Sujet : Re: Is there a way in Fortran to designate an integer value as integer*8 ?
De : lynnmcguire5 (at) *nospam* gmail.com (Lynn McGuire)
Groupes : comp.lang.fortranDate : 03. Oct 2024, 08:06:28
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vdlfpl$3l0f5$1@dont-email.me>
References : 1 2 3 4
User-Agent : Mozilla Thunderbird
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 ?
Thanks,
Lynn