Fortran: specification statement cannot appear in the executable section

In summary: C(NX,3) = DY(NX-1)/DX(NX-1) IF (NX .EQ. 2) GO TO 20 DO 18 J=2,NXM1 C(NX-J,3) = C(NX-J,2)-C(NX-J,1)*C(NX-J+1,3) 18 CONTINUE 20 C(NX,3) = DY(NX)/DX(NX)C SOLVE FOR THE QUADRATIC COEFFICIENTS 30 BPAR(3) = C(NX,3)*HALF IF (NX .EQ. 1) GO TO 40
  • #1
Peptid
4
0
Hi guys,

I am working on a Fortran code for a while. Might seem ridiculous but to better trace it, I split code into three separate parts and trying to compile both using the command 'ifort ss.f singpot.f SNGPOT.f -o ss'. With this command I receive the error

singpot.f(22): error #6236: A specification statement cannot appear in the executable section.
BLOCK DATA SPOT
------^
compilation aborted for singpot.f (code 1). Does anyone has an idea why would that happen?

By the way, since I still couldn't go further the compilation step, I am still not sure if the program works just in the way that I suppose.

This is singpot.f COMMON/VSNGLT/RNODE(88),VNODE(88)
COMMON/VSPLIN/BPAR(4),SPLC(87,3)

C
DO 1200 I=1,4
1200 BPAR(I)=0.d0
CONTINUE
CALL ICSICU(RNODE,VNODE,88,BPAR,SPLC,87,IER)
C
BLOCK DATA SPOT
C implicit real*8 (a-h,o-z)
c
C common/vsnglt/rnode(88),vnode(88)
data(rnode(i),i=1,88)/0.d0,0.4d0,0.45d0,0.5d0,
1 0.55d0,0.60d0,0.65d0,0.7d0,0.75d0,0.8d0,0.9d0,1.0d0,
2 1.1d0,1.2d0,1.3d0,1.35d0,1.39d0,1.4d0,1.401d0,1.4011d0,
31.41d0,1.45d0,1.5d0,1.6d0,1.7d0,1.8d0,1.9d0,2.0d0,
42.1d0,2.2d0,2.3d0,2.4d0,2.5d0,2.6d0,2.7d0,2.8d0,2.9d0,3.0d0,
53.1d0,3.2d0,3.3d0,3.4d0,3.5d0,3.6d0,3.7d0,3.8d0,3.9d0,4.0d0,
64.1d0,4.2d0,4.3d0,4.4d0,4.5d0,4.6d0,4.7d0,4.8d0,4.9d0,5.0d0,
75.1d0,5.2d0,5.3d0,5.4d0,5.5d0,5.6d0,5.7d0,5.8d0,5.9d0,6.0d0,
86.1d0,6.2d0,6.3d0,6.4d0,6.5d0,6.6d0,6.7d0,6.8d0,6.9d0,7.0d0,
97.2d0,7.4d0,7.6d0,7.8d0,8.d0,8.25d0,8.5d0,9.d0,9.5d0,10.d0/
data(vnode(i),i=1,77)/2.5d0,
10.8797972d0,0.6490718d0,0.4733730d0,0.3372293d0,
20.2303659d0,0.1456386d0,0.0779739d0,0.0236643d0,
3-0.0200556d0,-0.0836422d0,-0.1245385d0,-0.1500562d0,
4-0.1649342d0,-0.1723459d0,-0.1739627d0,-0.1744517d0,
5-0.1744744d0,-0.1744746d0,-0.1744746d0,-0.1744599d0,
6-0.1740558d0,-0.1728537d0,-0.1685799d0,-0.1624570d0,
7-0.1550670d0,-0.1468496d0,-0.1381312d0,-0.1291562d0,
8-0.1201233d0,-0.1111725d0,-0.1024127d0,-0.0939273d0,
9-0.0857810d0,-0.0780164d0,-0.0706700d0,-0.0637641d0,
#-0.0573118d0,-0.0513185d0,-0.0457832d0,-0.0407003d0,
#-0.0360578d0,-0.0318402d0,-0.0280272d0,-0.0245978d0,
#-0.0215297d0,-0.0187967d0,-0.0163689d0,-0.0142247d0,
#-0.0123371d0,-0.0106810d0,-0.0092303d0,-0.0079682d0,
#-0.0068703d0,-0.0059178d0,-0.0050923d0,-0.0043782d0,
#-0.0037626d0,-0.0032309d0,-0.0027740d0,-0.0023800d0,
#-0.0020423d0,-0.0017521d0,-0.0015030d0,-0.0012899d0,
#-0.0011069d0,-0.0009498d0,-0.0008150d0,-0.0007002d0,
#-0.0006030d0,-0.0005162d0,-0.0004466d0,-0.0003864d0,
#-0.0003328d0,-0.0002906d0,-0.0002466d0,-0.0002154d0/
data (vnode(i),i=78,88)/-0.001889d0,-0.0001434d0,
1-0.0001086d0,-0.0000868d0,-0.0000682d0,-0.0000528d0,
2-0.0000404d0,-0.0000314d0,-0.0000185d0,-0.0000121d0,
3-0.0000091d0/
End

