获取最紧密的边界框

Question

我认为代码是不言自明的：

// Set the bbox min and max of the lower node
current_min_bound = lower_get_bbox_min(buf, lower_handle);
current_max_bound = lower_get_bbox_max(buf, lower_handle);
// Set lower node bbox 
if(current_min_bound.x > leaf_node->bbox_min.x) {
    current_min_bound.x = leaf_node->bbox_min.x;
} 
if(current_min_bound.y > leaf_node->bbox_min.y) {
    current_min_bound.y = leaf_node->bbox_min.y;
}
if(current_min_bound.z > leaf_node->bbox_min.z) {
    current_min_bound.z = leaf_node->bbox_min.z;
}
if(current_max_bound.x < leaf_node->origin.x + highest_voxel_coord.x) {
    current_max_bound.x = leaf_node->origin.x + highest_voxel_coord.x;
}
if(current_max_bound.y < leaf_node->origin.y + highest_voxel_coord.y) {
    current_max_bound.y = leaf_node->origin.y + highest_voxel_coord.y;
}
if(current_max_bound.z < leaf_node->origin.z + highest_voxel_coord.z) {
    current_max_bound.z = leaf_node->origin.z + highest_voxel_coord.z;
}
lower_set_bbox_min(buf, lower_handle, current_min_bound);
lower_set_bbox_max(buf, lower_handle, current_max_bound);

所以我想知道是否有更快的方法来比较和获得给定 3D 坐标系统中最紧密的边界框？我认为太多的比较正在扼杀我的 CPU 性能。数据结构是一棵树，但我认为这对这个问题无关紧要。

干杯。

是否可以编写一种更快的方法来计算数学问题。

编辑：

看起来使用三元条件运算符允许编译器优化代码：

                // Set lower node bbox min
                current_min_bound.x = (current_min_bound.x > leaf_node->origin.x + lowest_voxel_coord.x) ? leaf_node->origin.x + lowest_voxel_coord.x : current_min_bound.x;
                current_min_bound.y = (current_min_bound.y > leaf_node->origin.y + lowest_voxel_coord.y) ? leaf_node->origin.y + lowest_voxel_coord.y : current_min_bound.y;
                current_min_bound.z = (current_min_bound.z > leaf_node->origin.z + lowest_voxel_coord.z) ? leaf_node->origin.z + lowest_voxel_coord.z : current_min_bound.z;
                // Set lower node bbox max
                current_max_bound.x = (current_max_bound.x < leaf_node->origin.x + highest_voxel_coord.x) ? leaf_node->origin.x + highest_voxel_coord.x : current_max_bound.x;
                current_max_bound.y = (current_max_bound.y < leaf_node->origin.y + highest_voxel_coord.y) ? leaf_node->origin.y + highest_voxel_coord.y : current_max_bound.y;
                current_max_bound.z = (current_max_bound.z < leaf_node->origin.z + highest_voxel_coord.z) ? leaf_node->origin.z + highest_voxel_coord.z : current_max_bound.z;

迭代次数：27 546

先前代码的相同迭代：0.860708 秒 相同的迭代当前代码：0.717957 秒

Answer 1

分支预测失败会减慢速度。

您可以尝试“无分支”比较：

static int min( int x, int y ) { return y ^ ((x^y) & -(x<y)); }
static int max( int x, int y ) { return y ^ ((x^y) & -(x>y)); }

也许，将这些转换为宏的任务留给您。