[mercury-users] number of args for bag.init exported to C

Julien Fischer juliensf at csse.unimelb.edu.au
Tue Jun 14 01:45:49 AEST 2011


On Thu, 9 Jun 2011, Guillaume Yziquel wrote:

> OK. I think I got this one...
> On the same topic, I believe that the C interface to gprolog allows you
> to run successively a query, hereby exhausting the answers that gprolog
> finds for you. That's quite nice. I see that there's the solutions
> predicate, but it finds everything in one go, and thus doesn't allow you
> to interleave the execution of calling code and mercury code as gprolog
> allows you to do. That would be really a nice feature to have to be able
> to integrate that with Lwt monadic green threads in OCaml. No simple way
> to do that?

It is possible to do something of the sort using the do_while/4
predicate in the standard library's solutions module.  (The basic idea
is to use the Filter argument to make calls back to the foreign language
after each solution is generated.)


> Well, I'm now doing almost fine with the following code. Almost fine
> because if I uncomment the two '%', I get the following error message:
> 	./test.native
> 	Mercury runtime: memory zone nondetstack#0 overflowed
> 	This may have been caused by a stack overflow, due to unbounded
> 	recursion.
> 	make[1]: *** [test] Erreur 1
> Here's the code:
> 	:- type prop --->
> 	        term_definition(term, term)     ;
> 	        term_typed(term, typ)           .
> 	:- func no_props = bag(prop).
> 	        no_props = bag.init.
> 	:- pragma foreign_export("C", no_props = out, "no_props").
> 	:- func define_term(term, term, bag(prop)) = bag(prop).
> 	        define_term(Term1, Term2, Props) =
> 			bag.insert(Props, term_definition(Term1, Term2)).
> 	:- pragma foreign_export("C", define_term(in, in, in) = out, "define_term").
> 	:- func type_term(term, typ, bag(prop)) = bag(prop).
> 	        type_term(Term, Type, Props) = bag.insert(Props, term_typed(Term, Type)).
> 	:- pragma foreign_export("C", type_term(in, in, in) = out, "type_term").
> 	:- pred search_type_annotation(term::in, typ::out, bag(prop)::in).
> 	        search_type_annotation(Term, Type, Props) :-
> 	                bag.member(Prop, Props, Remaining),
> 	                (       if      Prop = term_typed(Term, T)
> 	                        then    Type = T
> 	                        else    search_type_annotation(Term, Type, Remaining)
> 	                ).
> 	:- pred search_equivalent_term(term::in, term::out, bag(prop)::in).
> 	%       search_equivalent_term(Term, EqTerm, _) :-
> 	%                       Term = EqTerm.
> 	        search_equivalent_term(Term, EqTerm, Props) :-
> 	                bag.member(Prop, Props, Remaining),
> 	                (       if      Prop = term_definition(Term, Eq)
> 	                        then    EqTerm = Eq
> 	                        else    search_equivalent_term(Term, EqTerm, Remaining)
> 	                ).
> 	:- pred typecheck(term::in, typ::out, bag(prop)::in).
> 	        typecheck(Term, Type, Props) :-
> 	                search_type_annotation(Term, Type, Props).
> 	        typecheck(Term, Type, Props) :-
> 	                (       search_equivalent_term(Term, EqTerm, Props),
> 	                        typecheck(EqTerm, Type, Props)
> 	                ).
> 	:- pred typeinfer(term::in, typ::out, bag(prop)::in).
> 	:- pragma foreign_export("C", typeinfer(in, out, in), "typeinfer").
> 	        typeinfer(Term, Type, Props) :-
> 	                promise_equivalent_solutions [Type]
> 	                typecheck(Term, Type, Props).
> I do not really understand where the unbounded recursion comes from.
> When commented, the fact that the search exhausts the bag gives some
> kind of termination argument. But I do not see why the uncommented code
> loops indefinitely...

Here's one way in which the above code could go into an infinite loop
when that clause is uncommented.  Consider what happens when
typeinfer/3 is called with the Props argument set to the empty bag.

(1) We call typecheck(Term, Type, Props)
(2) The first clause of typecheck/3 will fail since the call to
     bag.member/3 in search_type_annotation/3 will fail.
(3) Trying the second clause of typecheck/3, we will first call
     search_equivalent_term/3.  The first clause (the commented out one)
     will succeed (it's deterministic), and the value returend in EqTerm
     will the same as the value Term.
(4) The recursive call to typecheck/3 has the same arguments as in (1)

The same could happen with non-empty bags if the unification "Type = T" in
search_type_annotation/3 always fails.

(The above depends on the order of the disjunctions, but I don't think
the compiler has much of a reason to reoder them in this case.)

> Another thing I do not understand is why the Mercury compiler infers the
> determinism category of det for typeinfer. As far as I can see, with
> promise_equivalent_solutions, it should be semidet as I do not see a
> guarantee that the search will yield a solution.

It infers the determinism of det for typeinfer/3 because it infers the
determinism of multi for typecheck/3.  If the solutions of a multi goal
are promised to be equivalent then it is det.

Why does it infer a determinsim of multi for typecheck/3? 
The first clause:

    typecheck(Term, Type, Props) :-
        search_type_annotation(Term, Type, Props).

is nondet, while the second:

    typecheck(Term, Type, Props) :-
        search_equivalent_term(Term, EqTerm, Props),
        typecheck(EqTerm, Type, Props)

is inferred to be multi.  The overall determinism is therefore
multi, since even if the first clause fails the second cannot fail.

Rather than relying on determinism inference I would suggest that you
include determinism annotations in your predicate declarations.  The
compiler would then be able to tell you about cases where the inferred
determinism differed from your expectations.

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