error #5082: Syntax error, found ',' when expecting one of: )

gumbira__gugum · ‎03-16-2018

Hi, I got this error when I write my own code. It's look strange because the syntax I think is correct. The error exist in this part

auE_h=(-uCe,(uDx-uCe)/2,0)

Any help will be great thank you and here I posted the whole code

PROGRAM BACKWARD

IMPLICIT NONE

! Declaration parameter

    ! INTEGER TYPE
INTEGER, PARAMETER :: NX=300
INTEGER, PARAMETER :: NY=30
INTEGER dx, dy, i,j, maxIter, maxNMiter, n

    ! REAL TYPE
REAL, PARAMETER :: L=15
REAL, PARAMETER :: H=1
REAL(8) alpha_u, alpha_p, Cs, err, Ho, Hs, nu, Re, uMax


REAL(8) uCe, uCw, uCn, uCs     !CONVECTIVE TERM PARAMETER FOR u 
REAL(8) vCe, vCw, vCn, vCs     !CONVECTIVE TERM PARAMETER FOR v 
REAL(8) dudy, dudx, nutu       !SGS CLASSIC SMAGORINSKY PARAMETER for u
REAL(8) dvdy, dvdx, nutv       !SGS CLASSIC SMAGORINSKY PARAMETER for v
REAL(8) uDx, uDy               !DIFFUISIVE TERM PARAMETER for u
REAL(8) vDx, vDy               !DIFFUISIVE TERM PARAMETER for v


    ! REAL TYPE WITH DIMENSION
REAL, DIMENSION(NX+1,NY+2) :: u, uold, ustar, uprime
REAL, DIMENSION(NX+2,NY+1) :: v, vold, vstar, vprime
REAL, DIMENSION(NX+2,NY+2) :: p, pstar, pprime        
REAL, DIMENSION(1,NY+2) :: uLeft
REAL, DIMENSION(1,NY+2) :: vLeft

REAL(8) auE_h, auW_h, auN_h, auS_h
REAL, DIMENSION(NX,NY) ::auE
REAL, DIMENSION(NX,NY) ::auW
REAL, DIMENSION(NX,NY) ::auN
REAL, DIMENSION(NX,NY) ::auS
REAL, DIMENSION(NX,NY) ::auP
REAL, DIMENSION(NX,NY) ::dU
REAL, DIMENSION(NX,NY) ::dV

REAL(8) avE_h, avW_h, avN_h, avS_h
REAL, DIMENSION(NX,NY) ::avE
REAL, DIMENSION(NX,NY) ::avW
REAL, DIMENSION(NX,NY) ::avN
REAL, DIMENSION(NX,NY) ::avS
REAL, DIMENSION(NX,NY) ::avP

REAL(8) apE_h, apW_h, apN_h, apS_h
REAL, DIMENSION(NX,NY) ::apE
REAL, DIMENSION(NX,NY) ::apW
REAL, DIMENSION(NX,NY) ::apN
REAL, DIMENSION(NX,NY) ::apS
REAL, DIMENSION(NX,NY) ::apP
REAL, DIMENSION(NX,NY) ::bpP

REAL, DIMENSION(NX,NY) ::ux            ! Convergence coeff
REAL, DIMENSION(NX,NY) ::uxold
REAL, DIMENSION(NX,NY) ::vx
REAL, DIMENSION(NX,NY) ::vxold 
REAL(8) cmax_1, cmax_2, convergence       !CONVERGENCE


REAL, DIMENSION(1,NY) ::Y 

! FILL THE VALUE
maxIter=50000       !INTEGER VALUE
maxNMiter=1

alpha_u=0.3         !REAL VALUE
alpha_p=0.7
Cs=0.07
err=1e-5
Ho=1.0
Re=100
nu=2e-3 

dx=L/NX
dy=H/NY  !Step size
Hs=H-Ho
uMax=Re*nu/Hs
! ALLOCATE THE GUESS VALUE AND INITIAL CONDITION

u(:,:)=0
uold(:,:)=0     !INITIAL CONDITION
ustar(:,:)=0
uprime(:,:)=0
dU(:,:)=0

v(:,:)=0
vold(:,:)=0     !INITIAL CONDITION
vstar(:,:)=0
vprime(:,:)=0
dV(:,:)=0

p(:,:)=0
pstar(:,:)=0
pprime(:,:)=0
uLeft(:,:)=0

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!    SIMPLE SCHEME APPLY FROM HERE &
!!!!        CLASSIC SMAGORINSKY SGS
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

