mjhjmtu

I wrote an if statement that checks if an iteratively altered vector "crosses" the up vector (0,1,0) or the down vector (0,-1,0).

if ((lastDiff.x() > 0 && diff.x() < 0 && lastDiff.z() >= 0 && diff.z() <= 0) || 
 (lastDiff.x() < 0 && diff.x() > 0 && lastDiff.z() >= 0 && diff.z() <= 0) ||
 (lastDiff.x() > 0 && diff.x() < 0 && lastDiff.z() <= 0 && diff.z() >= 0) ||
 (lastDiff.x() < 0 && diff.x() > 0 && lastDiff.z() <= 0 && diff.z() >= 0) ||
 (lastDiff.z() > 0 && diff.z() < 0 && lastDiff.x() >= 0 && diff.x() <= 0) ||
 (lastDiff.z() < 0 && diff.z() > 0 && lastDiff.x() >= 0 && diff.x() <= 0) ||
 (lastDiff.z() > 0 && diff.z() < 0 && lastDiff.x() <= 0 && diff.x() >= 0) ||
 (lastDiff.z() < 0 && diff.z() > 0 && lastDiff.x() <= 0 && diff.x() >= 0))

I feel like there is a simpler or better way to implement this if guard but I can't find it.

You have four groups of repeating conditions:

bool crossX = (lastDiff.x() < 0 && diff.x() > 0) || (lastDiff.x() > 0 && diff.x() < 0);
bool crossZ = (lastDiff.z() < 0 && diff.z() > 0) || (lastDiff.z() > 0 && diff.z() < 0);
bool crossTouchX = (lastDiff.x() <= 0 && diff.x() >= 0) || (lastDiff.x() >= 0 && diff.x() <= 0);
bool crossTouchZ = (lastDiff.z() <= 0 && diff.z() >= 0) || (lastDiff.z() >= 0 && diff.z() <= 0);

Using these the original condition can be simplified to:

if ((crossX && crossTouchZ) || (crossZ && crossTouchX))

 // do stuff

Code review:

We're checking if the signs of the x and z components have both changed. So we can simplify quite a lot:

auto sign = (float f) return (f > 0.f) ? true : false; 
auto const& a = lastDiff;
auto const& b = diff;

if (sign(a.x()) != sign(b.x()) && sign(a.z()) != sign(b.z())) ... 

An alternative:

Set the y coordinate of both vectors to zero. Normalize them. Take the dot product. If the result (which is the cosine of the angle between them) is < 0 they're on opposite sides. If the result is > 0 they're on the same side.

(Make sure to check for zero vectors after that first step).