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'.