Re: remark on defining size of basic types

Michael S <already5chosen@yahoo.com> writes:

On Fri, 5 Apr 2024 07:18:09 -0000 (UTC)
Lawrence D'Oliveiro <ldo@nz.invalid> wrote:
>

On Fri, 5 Apr 2024 08:30:16 +0200, Janis Papanagnou wrote:
 

For the first question I thought you were referring to old Fortran
versions ... 

 
I did say “has”, not “had”.
 

Yes, but I read that 'KIND' specification as a readable variant of
the old syntax ... 

 
It’s parameterizable. That means it is easy to change the precision
of the type right through your code just by altering one definition.
 
Also, KIND participates in function overloading. That requires it to
be compile-time constant. There are also LEN parameters, which can be
dynamic, but can only be used to size arrays and strings.

>
I still think that Janis asked correct questions and that you answered
misleading answers.
I had never read the Fortran-2003 Standard, however I will be very
surprised if it mandates support for integer types wider than 64 bits
(or 18 decimal digits). I would not be surprised if hard mandate is
even lower, may be only 9 digits.
Also I would be surprised if there are implementations that support
integers that are wider than 128 bits (38 digits).

I think this was the source of some confusion :

    On 05.04.2024 01:38, Lawrence D'Oliveiro wrote:
    > On Thu, 4 Apr 2024 15:15:26 +0200, Janis Papanagnou wrote:
    >
    >> Sometimes it's useful to have an unbounded or parameterized integral
    >> data type available ...
    >
    > Interestingly, Fortran (of all things) has that.

As far as I can tell, Fortran has parameterized integral types, but not
unbounded integral types.

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
Working, but not speaking, for Medtronic
void Void(void) { Void(); } /* The recursive call of the void */