :: f
f = x**2-cos(x)
end function f
function df(x)
double precision, intent(in) :: x
double precision :: df
df = 2.0D0*x+sin(x)
end function df
end program
If you change your program as follows then it will compile:
program prob_1
implicit real*8(a-h, o-z)
f(x) = x**2-cos(x)
df(x) = 2*x+sin(x)
x0 = 0; x1 = 1
do i = 1, 3
if (f((x0+x1)/2) < 0) then
x0 = (x0+x1)/2
else
x1 = (x0+x1)/2
endif
end do
print *,"x = ", x
end program
As mentionned in the comments, you have to add the semicolon ; to separate statements in one line and you have to add the then as well as endif to your if condition.
Hope it helps.
I want to integrate a 3d array over space in r, theta and phi (spherical polar). For 1d I use Simpson's 1/3rd rule but I am confused about that for 3d. Also, would you like to suggest any other method for integration or subroutine? I am using Fortran 95.
I have written the Fortran code for integration in 3d, I thought I should share with you.
The code for calculating integration of a function is 3 dimension is:
!This program uses Simpson's 1/3 method to calulate volume
integral in r,theta & phi.
program SimpsonInteg3d
implicit none
integer::i,j,k
integer, parameter :: N=10,M=360,L=180
integer, parameter:: rmin=0,rmax=N,phimin=0,phimax=M,&
thetamin=0,thetamax=L
double precision,&
dimension(rmin:rmax,thetamin:thetamax,phimin:phimax)::U
real*8, parameter :: pi = 4*atan(1.0),dr=1./N,&
dtheta=pi/(L),dphi=2*pi/M
real*8 :: r(rmin:rmax)=(/(i*dr,i=rmin,rmax)/),&
theta(thetamin:thetamax)=(/(j*dtheta,j=thetamin,thetamax)/),&
p(phimin:phimax)=(/(k*dphi,k=phimin,phimax)/)
real*8::intg
do i=rmin,rmax
do j=thetamin, thetamax
do k=phimin,phimax
!The function which has to be integrated.
U(i,j,k)=r(i)* (sin((p(k)))**2) *sin(theta(j))
enddo
enddo
enddo
call Integration(Intg,U,r,theta,p)
print*,"Integration of function U using simpson's 1/3=", Intg
end program
!===============================================================!
!Subroutine for calculating integral of a function in 3d.
subroutine Integration(Intg,U,r,theta,p)
implicit none
integer::i,j,k
integer, parameter :: N=10,M=360,L=180
integer, parameter ::rmin=0,rmax=N,&
phimin=0,phimax=M,thetamin=0,thetamax=L
double precision,&
dimension(rmin:rmax,thetamin:thetamax,phimin:phimax):: U
real*8::
r(rmin:rmax),theta(thetamin:thetamax),p(phimin:phimax),Intg,Ia
double precision,dimension(rmin:rmax)::Itheta
real*8, parameter :: pi = 4*atan(1.0),dr=1./N,&
dtheta=pi/(L),dphi=2*pi/M
Intg=0
Ia=0
do i=rmin+1,rmax-1
call Integtheta(Itheta,i,U,r,theta,p)
if(mod(i,2).eq.0) then
Ia = Ia + 2*Itheta(i)*r(i)**2
else
Ia = Ia + 4*Itheta(i)*r(i)**2
endif
end do
call Integtheta(Itheta,rmin,U,r,theta,p)
call Integtheta(Itheta,rmax,U,r,theta,p)
Intg=(dr/3)*(Itheta(rmin)+Itheta(rmax)+ Ia)
end subroutine Integration
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!Subroutine for calculating integral of U along theta and phi
subroutine Integtheta(Itheta,i,U,r,theta,p)
implicit none
integer::i,j,k
integer, parameter :: N=10,M=360,L=180
integer, parameter ::rmin=0,rmax=N,&
phimin=0,phimax=M,thetamin=0,thetamax=L
double precision,&
dimension(rmin:rmax,thetamin:thetamax,phimin:phimax)::U
real*8:: r(rmin:rmax),theta(thetamin:thetamax),p(phimin:phimax)
