Re: [Caml-list] Suppress warning from within Ocaml source file

[Hendrik Boom <hendrik@topoi.pooq.com>]

On Thu, Nov 12, 2015 at 08:21:46PM +0100, Gerd Stolpmann wrote:
> Am Donnerstag, den 12.11.2015, 19:39 +0100 schrieb Virgile Prevosto:
> > 2015-11-12 18:26 GMT+01:00 Daniel Bünzli <daniel.buenzli@erratique.ch>:
> > > Le jeudi, 12 novembre 2015 à 16:37, Leo White a écrit :
> > >> > This misses the point. By using `with` I can instruct the compiler that I don't want my program not to compile if I add or remove fields. In the current system I can do this with any update I want to make to a record except if for some strange reason that update happen to touch all the current fields of the record.
> > >>
> > >> I don't particularly wish to argue for or against using this warning, but I think it is worth noting
> > >> that this is not really any different from:
> > >
> > > […]
> > >
> > > It's quite different: you have dead code in your example.
> > >
> > 
> > But in the record case, the expr in { expr with ... } will indeed be
> > evaluated (thus is not dead), but its result will never be used, which
> > is barely better. In the (admittedly contrived) example below,
> > investigating the cause of the warning would save you from a runtime
> > exception:
> > 
> > type foo = { a : int; }
> > let rec x = lazy { a = (Lazy.force x).a }
> > let y = { (Lazy.force x) with a = 42 }
> > 
> > Now that I think of it, it is more like warning 26, with the exception
> > that in the present case, the expression is usually a variable and
> > thus already evaluated.
> 
> Maybe print the warning only in the non-variable case? And rephrase the
> warning to "the expression before 'with' is not evaluated"?

Isn't this message also misleading?  Isn't the expression evaluated even 
though its value ignored?  This could show up f it has side effects.  

-- hendr9ik