Op 18-02-13 14:00, Willem van Asperen
schreef:
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?
I think so too.
It looks like the : operator has a higher precedence then + (and
therefor >= too), which is a bit surprising indeed.
Personally, I never mix the : operator with other operators, to
avoid having to worry about the precendence.
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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