subroutine ICSICU (X,Y,NX,BPAR,C,IC,IER)
INTEGER NX,IC,IER
real*8 X(NX),Y(NX),BPAR(4),C(IC,3)
INTEGER I,J,NXM1
real*8 DX,DXJ,DXJP1,DXP,DYJ,DYJP1,HALF,ONE,PJ,
1 SIX,SIXI,TWO,YPPA,YPPB,ZERO
EQUIVALENCE (DXJ,YPPB),(PJ,SIXI),(DXJP1,YPPA)
DATA ZERO/0.0d0/,HALF/0.5d0/,ONE/1.0d0/,
1 TWO/2.0d0/,SIX/6.0d0/
C FIRST EXECUTABLE STATEMENT
IER = 0
C CHECK ERROR CONDITIONS
NXM1 = NX-1
IF (IC .LT. NXM1) GO TO 30
IF (NX .LT. 2) GO TO 35
IF (NX .EQ. 2) GO TO 10
C COMPUTE COEFFICIENTS AND RIGHT
C HAND SIDE OF THE TRIDIAGONAL
C SYSTEM DEFINING THE SECOND
C DERIVATIVES OF THE SPLINE
C INTERPOLANT FOR (X,Y)
C C(J,1) = LAMBDA(J)
C C(J,2) = MU(J)
C C(J,3) = D(J)
DXJ = X(2)-X(1)
IF (DXJ .LE. ZERO) GO TO 40
DYJ = Y(2)-Y(1)
DO 5 J=2,NXM1
DXJP1 = X(J+1)-X(J)
IF (DXJP1 .LE. ZERO) GO TO 40
DYJP1 = Y(J+1)-Y(J)
DXP = DXJ+DXJP1
C(J,1) = DXJP1/DXP
C(J,2) = ONE-C(J,1)
C(J,3) = SIX*(DYJP1/DXJP1-DYJ/DXJ)/DXP
DXJ = DXJP1
DYJ = DYJP1
5 CONTINUE
C FACTOR THE TRIDIAGONAL MATRIX
C AND SOLVE FOR U
C C(J,2) = U(J)
C C(J,1) = Q(J)
C BPAR(1) = LAMBDA(1)
C BPAR(2) = D(1)
C BPAR(3) = MU(NX)
C BPAR(4) = D(NX)
10 C(1,1) = -BPAR(1)*HALF
C(1,2) = BPAR(2)*HALF
IF (NX .EQ. 2) GO TO 20
DO 15 J=2,NXM1
PJ = C(J,2)*C(J-1,1)+TWO
C(J,1) = -C(J,1)/PJ
C(J,2) = (C(J,3)-C(J,2)*C(J-1,2))/PJ
15 CONTINUE
C SOLVE FOR CUBIC COEFFICIENTS
C OF SPLINE INTERPOLANT
C C(J,1), C(J,2), AND C(J,3)
20 YPPB = (BPAR(4)-BPAR(3)*C(NXM1,2))/(BPAR(3)*C(NXM1,1)+TWO)
SIXI = ONE/SIX
DO 25 I=1,NXM1
J = NX-I
YPPA = C(J,1)*YPPB+C(J,2)
DX = X(J+1)-X(J)
C(J,3) = SIXI*(YPPB-YPPA)/DX
C(J,2) = HALF*YPPA
C(J,1) = (Y(J+1)-Y(J))/DX-(C(J,2)+C(J,3)*DX)*DX
YPPB = YPPA
25 CONTINUE
GO TO 9005
30 IER = 129
GO TO 9000
35 IER = 130
GO TO 9000
40 IER = 131
9000 CONTINUE
write(6,29888) ier
29888 format(1x,'imsl err.from ICSICU. ier=',i4)
9005 RETURN
END

