Re: Is there a way in Fortran to designate an integer value as integer*8 ?

On Thu, 03 Oct 2024 14:32:28 -0500, Lynn McGuire wrote:

On 10/3/2024 10:02 AM, Steven G. Kargl wrote:

On Thu, 03 Oct 2024 02:06:28 -0500, Lynn McGuire wrote:
 

On 10/2/2024 11:27 PM, Steven G. Kargl wrote:
 

As the person who gave gfortran the -fdefault-integer-8 option, I hope
your XXX kloc of code uses neither equivalence nor common blocks.

 
I have 197 common blocks included from dedicated files and a massive
number of equivalences all over the place.  Several of the equivalences
are actually in the common block files.  The equivalences have made the
eventual C++ conversion of the Fortran code tricky.
 
This code is 850,000 lines of F77 code and 50,000 lines of C++ code that
dates back to 1965 or so.  Half of the code is Fortran IV and half is
F77.  It has been ported to 12 ? different platforms, mostly mainframes
in the 1960s, 1970s, 1980s, and 1990s.

The gfortran's -fdefault-* options break storage association.
This may have some interesting alignment issues, which can
lead to hard to find bugs.  If you think to needed the
-fdefault-integer-8 option, then I'll kindly suggest that you
likely want to use -finteger-4-integer-8 option instead.

My personal recommendation would be to do a proper porting from
integer (aka integer*4) to integer(kind=8).  And, yes, 8 in the
'kind=8' is not portable.

--
steve