I used a profiler to look over some code which does not yet run fast enough. It found that the following function took most of the time, and half of the time in this function was spent in floor. Now, there are two possibilities: optimizing this function or going one level above and reducing the calls to this function. I wonder, if the first one is possible.
int Sph::gridIndex (Vector3 position) const {
int mx = ((int)floor(position.x / _gridIntervalSize) % _gridSize);
int my = ((int)floor(position.y / _gridIntervalSize) % _gridSize);
int mz = ((int)floor(position.z / _gridIntervalSize) % _gridSize);
if (mx < 0) {
mx += _gridSize;
}
if (my < 0) {
my += _gridSize;
}
if (mz < 0) {
mz += _gridSize;
}
int x = mx * _gridSize * _gridSize;
int y = my * _gridSize;
int z = mz * 1;
return x + y + z;
}
Vector3 is just some simple class which stores three floats and provides some overloaded operators. _gridSize is of type int and _gridIntervalSize is a float. There are _gridSize ^ 3 buckets.
The purpose of the function is to provide hash table support. Every 3d-point is mapped to an index, and points which lie in the same voxel of size _gridIntervalSize ^ 3 should land in the same bucket.
First rule of optimization when there is math involved: Eliminate division, square roots, and trig functions.
inverse_size = 1 / _gridIntervalSize;
....that should be done only once, not once per call.
int mx = ((int)floor(position.x * inverse_size) % _gridSize);
int my = ((int)floor(position.y * inverse_size) % _gridSize);
int mz = ((int)floor(position.z * inverse_size) % _gridSize);
I would also recommend dropping the mod operation because that's another division - if your grid size is a power of 2 you can use & (gridsize-1) which will also allow you to delete the conditional code at the bottom which is another big savings.
On another note, using overloaded operators may be hurting you. This is a touchy subject here so I'll let you experiment with it and decide for yourself.