DO n=1,maxIter

! STEP 1a, solve x-momentum as ustar
DO i=2,NX
DO j=2,NY+1
! Convective flux terms use CENTRAL DIFFERENCE
uCe=dy*0.5*(uold(i,j)+uold(i+1,j))
uCw=dy*0.5*(uold(i,j)+uold(i-1,j))
uCn=dx*0.5*(vold(i,j)+vold(i,j+1))
uCs=dx*0.5*(vold(i,j-1)+vold(i+1,j-1))
! Calculation for rate of strain tensor
dudy=(0.5*(uold(i,j)+uold(i,j+1)))/dy
dudy=(0.5*(uold(i,j)+uold(i+1,j)))/dx
nutu=(Cs*FLOAT(dx))**2*((0.5*(dudx+dudy)**2)**0.5)*(dudx+dudy)
! Diffusive flux term
uDx=(FLOAT(dy)/FLOAT(dx)*((1/Re)+nutu)) 
uDy=(FLOAT(dx)/FLOAT(dy))*((1/Re)+nutu)
! ASSEMBLY COEFFICIENT FROM HYBRID SCHEME
auE_h=(-uCe,(uDx-uCe)/2,0)
auW_h=(uCw, (uDx+0.5*uCw), 0)
auN_h=(-uCn, (uDy-0.5*uCn), 0)
auS_h=(uCs, (uDy+0.5*uCs), 0)
auE(i,j)=MAX(auE_h)
auW(i,j)=MAX(uCw, (uDx+0.5*uCw), 0)
auN(i,j)=MAX(-uCn, (uDy-0.5*uCn), 0)
auS(i,j)=MAX(uCs, (uDy+0.5*uCs), 0)
!Central coeff & under relaxation 
auP(i,j)=(auE(i,j)+auW(i,j)+auN(i,j)+auS(i,j))/alpha_u
!Velocity component for PCE
dU(i,j)=FLOAT(dy)/auP(i,j)
END
END
!Use uold for calculate ustar
ustar=uold
DO iter=maxNMiter
DO i=2, NX
DO j=2, NY+1
ustar(i,j)=(auE(i,j)*ustar(i+1,j)+auW(i,j)*ustar(i-1,j)+auN(i,j)*ustar(i,j+1)+auS(i,j)*ustar(i,j-1)-FLOAT(dy)*(pstar(i+1,j)-pstar(i,j))+(1-alpha_u)*auP(i,j)*uold(i,j))/auP(i,j)
END
END
END
! ustar FOR X-MOMENTUM HAS BEEN SOLVE
!!!!!!!!!! STEP 1B BEGIN
! Solve y-momentum as vstar
DO i=2, NX+1
DO j=2, NY
! Convective flux terms use CENTRAL DIFFERENCE
vCe=dy*0.5*(uold(i,j)+uold(i,j+1))
vCw=dy*0.5*(uold(i-1,j)+uold(i-1,j+1))
vCn=dx*0.5*(vold(i,j)+vold(i,j+1))
vCs=dx*0.5*(vold(i,j)+vold(i,j-1))
! Calculation for rate of strain tensor
dvdy=(0.5*(vold(i,j)+vold(i,j+1)))/dy
dvdy=(0.5*(vold(i,j)+vold(i+1,j)))/dx
nutv=(Cs*FLOAT(dx))**2*((0.5*(dvdx+dvdy)**2)**0.5)*(dvdx+dvdy)
! Diffusive flux term
vDx=(FLOAT(dy)/FLOAT(dx))*((1/Re)+nutv) 
vDy=(FLOAT(dx)/FLOAT(dy))*((1/Re)+nutv)
! ASSEMBLY COEFFICIENT FROM HYBRID SCHEME
avE_h=(-vCe, (vDx-0.5*vCe), 0)
avW_h=(vCw, (vDx+0.5*vCw), 0)
avN_h=(-vCn, (vDy-0.5*vCn), 0)
avS_h=(vCs, (vDy+0.5*vCs), 0)
avE(i,j)=MAX(avE_h)
avW(i,j)=MAX(avW_h)
avN(i,j)=MAX(avN_h)
avS(i,j)=MAX(avS_h)
!Central coeff & under relaxation 
avP(i,j)=(avE(i,j)+avW(i,j)+avN(i,j)+avS(i,j))/alpha_u
!Velocity component for PCE
dV(i,j)=FLOAT(dx)/avP(i,j)
END
END
! Calculate vstar
vstar=vold
DO iter=maxNMiter
DO i=2, NX+1
DO j=2, NY
vstar(i,j)=(avE(i,j)*vstar(i+1,j)+avW(i,j)*vstar(i-1,j)+avN(i,j)*vstar(i,j+1)+avS(i,j)*vstar(i,j-1)-FLOAT(dy)*(pstar(i+1,j)-pstar(i,j))+(1-alpha_u)*avP(i,j)*vold(i,j))/avP(i,j)
END
END
END
! vstar has been calculated, move to next step

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!! STEP 2 CALCULATE PRESSURE CORRECTION

