[mercury-users] Deterministic or not? (PART 2)

[David Overton]

On Mon, Apr 19, 2004 at 07:57:21PM +0100, Sergio Rafael Trindade Marques wrote:
> Hi,
> I'm having a little problem with this code:
> 
> 
> /:- pred get_ys(list(string),list(string)).
> :- mode get_ys(in,out) is nondet.

This should be `is det'.

> 
> get_ys(Vars,R):-
>    (Vars=[],
>        R=[]
>    ;Vars=[E|Es],
>        count_element(E,Es,0,N),
>        (N=0,
>            get_ys(Es,R)
>        ;N\=0,
>            R=[E|Rs],
>            get_ys(Es,Rs)
>        )
>    )./

The Mercury determinism analysis algorithm is not able to infer that the
inner disjunction will not fail, and that it can only succeed once.
Instead you need to use an if-then-else:

	( if N = 0 then
		get_ys(Es, R)
	else
		R = [E | Rs],
		get_ys(Es, Rs)
	)

> 
> /:- pred count_element(string,list(string),int,int).
> :- mode count_element(in,in,in,out) is nondet.
> 
> count_element(E,Elements,NI,NF):-
>    (Elements=[],
>        NF=NI
>    ;Elements=[E1|Es],
>        (E=E1,
>            N1 is NI+1,
>            count_element(E,Es,N1,NF)
>        ;E\=E1,
>            count_element(E,Es,NI,NF)
>        )
>    ).

You will need to make the same change here.

> 
> /And it's returning the following error:
> 
> / In `get_pred(in, in, out)':
> comp.m:106:   error: determinism declaration not satisfied.
> comp.m:106:   Declared `det', inferred `nondet'.
> comp.m:135:   call to `get_ys(in, out)' can fail.
> comp.m:135:   call to `get_ys(in, out)' can succeed more than once.
> 
> /I have a predicate called get_pred that it's defined as det.
> How can i resolve the previous error and after that how can i call the 
> predicate get_ys in get_pred?
> 
> Thanks in advance,
> Sergio Marques
> 
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-- 
David Overton
WWW: http://www.overtons.id.au/
Mobile Phone (UK): +44 7799 344 322
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