Re: converting a 700,000+ line Fortran 77 plus 50,000+ line C++ program to C++, part 2

On Wed, 16 Apr 2025 07:44:08 +0000, Lawrence D'Oliveiro wrote:

On Tue, 15 Apr 2025 18:28:31 -0500, Lynn McGuire wrote:
 

On 4/15/2025 6:14 PM, Lawrence D'Oliveiro wrote:

>
On Mon, 14 Apr 2025 23:50:32 -0500, Lynn McGuire wrote:
 

Got rid of a few nasty bugs like:
>
        iword = 6Habcdef

 
Surely whether that’s a bug or not would depend on the type of “iword”
...

 
iword is a implicit 4 byte integer capable of storing 4 characters.

 
I thought you got rid of all the implicit typing.

Implicit typing has nothing do with numeric storage size.

program foo
   use iso_fortran_env, only : numeric_storage_size
   integer :: j = 0
   i = 6Habcdef    ! i has an implicit type of default integer kind
   j = 6Habcdef    ! j has an explicit type of default integer kind
   print *, i, j
   print *, numeric_storage_size
end program foo

% gfcx -std=legacy -o z a.f90
% ./z
  1684234849  1684234849
          32

Many (most? all?) systems today likely have a 4-byte internal
representation for a default integer.  The Fortran standard simply
states that 'i' and 'j' occupy one numeric storage unit.  Here,
the size of that unit is 32 bits.

PS: The gfortran's -Wall option reports a few warnings with
the above code.

% gfcx -std=legacy -o z -Wall a.f90
a.f90:26:7:

   26 |    i = 6Habcdef
      |       1
Warning: Conversion from HOLLERITH to INTEGER(4) at (1)
[-Wconversion]
a.f90:26:7:

   26 |    i = 6Habcdef
      |       1
Warning: The Hollerith constant at (1) is truncated in
conversion to 'INTEGER(4' [-Wcharacter-truncation]

and similar warnings for 'j'.

--
steve