double precision,dimension(rmin:rmax)::Itheta,Itha
double precision,dimension(rmin:rmax,thetamin:thetamax)::Ip
real*8, parameter :: pi = 4*atan(1.0),dr=1./N,&
dtheta=pi/(L),dphi=2*pi/M
Itheta(i)=0
Itha(i)=0
do j=thetamin+1,thetamax-1
call Integphi(Ip,i,j,U,r,theta,p)
if(mod(j,2).eq.0) then
Itha(i) = Itha(i) + 2*Ip(i,j)*sin(theta(j))
else
Itha(i) = Itha(i) + 4*Ip(i,j)*sin(theta(j))
endif
end do
call Integphi(Ip,i,thetamin,U,r,theta,p)
call Integphi(Ip,i,thetamax,U,r,theta,p)
Itheta(i)=(dtheta/3)*(Ip(i,thetamin)+Ip(i,thetamax)+ Itha(i))
end subroutine Integtheta
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!Subroutine for calculating integral of U along phi
subroutine Integphi(Ip,i,j,U,r,theta,p)
implicit none
integer::i,j,k
integer, parameter :: N=10,M=360,L=180
integer, parameter ::rmin=0,rmax=N,&
phimin=0,phimax=M,thetamin=0,thetamax=L
double precision,&
dimension(rmin:rmax,thetamin:thetamax,phimin:phimax)::U
real*8:: r(rmin:rmax),theta(thetamin:thetamax),p(phimin:phimax)
double precision,dimension(rmin:rmax,thetamin:thetamax)::Ip,Ipa
real*8, parameter :: pi = 4*atan(1.0),dr=1./N,&
dtheta=pi/(L),dphi=2*pi/M
Ipa(i,j)=0
do k=phimin+1,phimax-1
if(mod(k,2).eq.0) then
Ipa(i,j) = Ipa(i,j) + 2*U(i,j,k)
else
Ipa(i,j)= Ipa(i,j) + 4*U(i,j,k)
endif
end do
Ip(i,j)=(dphi/3)*(U(i,j,phimin)+U(i,j,phimax)+ Ipa(i,j))
end subroutine Integphi
It calculates the integration of the function U along phi first and then uses the function Ip to calculate integral along theta. Then finally the function Itheta is used to calculate integration along r.
The gfortran compiler gives wrong answer, when I run a parallel program using OpenMP. In the same time, ifort gives exact result.
This is the whole compilable code.
!_______________________________________________________________ !
!____________MODULE SECTION_____________________________________ !
MODULE MATRIC
IMPLICIT NONE
INTEGER , PARAMETER :: NG = 40
DOUBLE PRECISION,SAVE :: Z , PA , PB ,CMU
DOUBLE PRECISION , PARAMETER :: PI=2.0D0*ACOS(0.0D0) , &
FPI=4.0D0*PI , SQFPI = SQRT(FPI), DLAM=1.0D0
DOUBLE PRECISION , DIMENSION(450) :: DEL1, DEL2, X, R , SNLO
DOUBLE PRECISION :: XG(60) , WG(60)
END MODULE MATRIC
!_________________________________________________________________________!
! MODULE SECTION
!__________________________________________________________________________!
MODULE POTDATA
IMPLICIT NONE
INTEGER :: IA , IB , ID
DOUBLE PRECISION :: RA , RB , R1s(450)
END MODULE POTDATA
!__________________________________________________________________________!
!______________________________________________________________________!
program check
use matric
use potdata
implicit double precision(a-h,o-z)
pa = 0.72D0 ; pb = 0.19D0
mesh = 441 ; noint= 40 ; z = 2.0d0
CALL GAULEG(-1.d0,1.d0)
NB = MESH/NOINT
I = 1
X(I) = 0.0D+00
DELTAX = 0.0025D+00*40.0D+00/DBLE(NOINT)
DO J=1,NB
IMK = (J-1)*NOINT + 1
DO K=1,NOINT
AK=K
I=I+1
X(I)=X(IMK)+AK*DELTAX
END DO
DELTAX=2.0D+00*DELTAX
END DO
CMU=(9.0D00*PI*PI/(128.0D00*Z))**THIRD
R(1)=0.0D+00 ; SNLO(1) = 0.D00
DO I=2,MESH
R(I)=CMU*X(I)
SNLO(I) = R(I)*dexp(-Z*R(I))
R1S(I) = SNLO(I)/(SQFPI*R(I))
END DO
call EFFPOT(MESH,NOINT)
end program check
subroutine EFFPOT(MESH,NOINT)
USE OMP_LIB
USE MATRIC
