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   comp.lang.fortran      Putting John Backus on a giant pedestal      5,127 messages   

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   Message 4,699 of 5,127   
   Steven G. Kargl to Lynn McGuire   
   Re: Is there a way in Fortran to designa   
   03 Oct 24 15:02:45   
   
   From: sgk@REMOVEtroutmask.apl.washington.edu   
      
   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   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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