[rules-users] Easy mistake...

[Willem van Asperen]

On 02/18/2013 11:11 AM, Willem van Asperen wrote:
> Hi All,
>
> Just wanted to share with you an easy mistake. This is done running
> Drools 5.5.0.Final.
>
> Imagine a fact
>
> |Cheese (||
> ||    creationTick : double||
> ||    preservationTicks : double||
> ||)|
>
> Then the following does not do what you expect:
>
> |rule "dump old cheese"||
> ||    when||
> ||        $c : Cheese( $sbd : creationTick + preservationTicks )||
> ||        CurrentTime( tick >= $sbd )||
> ||    then||
> ||        System.out.println("||we have to dump "+$c);||
> ||end|
>
> You would expect that $c is dumped when current time has passed
> creationTick + preservationTicks. But no. The variable $sbd is bound
> to creationTick /before/ the preservationTicks are added!! I must say
> that I do not quite understand how ($sbd : creationTick) +
> preservationTicks resolves to "true" to make the premise succeed...
> Maybe because it is != 0?
>
> I found that it should be:
>
> |rule "dump old cheese"||
> ||    when||
> ||        $c : Cheese( $sbd : (creationTick + preservationTicks) )||
> ||        CurrentTime( tick >= $sbd )||
> ||    then||
> ||        System.out.println("we have to dump "+$c);||
> ||end|
>
> This makes sense. Now $sbd is bound to the result of the addition. But
> it is an easy trap!!
>
> Regards,
> Willem
Hi All,

Coming back to this... What will be evaluated in the following snippet:

rule "dump old cheese"
    when
        CurrentTime( $tick : tick )
        $c : Cheese( $tick >= creationTick + preservationTicks )
    then
        System.out.println("we need to dump "+$c);
end

I assume this evaluates if $tick is larger or equal to (creationTick +
preservationTicks), right? Otherwise it would say

step 1:    $tick > creationTick + preservationTicks
step 2:    true + preservationTicks

which would fail, obviously because you cannot add a double to a boolean.

Regards,
Willem
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