DO i=2, NX+1
DO j=2, NY+1
! Boundary coefficient
apE(i,j) = dU(i,j)*dy
apW(i,j) = dU(i-1,j)*dy
apN(i,j) = dV(i,j)*dx
apS(i,j) = dV(i,j-1)*dx
! Central coefficient
apP(i,j) = apE(i,j)+apW(i,j)+apN(i,j)+apS(i,j)
! RHS value
bpP(i,j)= (ustar(i-1,j)-ustar(i,j))*dy+(vstar(i,j-1)-vstar(i,j))*dx
END
END
!! APPLY BOUNDARY CONDITION FOR PRESSURE
IF (n==1 .AND. i=2 .AND. j=2) THEN
apP(i,j)=1.0
apE(i,j)=0
apW(i,j)=0
apN(i,j)=0
apS(i,j)=0
bpP(i,j)=0
END
!!Calculate pprime
pprime(:,:)=0
DO iter=maxNMiter
DO i=2, NX+1
DO j=2, NY+1
pprime(i,j)=
pprime(i,j)=(apE(i,j)*pprime(i+1,j)+apW(i,j)*pprime(i-1,j)+apN(i,j)*pprime(i,j+1)+apS(i,j)*pprime(i,j-1)+bpP(i,j)/apP(i,j)
END
END
END
!!!!!!!!!! pprime has been calculated move to next section
!!!!!!!!!! STEP 3 CALCULATE VELOCITY CORRECTION
!! u-velocity correction
DO i=2, NX
DO j=2, NY+1
uprime(i,j)=dU(i,j)*(pprime(i,j)-pprime(i+1,j))
END
END
!! v-velocity correction 
DO i=2, NX+1
DO j=2, NY
vprime(i,j)=dV(i,j)*(pprime(i,j)-pprime(i,j+1))
END
END
!!!!!! STEP 3 FINISH MOVE TO STEP 4


!!!!!! STEP 4 CALCULATE REAL u,v, and p
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!! CALCULATE REAL u
Do i=2, NX
Do j=2, NY+1
u(i,j)=ustar(i,j)+uprime(i,j)
END
END
!! CALCULATE REAL v
Do i=2, NX+1
Do j=2, NY
v(i,j)=vstar(i,j)+vprime(i,j)
END
END
!! CALCULATE REAL p
Do i=2, NX+1
Do j=2, NY+1
p(i,j)=pstar(i,j)+pprime(i,j)*alpha_p
END
END
!!!!!!! STEP 4 FINISH !!!!!!!!!!

!!!!!!! STEP 5 ITERATION UNTIL CONVERGENCE
!!!!!!! 
ux=u(1:NX+1, 1:NY+1)
uxold=uold(1:NX+1, 1:NY+1)
vx=v(1:NX+1, 1:NY+1)
vxold=vold(1:NX+1, 1:NY+1)
cmax1=MAX(MAX(ux-uxold))
cmax2=MAX(MAX(vx-vxold))
convergence=MAX(cmax1, cmax2)
IF convergence<err THEN
STOP
ELSE 
uold=u
vold=v
pstar=p
!!!!!!!! CONVERGENCE CRITERIA IS SET
!!!!!!!! STEP 6 PROBLEM SET UP 

!!!!!! APPLY INLET VELOCITY FUNCTION
DO j=NY+1, 2, -1
Y(1,j)=(1, H:-dy:0)
IF Y(1,j)<Ho THEN
uLeft(1,j)=4*umax*(Y(i,j)/Ho)*(1-(Y(1,j)/Ho))
vLeft(1,j)=0
ELSE IF Y(1,j)>Ho THEN
uLeft(1,j)=0
vLeft(1,j)=0
END
END
!!!!!! APPLY BOUNDARY CONDITION FOR VELOCITY
uold(:,1)=0.0 !bottom bound
vold(:,1)=0.0
uOld(:,Ny+2) = 0.0 
vOld(:,Ny+1) = 0.0
uOld(1,:) = uLeft(1,:)
vOld(:,1) = vLeft(:,1)
uOld(Nx+1,:) = uOld(Nx,:)
vOld(Nx+2,:) = vOld(Nx+1,:)
END
END
END
END
END PROGRAM BACKWARD

Steve_Lionel · ‎03-16-2018

avE_h, for example, is a REAL(8) two-dimensional array. You have:

avE_h=(-vCe, (vDx-0.5*vCe), 0)

The syntax of values in parentheses separated by commas looks like a complex constant, but there are three values, so that doesn't fit. There's no other Fortran syntax that matches this.

What do you want this code to do? Were you trying for an array constructor?

jimdempseyatthecove · ‎03-16-2018

In looking at the rest of the code, I think there is a missing MAX or MIN to produce a scalar. But his is a speculation on my part.

IOW: What do you want this code to do?

Jim Dempsey

mecej4 · ‎03-16-2018

The code is not yet Fortran. For example, it has a number of places where an apparent DO block is terminated by END instead of END DO. Similarly, IF blocks have no matching ENDIF or ELSE. The lack of indentation makes the code hard to follow. Note that, as far as the compiler is concerned, an END statement terminates the program (on line 134, and many more times again!). Since that is not followed by a SUBROUTINE, FUNCTION or MODULE statement, everything after line 134 is invalid.

Line 122 starts with <s>, which is not meaningful in Fortran. Line 125 ends with </s>. Did you, perhaps, copy and paste the Fortran source code from a Web page? That might also explain why the parenthesis around the conditional clauses of IF statements were lost.