> Using the C preprocessor on COBOL code [...]

   Which "the C preprocessor" worries you?

> I'm not sure what the compiler will do [...]

   So, try it?  I don't have a COBOL compiler installed, so I know
nothing, but Fortran seems happy enough.  For example (not widely
tested):

alp $ type arch_test_c.for
      program arch_test_c

      character* 8 ARCH

#include "empty.fh"

#if defined( __alpha)
      parameter (ARCH = "Alpha")
#elif defined( __ia64)
      parameter (ARCH = "IA64")
#elif defined( __vax)
      parameter (ARCH = "VAX")
#else
      parameter (ARCH = "Unknown")
#endif

      type *, "ARCH = ", ARCH

      end
alp $

alp $ fortran arch_test_c.f
alp $ link arch_test_c
alp $ run arch_test_c
ARCH = Alpha
alp $ 

alp $ type arch_test_c.f
# 1 "ALP$DKC0:[SMS.ITRC]arch_test_c.for;2"
      program arch_test_c

      character* 8 ARCH


# 1 "ALP$DKC0:[SMS.ITRC]empty.fh;1"



# 6 "ALP$DKC0:[SMS.ITRC]arch_test_c.for;2"


      parameter (ARCH = "Alpha")








      type *, "ARCH = ", ARCH

      end
alp $ 


   I don't see "any #file or #line directives left in the output", and a
"#" in column one seems to be pretty Fortran-compatible.  If a COBOL
compiler is harder to satisfy, then additional options include writing
one's own preprocessor, and filtering the output of a sub-ideal
preprocessor.


alp $ cc /version
HP C V7.3-009 on OpenVMS Alpha V8.3

alp $ fortran /version
HP Fortran V8.2-104679-48H9K