subroutine ICSEVU (X,Y,NX,C,IC,U,S,M,IER)
INTEGER NX,IC,M,IER
real*8 X(NX),Y(NX),C(IC,3),U(M),S(M)
INTEGER I,JER,KER,NXM1,K
real*8 D,DD,ZERO
DATA I/1/,ZERO/0.0d0/
C FIRST EXECUTABLE STATEMENT
JER = 0
KER = 0
IF (M .LE. 0) GO TO 9005
NXM1 = NX-1
IF (I .GT. NXM1) I = 1
C EVALUATE SPLINE AT M POINTS
DO 40 K=1,M
C FIND THE PROPER INTERVAL
D = U(K)-X(I)
IF (D) 5,25,15
5 IF (I .EQ. 1) GO TO 30
I = I-1
D = U(K)-X(I)
IF (D) 5,25,20
10 I = I+1
D = DD
15 IF (I .GE. NX) GO TO 35
DD = U(K)-X(I+1)
IF (DD .GE. ZERO) GO TO 10
IF (D .EQ. ZERO) GO TO 25
C PERFORM EVALUATION
20 S(K) = ((C(I,3)*D+C(I,2))*D+C(I,1))*D+Y(I)
GO TO 40
25 S(K) = Y(I)
GO TO 40
C WARNING - U(I) .LT. X(1)
30 JER = 33
GO TO 20
C IF U(I) .GT. X(NX) - WARNING
35 IF (DD .GT. ZERO) KER = 34
D = U(K)-X(NXM1)
I = NXM1
GO TO 20
40 CONTINUE
IER = MAX0(JER,KER)
9000 CONTINUE
if(jer.gt.0) then
write(6,29666) jer
29666 format(1x,'err.mess. from ICSEVU jer=',i4)
end if
if(ker.gt.0) then
write(6,29777) ker
29777 format(1x,'err.mess. from ICSEVU ker=',i4)
end if
9005 RETURN
END

Here is ss.f

C TRYING SNGPOT
COMMON/VSNGLT/RNODE(88),VNODE(88)
COMMON/VSPLIN/BPAR(4),SPLC(87,3)
REAL B
B=SNGPOT(1)
PRINT *,B
END

and finally SNGPOT.f

FUNCTION SNGPOT(X)
IMPLICIT REAL*8 (a-h,o-z)
c
COMMON/VSNGLT/RNODE(88),VNODE(88)
COMMON/VSPLIN/BPAR(4),SPLC(87,3)
DIMENSION Y(1),VY(1)
C
Y(1)=X
CALL ICSEVU(RNODE,VNODE,88,SPLC,87,Y,VY,1,IER)
SNGPOT=VY(1)
RETURN
END

Thank you in advance.
 
Last edited:
Technology news on Phys.org
  • #2
SNGPOT.f starts with a line that starts with FUNCTION and ends with one that says END...that is fine; every file needs to start and end with that, or SUBROUTINE and END. Only one file, the one with the main program is to start with PROGRAM

...as it is, 2 of your files ( signpot.f and ss.f ) start with nothing at all, which implicitly means that it is the main procedure and equivalent to PROGRAM...in other words, you have two PROGRAMs and that cannot be...there should only one entry point to your program.
 
  • #3
Oh, and yes...the BLOCK needs to be at the beginning right after declarations...you cannot have it after executable statements like that DO loop.
 
