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

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

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

On Wed, 02 Oct 2024 14:30:48 -0500, Lynn McGuire wrote:
 

On 10/2/2024 2:00 AM, Lawrence D'Oliveiro wrote:

On Tue, 1 Oct 2024 21:58:40 -0500, Lynn McGuire wrote:
>

I need many of my integers to be integer*8 in my port to 64 bit.  In
C/C++ code, I can say 123456L to mean a long long value, generally 64
bit.  Is there a corresponding way to do this in Fortran ...

>
      integer(kind = 8), parameter :: bigval = 9223372036854775807_8
      print *, bigval
>
prints
>
      9223372036854775807

>
Thanks !
>
I was afraid of that.  I will have to put _8 in about 100,000 lines of
my F77 code.  And the future conversion to C++ will need special handling.
>

 
If you 100,000 lines of C++ without a trailing 'L', you would
need to add 'L' to get a long int.  You also only need to add
'_8' (or 'L') to those values that would exceed huge(1) in
magnitude as integer*4 is a proper subset of integer*8 and
Fortran does conversion when required.

 
If Fortran does an automatic conversion from I*4 to I*8, why does the
compiler gripe at me that the integer constant does not match the
subroutine argument type ?

Well, to begin, you were talking about numeric literal constants.
I doubt you add '_8' (or 'L') to all entities declared as 'integer*4'
(or long int).

integer*8 i ! 42 is integer*4 and automatically converted to integer*8
i = 42      ! on assignment.
i = 3_8 * 2 ! Mixed-mode math.  2 is magically converted to integer*8

The compiler is not complaining.  It is informing you of an mismatch
between an actual argument and the dummy argument.  If one is 'integer*4'
and the other 'integer*8', you have 32 undefined bits.

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

--
steve