[mercury-users] Newbie problem. :)
Fergus Henderson
fjh at cs.mu.OZ.AU
Sat Jun 12 19:30:40 AEST 1999
On 12-Jun-1999, Rob <zharradan at suffering.org> wrote:
> my_insert(X, Y, Z) :-
> (
> Y = [Hd | Tl],
> (
> X =< Hd, <-- Line 40
> Z = [X | [Hd | Tl]]
> ;
> X > Hd,
> my_insert(X, Tl, W),
> Z = [Hd | W]
> )
> ;
> Y = [],
> Z = [X]
> ).
...
> blah.m:018: In `my_insert(in, in, out)':
> blah.m:018: error: determinism declaration not satisfied.
> blah.m:018: Declared `det', inferred `nondet'.
> blah.m:040: call to `=<(in, in)' can fail.
> blah.m:043: call to `>(in, in)' can fail.
> blah.m:042: Disjunction has multiple clauses with solutions.
>
> I am obviously missing something here, but I thought this was the way
> things were supposed to work - part of the disjunction fails, and it
> checks against the rest..
>
> Help. :)
The compiler's determinism analysis follows a fairly simple algorithm;
this algorithm is not smart enough to figure out that the two different
tests X =< Hd and X > Hd are mutually exclusive and cover all cases,
because it doesn't have any knowledge about the semantics of particular
predicates such as `=<' and `>' other than that given by their type,
mode and determinism declarations. As far as the compiler's determinism
analysis is concerned, you might equally as well have written
(
p(X, Hd),
Z = [X | [Hd | Tl]]
;
q(X, Hd),
my_insert(X, Tl, W),
Z = [Hd | W]
)
...
:- pred p(int::in, int::in) is semidet.
:- pred q(int::in, int::in) is semidet.
and so the compiler has to assume that this disjunction might fail
or might have two solutions.
The solution in cases like this is to use an if-then-else rather than
a disjunction:
(if X =< Hd then
Z = [X | [Hd | Tl]]
else /* X > Hd */
my_insert(X, Tl, W),
Z = [Hd | W]
)
P.S.
This is a quite common question; it ought to be in the FAQ list...
--
Fergus Henderson <fjh at cs.mu.oz.au> | "I have always known that the pursuit
WWW: <http://www.cs.mu.oz.au/~fjh> | of excellence is a lethal habit"
PGP: finger fjh at 128.250.37.3 | -- the last words of T. S. Garp.
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