  • #4
dear gsal, I have made several changes according to your advice. here is the final result;

C SINGLET POTENTIAL OF H2
C by spline interpolation of Kolos-Woniewicz data
C
C Before calling SNGPOT we have to
C INITIALIZE FOR SPLINE INTERPOLATION (THIS IS USED
C ONLY IF KOLOS-WOLNIEWICZ DATA IS USED)
C calling program should have the following common blocks
COMMON/VSNGLT/RNODE(88),VNODE(88)
COMMON/VSPLIN/BPAR(4),SPLC(87,3)
REAL SONUC,X
BLOCK DATA SPOT
c real*8 (a-h,o-z)
c
c common/vsnglt/rnode(88),vnode(88)
data(rnode(i),i=1,88)/0.d0,0.4d0,0.45d0,0.5d0,
1 0.55d0,0.60d0,0.65d0,0.7d0,0.75d0,0.8d0,0.9d0,1.0d0,
2 1.1d0,1.2d0,1.3d0,1.35d0,1.39d0,1.4d0,1.401d0,1.4011d0,
3 31.41d0,1.45d0,1.5d0,1.6d0,1.7d0,1.8d0,1.9d0,2.0d0,
4 42.1d0,2.2d0,2.3d0,2.4d0,2.5d0,2.6d0,2.7d0,2.8d0,2.9d0,3.0d0,
5 53.1d0,3.2d0,3.3d0,3.4d0,3.5d0,3.6d0,3.7d0,3.8d0,3.9d0,4.0d0,
6 64.1d0,4.2d0,4.3d0,4.4d0,4.5d0,4.6d0,4.7d0,4.8d0,4.9d0,5.0d0,
7 75.1d0,5.2d0,5.3d0,5.4d0,5.5d0,5.6d0,5.7d0,5.8d0,5.9d0,6.0d0,
8 86.1d0,6.2d0,6.3d0,6.4d0,6.5d0,6.6d0,6.7d0,6.8d0,6.9d0,7.0d0,
9 97.2d0,7.4d0,7.6d0,7.8d0,8.d0,8.25d0,8.5d0,9.d0,9.5d0,10.d0/
data(vnode(i),i=1,77)/2.5d0,
10.8797972d0,0.6490718d0,0.4733730d0,0.3372293d0,
20.2303659d0,0.1456386d0,0.0779739d0,0.0236643d0,
3-0.0200556d0,-0.0836422d0,-0.1245385d0,-0.1500562d0,
4-0.1649342d0,-0.1723459d0,-0.1739627d0,-0.1744517d0,
5-0.1744744d0,-0.1744746d0,-0.1744746d0,-0.1744599d0,
6-0.1740558d0,-0.1728537d0,-0.1685799d0,-0.1624570d0,
7-0.1550670d0,-0.1468496d0,-0.1381312d0,-0.1291562d0,
8-0.1201233d0,-0.1111725d0,-0.1024127d0,-0.0939273d0,
9-0.0857810d0,-0.0780164d0,-0.0706700d0,-0.0637641d0,
#-0.0573118d0,-0.0513185d0,-0.0457832d0,-0.0407003d0,
#-0.0360578d0,-0.0318402d0,-0.0280272d0,-0.0245978d0,
#-0.0215297d0,-0.0187967d0,-0.0163689d0,-0.0142247d0,
#-0.0123371d0,-0.0106810d0,-0.0092303d0,-0.0079682d0,
#-0.0068703d0,-0.0059178d0,-0.0050923d0,-0.0043782d0,
#-0.0037626d0,-0.0032309d0,-0.0027740d0,-0.0023800d0,
#-0.0020423d0,-0.0017521d0,-0.0015030d0,-0.0012899d0,
#-0.0011069d0,-0.0009498d0,-0.0008150d0,-0.0007002d0,
#-0.0006030d0,-0.0005162d0,-0.0004466d0,-0.0003864d0,
#-0.0003328d0,-0.0002906d0,-0.0002466d0,-0.0002154d0/
data (vnode(i),i=78,88)/-0.001889d0,-0.0001434d0,
1-0.0001086d0,-0.0000868d0,-0.0000682d0,-0.0000528d0,
2-0.0000404d0,-0.0000314d0,-0.0000185d0,-0.0000121d0,
3-0.0000091d0/
End