USE POTDATA
implicit none
integer, intent(in) :: MESH, NOINT
double precision::anorm(450)
double precision, external :: funct
double precision :: asum, fac, cnorm
!$omp parallel do default(none) private(del1,ia,asum,ib,ra,rb,fac) &
!$omp shared(id,mesh,r,anorm,NOINT,del2,R1s)
do ia = 2,mesh
ra = r(ia)
if(R1s(ia).lt.1.D-7.and.R1s(ia).ge.1.D-8)id = ia
do ib = 2,mesh
rb = r(ib)
call QGAUSS(funct,-1.d0,1.d0,fac)
del1(ib) = rb**2*fac*R1s(ib)**2
end do
CALL NCDF(del1,ASUM,r(2),mesh,NOINT)
anorm(ia) = 2.0d0*pi*asum
del2(ia) = 2.0d0*pi*asum*(ra*R1s(ia))**2
end do
!$omp end parallel do
CALL NCDF(del2,ASUM,r(2),mesh,NOINT)
cnorm = 1.0/dsqrt(4.*pi*ASUM)
write(6,*)'cnorm =',cnorm
return
end
double precision function funct(x)
USE POTDATA , ONLY : RA , RB
USE MATRIC , ONLY : PA , PB , DLAM
implicit none
double precision, intent(in) :: x
double precision :: f1, f2, ramrb
ramrb = dsqrt(ra**2+rb**2-2.d0*ra*rb*x)
f1 = dcosh(pa*ra)+dcosh(pa*rb)
f2 = 1.d0+0.5*dlam*ramrb*dexp(-pb*ramrb)
funct = (f1*f2)**2
return
end
SUBROUTINE QGAUSS(func,aa,bb,ss)
USE OMP_LIB
USE MATRIC , ONLY : XG ,WG , NG
IMPLICIT DOUBLE PRECISION(A-H,O-Z)
external func
xm = 0.5d0*(bb+aa)
xl = 0.5d0*(bb-aa)
ss = 0.d0
do j=1,ng
dx = xl*xg(j)
ss = ss + wg(j)*(func(xm+dx)+func(xm-dx))
end do
ss = xl*ss/2.0
return
END
SUBROUTINE GAULEG(x1,x2)
USE MATRIC , ONLY : XG ,WG ,NG , PI
IMPLICIT DOUBLE PRECISION(A-H,O-Z)
eps = 1.d-14
m = (ng+1)/2
xm = 0.5D0*(x1+x2)
xl = 0.5D0*(x2-x1)
do i=1,m
z = dcos(pi*(dfloat(i)-0.25d0)/(dfloat(ng)+0.5d0))
1 continue
p1 = 1.d0
p2 = 0.d0
do j=1,ng
p3 = p2
p2 = p1
p1 = ((2.d0*dfloat(j)-1.d0)*z*p2 &
- (dfloat(j)-1.d0)*p3)/dfloat(j)
end do
pp = dfloat(ng)*(z*p1-p2)/(z*z-1.d0)
z1 = z
z = z1 - p1/pp
if (dabs(z-z1).gt.eps) go to 1
xg(i) = xm - xl*z
xg(ng+1-i) = xm + xl*z
wg(i) = 2.d0*xl/((1.d0-z*z)*pp*pp)
wg(ng+1-i) = wg(i)
end do
return
end
SUBROUTINE NCDF(F,ASUM,H,KKK,NOINT)
IMPLICIT DOUBLE PRECISION (A-H,O-Z)
DIMENSION F(450)
NBLOCK = (KKK-2)/NOINT + 1
C2HO45 = 2.0D+00*H/45.0D+00
ASUM = 0.0D+00
DO J=1,NBLOCK
ISTAR = NOINT*(J-1)+5
IEND = NOINT*J + 1
IEND = MIN0(KKK,IEND)
DO I=ISTAR,IEND,4
ASUM = ASUM + C2HO45*(7.0D+00*(F(I-4)+F(I)) &
+32.0D+00*(F(I-3)+F(I-1)) + 12.0D+00*F(I-2))
END DO
IF(IEND.EQ.KKK) GO TO 4
C2HO45 = 2.0D+00*C2HO45
4 END DO
RETURN
END
Thanks everybody specially #Vladimir who has taken interest in my problem. Finally i got the right answer by removing ra and rb from the module potdata and defined function as funct(x, ra, rb) and then removing ra and rb from the loop. Because i was writing ra, rb then reading their values in the above code so loop was having flow dependence. Now i get exact result from both compiler (which is 8.7933767516) parallelly, sequentially both. Exact way is this
subroutine EFFPOT(MESH,NOINT)
USE OMP_LIB
USE MATRIC
USE POTDATA
implicit none
integer, intent(in) :: MESH, NOINT
double precision::anorm(450)
double precision, external :: funct
double precision :: asum, fac, cnorm
!$omp parallel do default(none) private(del1,ia,asum,ib,fac) &
!$omp shared(id,mesh,r,anorm,NOINT,del2,R1s)
do ia = 2,mesh
if(R1s(ia).lt.1.D-7.and.R1s(ia).ge.1.D-8)id = ia
do ib = 2,mesh
call QGAUSS(funct,-1.d0,1.d0,fac,r(ia),r(ib))
del1(ib) = r(ib)**2*fac*R1s(ib)**2
end do
