Fortran: specification statement cannot appear in the executable section

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Discussion Overview

The discussion revolves around a compilation error encountered in Fortran code, specifically related to the placement of specification statements and the structure of the program files. Participants are exploring the implications of the Fortran language rules on code organization and execution flow.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant reports an error indicating that a specification statement cannot appear in the executable section, specifically mentioning the placement of a BLOCK DATA statement.
  • Another participant notes that each Fortran file should start with either a FUNCTION or SUBROUTINE statement, and emphasizes that only one file should serve as the main program entry point.
  • A further contribution suggests that the BLOCK DATA statement should be placed at the beginning of the file, immediately following declarations, rather than after executable statements like DO loops.

Areas of Agreement / Disagreement

Participants express differing views on the organization of the Fortran code, particularly regarding the structure of the files and the placement of statements. There is no consensus on the best approach to resolve the compilation error.

Contextual Notes

Participants highlight potential issues with the organization of the Fortran files, including the presence of multiple main program entries and the placement of BLOCK DATA statements, but do not resolve these issues definitively.

Peptid
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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:
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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.
 
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.
 
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:

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