READ *, XC
C DO 1200 I=1,4
C 1200 BPAR(I)=0.d0
CALL ICSICU(RNODE,VNODE,88,BPAR,SPLC,87,IER)
C
SONUC=SNGPOT(X)

subroutine ICSICU (X,Y,NX,BPAR,C,IC,IER)
INTEGER NX,IC,IER
real*8 X(NX),Y(NX),BPAR(4),C(IC,3)
INTEGER I,J,NXM1
real*8 DX,DXJ,DXJP1,DXP,DYJ,DYJP1,HALF,ONE,PJ,
1 SIX,SIXI,TWO,YPPA,YPPB,ZERO
EQUIVALENCE (DXJ,YPPB),(PJ,SIXI),(DXJP1,YPPA)
DATA ZERO/0.0d0/,HALF/0.5d0/,ONE/1.0d0/,
1 TWO/2.0d0/,SIX/6.0d0/
C FIRST EXECUTABLE STATEMENT
IER = 0
C CHECK ERROR CONDITIONS
NXM1 = NX-1
IF (IC .LT. NXM1) GO TO 30
IF (NX .LT. 2) GO TO 35
IF (NX .EQ. 2) GO TO 10
C COMPUTE COEFFICIENTS AND RIGHT
C HAND SIDE OF THE TRIDIAGONAL
C SYSTEM DEFINING THE SECOND
C DERIVATIVES OF THE SPLINE
C INTERPOLANT FOR (X,Y)
C C(J,1) = LAMBDA(J)
C C(J,2) = MU(J)
C C(J,3) = D(J)
DXJ = X(2)-X(1)
IF (DXJ .LE. ZERO) GO TO 40
DYJ = Y(2)-Y(1)
DO 5 J=2,NXM1
DXJP1 = X(J+1)-X(J)
IF (DXJP1 .LE. ZERO) GO TO 40
DYJP1 = Y(J+1)-Y(J)
DXP = DXJ+DXJP1
C(J,1) = DXJP1/DXP
C(J,2) = ONE-C(J,1)
C(J,3) = SIX*(DYJP1/DXJP1-DYJ/DXJ)/DXP
DXJ = DXJP1
DYJ = DYJP1
5 CONTINUE
C FACTOR THE TRIDIAGONAL MATRIX
C AND SOLVE FOR U
C C(J,2) = U(J)
C C(J,1) = Q(J)
C BPAR(1) = LAMBDA(1)
C BPAR(2) = D(1)
C BPAR(3) = MU(NX)
C BPAR(4) = D(NX)
10 C(1,1) = -BPAR(1)*HALF
C(1,2) = BPAR(2)*HALF
IF (NX .EQ. 2) GO TO 20
DO 15 J=2,NXM1
PJ = C(J,2)*C(J-1,1)+TWO
C(J,1) = -C(J,1)/PJ
C(J,2) = (C(J,3)-C(J,2)*C(J-1,2))/PJ
15 CONTINUE
C SOLVE FOR CUBIC COEFFICIENTS
C OF SPLINE INTERPOLANT
C C(J,1), C(J,2), AND C(J,3)
20 YPPB = (BPAR(4)-BPAR(3)*C(NXM1,2))/(BPAR(3)*C(NXM1,1)+TWO)
SIXI = ONE/SIX
DO 25 I=1,NXM1
J = NX-I
YPPA = C(J,1)*YPPB+C(J,2)
DX = X(J+1)-X(J)
C(J,3) = SIXI*(YPPB-YPPA)/DX
C(J,2) = HALF*YPPA
C(J,1) = (Y(J+1)-Y(J))/DX-(C(J,2)+C(J,3)*DX)*DX
YPPB = YPPA
25 CONTINUE
GO TO 9005
30 IER = 129
GO TO 9000
35 IER = 130
GO TO 9000
40 IER = 131
9000 CONTINUE
write(6,29888) ier
29888 format(1x,'imsl err.from ICSICU. ier=',i4)
9005 RETURN
END

