Problem compiling FORTRAN code

[FatLynn]

Hello, I received some very old cold, and I don't even know for which FORTRAN standard it was originally written, only that it has existed since at least 1993. I'm trying to compile it in gfortran, and I get the following error:

'Error: Expected a right parenthesis in expression at (1)'
1 / 1H( ,1H1 ,1HX ,1H, ,1H ,1H ,1HA ,1H ,1H ,1H) /
1

The code is shown below, and the line indicated in the error message is the fourth line of the code shown here. Anyone know why I am getting this issue?

INTEGER F(10),G(14),LUN(5)
DIMENSION MESSG(NMESSG)
DATA F(1),F(2),F(3),F(4),F(5),F(6),F(7),F(8),F(9),F(10)
1 / 1H( ,1H1 ,1HX ,1H, ,1H ,1H ,1HA ,1H ,1H ,1H) /
DATA G(1),G(2),G(3),G(4),G(5),G(6),G(7),G(8),G(9),G(10)
1 / 1H( ,1H1 ,1HX ,1H ,1H ,1H ,1H ,1H ,1H ,1H /
DATA G(11),G(12),G(13),G(14)
1 / 1H ,1H ,1H ,1H) /
DATA LA/1HA/,LCOM/1H,/,LBLANK/1H /
C PREPARE FORMAT FOR WHOLE LINES
C***FIRST EXECUTABLE STATEMENT XERPRT
NCHAR = I1MACH(6)
NFIELD = 72/NCHAR
CALL S88FMT(2,NFIELD,F(5))
CALL S88FMT(2,NCHAR,F(8))
C PREPARE FORMAT FOR LAST, PARTIAL LINE, IF NEEDED
NCHARL = NFIELD*NCHAR
NLINES = NMESSG/NCHARL
NWORD = NLINES*NFIELD
NCHREM = NMESSG - NLINES*NCHARL
IF (NCHREM.LE.0) GO TO 40
DO 10 I=4,13
10 G(I) = LBLANK
NFIELD = NCHREM/NCHAR
IF (NFIELD.LE.0) GO TO 20
C PREPARE WHOLE WORD FIELDS
G(4) = LCOM
CALL S88FMT(2,NFIELD,G(5))
G(7) = LA
CALL S88FMT(2,NCHAR,G(8))
20 CONTINUE
NCHLST = MOD(NCHREM,NCHAR)
IF (NCHLST.LE.0) GO TO 30
C PREPARE PARTIAL WORD FIELD
G(10) = LCOM
G(11) = LA
CALL S88FMT(2,NCHLST,G(12))
30 CONTINUE
40 CONTINUE
C PRINT THE MESSAGE
NWORD1 = NWORD+1
NWORD2 = (NMESSG+NCHAR-1)/NCHAR
CALL XGETUA(LUN,NUNIT)
DO 50 KUNIT = 1,NUNIT
IUNIT = LUN(KUNIT)
IF (IUNIT.EQ.0) IUNIT = I1MACH(4)
IF (NWORD.GT.0) WRITE (IUNIT,F) (MESSG(I),I=1,NWORD)
IF (NCHREM.GT.0) WRITE (IUNIT,G) (MESSG(I),I=NWORD1,NWORD2)
50 CONTINUE
RETURN
END

 

[Wrichik Basu]

I don't know Fortran, but from a general knowledge of programming, this seems to be a syntax error:

1H( ,1H1 ,1HX

You must have something after the opening bracket and before the first comma.

 

[Bill Simpson]

I believe that for some old FORTRAN compilers that is not a syntax error.

What you are seeing is called a Hollerith Constant, described here
which consist of a positive integer constant followed by H followed by
the integer constant number of characters, so 1H( is a one byte constant
containing the character (
and that would be followed by the character 1 followed by the character X

Documentation describing GNU FORTRAN support for Hollerith Constants
described here

 

[anorlunda]

What happened to the indentation?

Columns 1-5 are for statement number
C in column 1 makes the rest of the line a comment.
Non blank in column 6 means a continuation of the previous line.
Columns 7-71 are for the FORTRAN
Columns 72-80 are for the sequence number.

1 / 1H( ,1H1 ,1HX ,1H, ,1H ,1H ,1HA ,1H ,1H ,1H) /

I suspect that leading 1 is not part of the code, it should have been in column 6 to signal that line is a continuation of the previous line.I suspect that leading 1 was meant to be in column 6. It is not part of the code but a signal that the line is a continuation of the previous line.

If someone threw away the leading blanks on all lines they F*ed up the code.

 

[jedishrfu]

Yes I think @anorlunda nailed it. The program was probably written on punch cards and later saved as a file. Whoever did it, didnt realize the need to keep the columnar format of fortran.

 

[Klystron]

Also, using a modern or semi-modern compiler, best to explicitly indicate Hollerith characters.

 

[Vanadium 50]

What happened to the indentation?

Bingo. The 1 is almost certainly a continuation character, and used to be in column 6.