我为什么要模数减法？

Question

天哪，这一定很简单。我有一个范围（0,2π）的标题和两个点，我从中得到标题。我必须比较它们，看看是否在另一个范围内。到目前为止我得到的是。

//get the angle
float angle = atan(here.x - there.x, here.y - there.y);

//atan2 uses (-pi, pi) range, convert to (0, 2pi)
if(angle < 0) angle += 2*pi;

//subtract them pesky headings
float diff = angle - givenAngle;

//a difference of 350 degrees really is a difference of 10 degrees
if(diff > pi) diff = 2*pi - diff;

//a difference of -10 degrees really is a difference of 10 degrees
if(diff < 0) diff *= -1;

//check if the point is in range of givenAngle
if(diff > fov) do_magic(diff - fov);

但是，当两个角度都变为零时，我会遇到各种各样的问题，而且我在解决这个问题时浪费了太多的脑力。

我在哪里做错了？如何正确找到两个标题之间的差异？

Answer 1

我怀疑您的操作顺序可能略有错误：

//a difference of 350 degrees really is a difference of 10 degrees
if(diff > pi) diff = 2*pi - diff;

//a difference of -10 degrees really is a difference of 10 degrees
if(diff < 0) diff *= -1;

这不会导致-350的差异，但是如果你切换它的语句：

//a difference of -10 degrees really is a difference of 10 degrees
if(diff < 0) diff *= -1;

//a difference of ±350 degrees really is a difference of 10 degrees
if(diff > pi) diff = 2*pi - diff;

Answer 2

其中一个主要问题是：

float angle = atan(here.x - there.x, here.y - there.y);

注意到这一点有点棘手，但这不是atan的正确参数顺序 - 偶数the mathematical definition on wikipedia在之前采用y组件（正弦） x组件（余弦）。

float angle = atan(here.y - there.y, here.x - there.x);

第二个问题是，事实证明，angle关闭了180度。换句话说，我应该计算here - there。

，而不是计算there - here

float angle = atan(there.y - here.y, there.x - here.x);

在sverre's observation中添加有关错误操作顺序的内容，我们的内容会更好：

float angle = atan(there.y - here.y, there.x - here.x);
if(angle < 0) angle += pi * 2; //use (0, 2pi) range, the same as angle
float diff = abs(angle - givenAngle);
if(diff > pi) diff = 2*pi - diff;
if(diff > fov) do_magic(diff - fov);