问题描述
程序中的角度以0至2pi表示。我想要一种方法来添加两个角度，如果结果高于2pi，则将2pi缠绕为0。或者，如果我从某个角度减去一个角度并且该角度小于0，则它将环绕2pi。
Angles in my program are expressed in 0 to 2pi. I want a way to add two angles and have it wrap around the 2pi to 0 if the result is higher than 2pi. Or if I subtracted an angle from an angle and it was below 0 it would wrap around 2pi.
有没有办法做到这一点？
Is there a way to do this?
谢谢。
推荐答案
您要寻找的是模数。 fmod函数将不起作用，因为它会计算余数而不是算术系数。这样的东西应该可以工作：
What you are looking for is the modulus. The fmod function will not work because it calculates the remainder and not the arithmetic modulus. Something like this should work:
inline double wrapAngle( double angle )
{
double twoPi = 2.0 * 3.141592865358979;
return angle - twoPi * floor( angle / twoPi );
}
编辑：
余数通常定义为经过长时间除法后剩余的数（例如18/4的余数为2，因为 18 = 4 * 4 + 2 ）。当您有负数时，这变得毛茸茸。查找有符号除法的余数的常见方法是使余数与结果具有相同的符号（例如，-18/4的余数为-2，因为 -18 = -4 * 4 +- 2 ）。
The remainder is commonly defined as what is left over after long division (eg. the remainder of 18/4 is 2, because 18 = 4 * 4 + 2). This gets hairy when you have negative numbers. The common way to find the remainder of a signed division is for the remainder to have the same sign as the result (eg. the remainder of -18/4 is -2, because -18 = -4 * 4 + -2).
x模数y的定义是等式x = y * c + m中m的最小正值，给定c为整数。因此 18 mod 4 将为2（其中c = 4），但是 -18 mod 4 也将为2（其中c = -5）。
The definition of x modulus y is the smallest positive value of m in the equation x=y*c+m, given c is an integer. So 18 mod 4 would be 2 (where c=4), however -18 mod 4 would also be 2 (where c=-5).
x mod y 的最简单计算方法是 xy * floor（x / y），其中floor是小于该值的最大整数等于或等于输入。
The simplest calculation of x mod y is x-y*floor(x/y), where floor is the largest integer that is less than or equal to the input.
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