Forums before death by AOL, social media and spammers... "We can't have nice things"
|    comp.lang.fortran    |    Putting John Backus on a giant pedestal    |    5,127 messages    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
|    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)    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
(c) 1994, bbs@darkrealms.ca