CALL NCDF(del1,ASUM,r(2),mesh,NOINT)
anorm(ia) = 2.0d0*pi*asum
del2(ia) = 2.0d0*pi*asum*(r(ia)*R1s(ia))**2
end do
!$omp end parallel do
CALL NCDF(del2,ASUM,r(2),mesh,NOINT)
cnorm = 1.0/dsqrt(4.*pi*ASUM)
write(6,*)'cnorm =',cnorm
return
end
double precision function funct(x,ra,rb)
USE MATRIC , ONLY : PA , PB , DLAM
implicit none
double precision, intent(in) :: x, ra, rb
double precision :: f1, f2, ramrb
ramrb = dsqrt(ra**2+rb**2-2.d0*ra*rb*x)
f1 = dcosh(pa*ra)+dcosh(pa*rb)
f2 = 1.d0+0.5*dlam*ramrb*dexp(-pb*ramrb)
funct = (f1*f2)**2
return
end
SUBROUTINE QGAUSS(func,aa,bb,ss,ra,rb)
USE OMP_LIB
USE MATRIC , ONLY : XG ,WG , NG
IMPLICIT DOUBLE PRECISION(A-H,O-Z)
external func
xm = 0.5d0*(bb+aa)
xl = 0.5d0*(bb-aa)
ss = 0.d0
do j=1,ng
dx = xl*xg(j)
ss = ss + wg(j)*(func(xm+dx,ra,rb)+func(xm-dx,ra,rb))
end do
ss = xl*ss/2.0
return
END
The cause of your problem is that the OpenMP standard does not specify what happens if a private list item is accessed in the region but outside of the construct. See example private.2f (found on page 135 of the OpenMP standard supplement) for a short version of the same problem.
Specifically, the module variables ra and rb are declared private in the OpenMP parallel region inside EFFPOT and also accessed from funct. funct is in the dynamic scope of the parallel region but (lexically) outside of it and therefore it is not specified whether ra and rb referenced by funct are the original module variables or their private copies (most compilers would go for the original variables).
You have already found one of the solutions. The other one would be to declare ra and rb threadprivate since they are only used to pass data from EFFPOT to funct:
MODULE POTDATA
IMPLICIT NONE
INTEGER :: IA , IB , ID
DOUBLE PRECISION :: RA , RB , R1s(450)
!$OMP THREADPRIVATE(RA,RB)
END MODULE POTDATA
You should then also remove ra and rb from the list of the private clause of the parallel region within EFFPOT.
On some platforms, e.g. OS X, using threadprivate with GCC (i.e. gfortran) could be slower than actually passing around the two variables as arguments because of the emulated TLS.
Note that this semantic error is really hard to detect and many OpenMP tools won't actually spot it.
First of all, it is very difficult to say something specific without seeing the actual code. However, I do have some comments on your situation and the conclusions you are drawing.
The fact that your program runs fine both in parallel and sequential execution when compiled with "ifort" doesn't mean that your program is correct. Since compiler bugs leading to programs giving wrong answers are very rare, but on the other hand manual parallel programming is very error-prone, we should assume a problem with the way you parallelized your code. We are probably talking about a race condition.
And no, the problem you are having is not at all unusual. When you have a race condition, it happens often that the sequential execution works everywhere and your parallel execution works in some environments and fails in others. It's even common that your code gives different answers every time you call it (not only depending on the compiler, but on many other factors that can change over time).
