What?int pairIndex = int(pair - &m_overlappingPairArray[0]);

[teravus]

In btOverlappingPairCache.cpp in btHashedOverlappingPairCache::removeOverlappingPair(btBroadphaseProxy* proxy0, btBroadphaseProxy* proxy1,btDispatcher* dispatcher)

Can someone explain,

int pairIndex = int(pair - &m_overlappingPairArray[0]);

It seems to do a subtraction of the btBroadPhasePair, but I didn't see any info that included a description of what ends up in pairIndex. There doesn't seem to be a subtraction operator for btBroadPhasePair and m_overlappingPairArray[0] would seem to be the address of the first element in the btAlignedObjectArray<btBroadPhasePair>.

Are we subtracting addresses now? 0.o

Is this another way of saying, 'Find this object in m_overlappingPairArray'?

Thanks

T

[rrck]

You can apply some arithmetic operation with pointers, but results is not like with numbers.

pair - overlappingPairArray[0] returns how many objects(holded by pair array) is stored between first pair and given pair. That is equal to pair index.

You can try to search on pointers arithmetics topic.

[teravus]

Much appreciated, sir. Helps me understand what's going on there.

An aligned object array.. so the memory address of the pair - the memory address of the first element is the total number of elements between pair and the first element.



So a compatible but slower algorithm would be ;

int pairIndex = m_overlappingPairArray.findLinearSearch(pair) + 1;

the array index of pair - 0 is always pair, then add 1 because the array is zero based == the number of elements between the specific element and the 0 element.

[rrck]

+1 is wrong.
linearSearch must return actual index of pair.

[teravus]

Ok, that makes sense.

Thanks.