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C:如何将 float 包装到区间 [-pi, pi)

转载 作者:行者123 更新时间:2023-11-30 17:57:07 29 4
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我正在寻找一些可以有效完成的不错的 C 代码:

while (deltaPhase >= M_PI) deltaPhase -= M_TWOPI;
while (deltaPhase < -M_PI) deltaPhase += M_TWOPI;

我有什么选择?

最佳答案

2013 年 4 月 19 日编辑:

更新了模函数以处理 aka.nice 和 arr_sea 指出的边界情况:

static const double     _PI= 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348;
static const double _TWO_PI= 6.2831853071795864769252867665590057683943387987502116419498891846156328125724179972560696;

// Floating-point modulo
// The result (the remainder) has same sign as the divisor.
// Similar to matlab's mod(); Not similar to fmod() - Mod(-3,4)= 1 fmod(-3,4)= -3
template<typename T>
T Mod(T x, T y)
{
static_assert(!std::numeric_limits<T>::is_exact , "Mod: floating-point type expected");

if (0. == y)
return x;

double m= x - y * floor(x/y);

// handle boundary cases resulted from floating-point cut off:

if (y > 0) // modulo range: [0..y)
{
if (m>=y) // Mod(-1e-16 , 360. ): m= 360.
return 0;

if (m<0 )
{
if (y+m == y)
return 0 ; // just in case...
else
return y+m; // Mod(106.81415022205296 , _TWO_PI ): m= -1.421e-14
}
}
else // modulo range: (y..0]
{
if (m<=y) // Mod(1e-16 , -360. ): m= -360.
return 0;

if (m>0 )
{
if (y+m == y)
return 0 ; // just in case...
else
return y+m; // Mod(-106.81415022205296, -_TWO_PI): m= 1.421e-14
}
}

return m;
}

// wrap [rad] angle to [-PI..PI)
inline double WrapPosNegPI(double fAng)
{
return Mod(fAng + _PI, _TWO_PI) - _PI;
}

// wrap [rad] angle to [0..TWO_PI)
inline double WrapTwoPI(double fAng)
{
return Mod(fAng, _TWO_PI);
}

// wrap [deg] angle to [-180..180)
inline double WrapPosNeg180(double fAng)
{
return Mod(fAng + 180., 360.) - 180.;
}

// wrap [deg] angle to [0..360)
inline double Wrap360(double fAng)
{
return Mod(fAng ,360.);
}

关于C:如何将 float 包装到区间 [-pi, pi),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13002739/

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