What I suggest you should do, is to get a parallel debugger, like for example TotalView that will help you keep track of the various threads and their states. Try to find a simple test environment (as few threads as possible) that fails reliably.
i try to solve this simple three body problem with the following code:
Program Main
Implicit real*8 (A-H,O-Z)
real*8 ome,mu, rho, R
duepi=8*datan(1.d0)
ome=1
mu=0.001
T_per=duepi/ome
rho=0.1
R=1.0
N_step=100
c Open the file
OPEN(unit=11, file="prova1.txt")
c Nested do loops
do iy0=1,100
do iP0=1,100
c Calc value for 0
y0 = real(iy0)/100.
x0 = 0
c Calc value for py0
py0 = real(iP0)/100.
px0 = 0
x=x0
y=y0
px=px0
py=py0
dt=T_per/N_step
E0=H(x,y,px,py)
k_max=100*N_step
k=0
t=0
errh=0
c---------
c start integration loop
c--------
do k=1,k_max
call sym4(x,y,px,py,dt)
E= H(x,y,px,py)
errh=abs(E-E0)
t=k*dt
enddo
do k=1,k_max
call sym4(x,y,px,py,-dt)
E= H(x,y,px,py)
errh=abs(E-E0)
t=t-dt
enddo
write(11,*) y0, py0, errh
enddo ! iP0
enddo ! iy0
close(11)
end
subroutine sym1(x,y,px,py,dt)
Implicit real*8 (A-H,O-Z)
c
call f(x,y,fx,fy)
pxnew=px+dt*fx
pynew=py+dt*fy
xnew=x+dt*pxnew
ynew=y+dt
c
x=xnew
y=ynew
px=pxnew
py=pynew
end
subroutine sym1_B(x,y,px,py,dt)
Implicit real*8 (A-H,O-Z)
c
xnew=x+dt*px
ynew=y+dt
call f(xnew,ynew,fxnew,fynew)
pxnew=px+dt*fxnew
pynew=py+dt*fynew
c
x=xnew
y=ynew
px=pxnew
py=pynew
end
subroutine f(x,y,fx,fy)
Implicit real*8 (A-H,O-Z)
real*8 ome,mu,rho,R
fx = ((1-mu)*(rho+x))/((rho*rho+2*rho*x+y*y)**(1.5)) -
& (mu*(R+x))/((R**2-2*R*x+x*x+y*y)**(1.5))
fy = ((1-mu)*(rho+x))/((rho**2+x**2+2*rho*x+y**2)**(1.5))/
& + (mu*y)/((R**2-2*R*x+x**2+y**2)**(1.5))
return
end
real*8 function H(x,y,px,py)
Implicit real*8 (A-H,O-Z)
real*8 ome,mu,rho,R
c h=px*px/2.d0+ py +(1+eps*cos(ome*y))*x*x/2
c h=px*px/2.d0+ py -(1+eps*cos(ome*y))*cos(x)
r12 = sqrt( ( (x*cos(ome*y)-y*sin(ome*y))+rho*cos(ome*y) )**2
& + ( x*sin(ome*y)+y*cos(ome*y) + rho*sin(ome*y) )**2 )
r13 = sqrt( ( (x*cos(ome*y)-y*sin(ome*y))-R*cos(ome*y) )**2
& + ( x*sin(ome*y)+y*cos(ome*y) - R*sin(ome*y) )**2 )
h=(px**2+py**2)/2.d0 - (1-mu)/r12 - mu/r13 + py
return
end
subroutine sym2(x,y,px,py,dt)
Implicit real*8 (A-H,O-Z)
call f(x,y,fx,fy)
xnew= x+ px*dt + fx*dt**2/2.d0
ynew= y+ dt ! così è giusto
call f(xnew,ynew,fxnew,fynew)
pxnew= px+ dt*(fx+fxnew )/2.d0
pynew= py+ dt*(fy+fynew )/2.d0
x=xnew
y=ynew
px=pxnew
py=pynew
end
subroutine sym4(x,y,px,py,dt)
Implicit real*8 (A-H,O-Z)
sq2=2**(1.d0/3.d0)
alpha= 1.d0/(2-sq2)
beta= sq2/(2-sq2)
dt1= dt*alpha
dt2=-dt*beta
call sym2(x,y,px,py,dt1)
call sym2(x,y,px,py,dt2)
call sym2(x,y,px,py,dt1)
return
end
the code calls sympletic integrators and solve the 3body problem. But when i try to run it, there aren't compiling errors, the output.txt file show only the initial grid and not the errh this column give me only NaN, can someone help me?
Is maybe an initial condition errors (strange initial conditions for velocities or positions, omega is wrong ...)?
As stated already in the comments, your program has many code style problems (wrong intentation, ...) as well as not using best Fortran practises (e.g. implicit none). Your problem can be solved by trivial usage of a debugger.
An obvious problem is that you are using uninitialized variables in the functions:
subroutine f(x,y,fx,fy) : rho, mu, R in calculation of fx and fy, which produces NaN
function H() : ome, rho, mu, ...
and similarly on other places