I spent some time yesterday looking at the fortran holdings at my nearest
university library. Among the books I leafed through was a book on the F
syntax. I used a quicksort routine that claimed F compliance, and it
worked wonderfully for me as fortran. Are the following conjectures true:

1) Legal F is also legal fortran90 and fortran95.

2) The fortran that is not F is deprecated.

3) F is in a good position to handle newer fortran features like
unlimittted polymorphic entities.

Is there any mention of F in the standard?

Hello,

On 2008-09-11 16:45:30 -0400, Ron Ford said:

Yes.

No.

I don't know why not; I've heard of no plans to do so
(not that I would).

No.

Dan Nagle wrote:
...

...

Specifically, the F dialect is a subset of Fortran (w/ some extensions
as well) by a single vendor.

--

Others have noted that the short answer to this is "no". The longer
answer is that F's authors have tried - as I understand it - to select
for this subset of F90 - or is it F95? - those features that make it a
"modern" language and to deselect, or make unavailable if you will, the
"legacy features" that are still in the language standard but that are
no longer required, because better (for some measure of "better")
methods exist to achieve the same goal. For instance, explicit
interfaces using modules are always required, all forms of type punning
are disabled, and so on.

While I agree with the general principle, and find it a worthwhile
exercise, I do remember from the past that, IMO, the definition is a bit
too restrictive: some features have been removed that make life
unnecessarily difficult in certain situations, with little gain from the
point of view of programming safety and security.

In any case, you could try the F mode of, well - gfortran or g95? I
don't remember exactly.

Jan