我需要在C ++模

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我有一个实数x，我想得到数字（x mod 2 * PI）= y，其中y在0和2 * PI之间。

我可以做一个循环，如果x是负的，我只是继续添加2 * pi，直到我的数字在[0,2 * pi]的范围内，或者如果x大于2 * pi我可以只是保持减去。

我试过这个：

fmod（-0.3434073,2 * M_PI）;

但这仍然是-0.3434073。

解决方案

编辑：填入Notinlist以便更快地回答。 p>

在C99标准中，7.12.10.1节可以这样说：

fmod函数返回值x-ny，对于一些整数n，使得如果y不为零，则结果具有与x相同的符号，并且幅度小于y的幅度。如果y为零，则无论发生域错误还是fmod函数返回零都是实现定义的。

double fmod_positive（double x，double y）{ double tmp = fmod（x，y）; return x< 0？ tmp + y：tmp; }

I have a real number x and I want to get the number (x mod 2*PI) =y, where y is between 0 and 2*PI.

I could do it with a loop and if x is negative I just keep adding 2*pi until my number is in range of [0,2*pi] or if x is greater the 2*pi I could just keep subtracting. However, I hope there is a more elegant way to do this.

I tried this:

fmod(-0.3434073,2*M_PI);

but this remains -0.3434073. Why and how can I get it working as I want it to?

解决方案

Edit: Credits to Notinlist for answering faster.

In the C99 standard, section 7.12.10.1 has this to say:

The fmod functions return the value x − ny, for some integer n such that, if y is nonzero, the result has the same sign as x and magnitude less than the magnitude of y. If y is zero, whether a domain error occurs or the fmod functions return zero is implementation- defined.

Going by this, you need to add y if you want the proper sign.

All in all:

double fmod_positive(double x, double y){ double tmp = fmod(x, y); return x < 0 ? tmp + y : tmp; }