实体路径跟踪算法

Question

在我的游戏中有许多实体，每个实体都有分配给它的路径(Array<Vector2>)。没有要遵循的路径(指示实体应该移动)是由空路径指示的(路径永远不是null)。updateMoveAttributes被称为每个实体的每一个帧，因此它的效率是非常重要的。

注意，我的实体的方向是弧度，范围从- Math.PI到Math.PI，因为这是Math.atan2返回的内容。

下面的代码片段包含updateMoveAttributes及其支持方法。

private final Array<Vector2> path = new Array<>();
private int pathIndex = 0;

private static final float EPSILON = 1e-3f;

private float getXDifference(Vector2 pathComponent) {
    float xDifference = pathComponent.x - getSprite().getX();

    xDifference = Math.abs(xDifference) < EPSILON ? 0 : xDifference;

    return xDifference;
}

private float getYDifference(Vector2 pathComponent) {
    float yDifference = pathComponent.y - getSprite().getY();

    yDifference = Math.abs(yDifference) < EPSILON ? 0 : yDifference;

    return yDifference;
}

public float orient(Vector2 pathComponent) {
    float xDiff = getXDifference(pathComponent);
    float yDiff = getYDifference(pathComponent);

    return (float) Math.atan2(yDiff, xDiff);
}

void updateMoveAttributes() {
    // No path assigned
    if (getPath().size == 0)
        return;

    // No more path to travel
    if (getPathIndex() == getPath().size) {
        setSpeed(0);

        return;
    }

    Vector2 nextPathComponent = getPath().get(getPathIndex());
    float angleToComponent = orient(nextPathComponent);

    setDirection(angleToComponent);

    float xDiff = getXDifference(nextPathComponent);
    float yDiff = getYDifference(nextPathComponent);

    boolean angleInPositiveXQuadrants = (-Math.PI / 2 <= angleToComponent && angleToComponent <= Math.PI / 2);
    boolean pastX = angleInPositiveXQuadrants ? (xDiff <= 0) : (xDiff >= 0);

    boolean angleInPositiveYQuadrants = (0 <= angleToComponent) && (angleToComponent <= Math.PI);
    boolean pastY = angleInPositiveYQuadrants ? (yDiff <= 0) : (yDiff >= 0);

    // Indicates that we have passed the path component we are on route to, adjust course with respect
    // to next available path component.
    if (pastY && pastX)
        setPathIndex(getPathIndex() + 1);

    // In case user upgrades entity during travel of entity -> movement speed changes -> update required for actual speed
    setSpeed(getMovementSpeed());
}

public Array<Vector2> getPath() {
    return path;
}

Answer 1

1. 方法getXDifference和getYDifference非常相似，您可以将其转换为接受两个变量的单个方法。
2. 您还可以内联返回语句，因为方法名称足以判断返回的是什么。
3. 您可以编写方法返回计算值，而不是计算pastX和pastY。布尔isPastX(变量.){ //逻辑来计算和返回pastX。}，if语句将转换为if( isPastX() && isPastY())。这也将有助于在计算pastX或pastY非常昂贵的情况下缩短条件(在本例中不是，只是一个建议)。
4. 大括号{}应该总是在条件语句或循环之后使用，即使它只是一行。它可能在某些情况下避免混淆。。如果(getPath().size == 0){返回；}