subroutine ICSEVU (X,Y,NX,C,IC,U,S,M,IER)
INTEGER NX,IC,M,IER
real*8 X(NX),Y(NX),C(IC,3),U(M),S(M)
INTEGER I,JER,KER,NXM1,K
real*8 D,DD,ZERO
DATA I/1/,ZERO/0.0d0/
C FIRST EXECUTABLE STATEMENT
JER = 0
KER = 0
IF (M .LE. 0) GO TO 9005
NXM1 = NX-1
IF (I .GT. NXM1) I = 1
C EVALUATE SPLINE AT M POINTS
DO 40 K=1,M
C FIND THE PROPER INTERVAL
D = U(K)-X(I)
IF (D) 5,25,15
5 IF (I .EQ. 1) GO TO 30
I = I-1
D = U(K)-X(I)
IF (D) 5,25,20
10 I = I+1
D = DD
15 IF (I .GE. NX) GO TO 35
DD = U(K)-X(I+1)
IF (DD .GE. ZERO) GO TO 10
IF (D .EQ. ZERO) GO TO 25
C PERFORM EVALUATION
20 S(K) = ((C(I,3)*D+C(I,2))*D+C(I,1))*D+Y(I)
GO TO 40
25 S(K) = Y(I)
GO TO 40
C WARNING - U(I) .LT. X(1)
30 JER = 33
GO TO 20
C IF U(I) .GT. X(NX) - WARNING
35 IF (DD .GT. ZERO) KER = 34
D = U(K)-X(NXM1)
I = NXM1
GO TO 20
40 CONTINUE
IER = MAX0(JER,KER)
9000 CONTINUE
if(jer.gt.0) then
write(6,29666) jer
29666 format(1x,'err.mess. from ICSEVU jer=',i4)
end if
if(ker.gt.0) then
write(6,29777) ker
29777 format(1x,'err.mess. from ICSEVU ker=',i4)
end if
9005 RETURN
END

and for the function SNGPOT.f

FUNCTION SNGPOT(X)
implicit real*8 (a-h,o-z)
c
COMMON/VSNGLT/RNODE(88),VNODE(88)
COMMON/VSPLIN/BPAR(4),SPLC(87,3)
DIMENSION Y(1),VY(1)
C
Y(1)=X
CALL ICSEVU(RNODE,VNODE,88,SPLC,87,Y,VY,1,IER)
SNGPOT=VY(1)
RETURN
END

I placed block data right after the declarations, but I still I get the error;

sngdeneme.f:8.72:

COMMON/VSNGLT/RNODE(88),VNODE(88)
1
sngdeneme.f:50.72:

REAL SONUC,X
2
Error: Two main PROGRAMs at (1) and (2)

Moreover, 50th line is only an executable, so I don't understand why it stands for a new main program.
 
Last edited:
  • #5

It appears that the error is occurring because you are trying to use a specification statement (BLOCK DATA) in the executable section of your code. In Fortran, specification statements must be placed before the executable section, as they define variables and other elements that will be used in the code. To fix this error, move the BLOCK DATA statement to the top of your code, before any executable statements. Additionally, make sure that all specification statements are placed before any executable statements in your code. This should allow your code to compile properly.
 

Related to Fortran: specification statement cannot appear in the executable section

1. What does the error "Fortran: specification statement cannot appear in the executable section" mean?

This error means that a specification statement, which defines the characteristics of a variable or function, was placed in the executable section of the code. This is not allowed in Fortran as specification statements should only appear in the specification section.

2. What is the difference between the specification section and the executable section in Fortran?

The specification section is where declarations and definitions of variables, functions, and other entities are made, while the executable section is where the actual code is written. Specification statements should only appear in the specification section, while executable statements should only appear in the executable section.

3. How can I fix the "Fortran: specification statement cannot appear in the executable section" error?

To fix this error, you need to move the specification statement from the executable section to the specification section. This will ensure that the statement is declared and defined properly before it is used in the executable section.

4. Are there any exceptions to the rule that specification statements cannot appear in the executable section?

Yes, there are some cases where specification statements can appear in the executable section. This includes certain intrinsic functions, such as LEN and ALLOCATE, which can be used in the executable section without prior declaration in the specification section.

5. Is it necessary to strictly follow the rule of separating specification and executable statements in Fortran?

Yes, it is important to follow this rule as it helps to organize the code and make it more readable and maintainable. It also ensures that all variables and functions are properly declared and defined before being used in the executable section, reducing the chances of errors and bugs in the code.

Similar threads

Fortran Help with Fortran 77 Programming for Two-Dimensional Axisymmetric Flow Study
    • Programming and Computer Science
Replies
6
Views
3K
Fortran A specification statement cannot appear in the executable section (fortran)
    • Programming and Computer Science
Replies
11
Views
14K
Comp Sci Errors during executing the program (Compaq Visual Fortran)
    • Engineering and Comp Sci Homework Help
Replies
5
Views
1K

Hot Threads

Recent Insights

Back
Top