Question

我正在使用Code :: Blocks 10.05编译这个程序，但通常我会在每个输出开始生成Nan之前完成大约10次迭代。我想知道这是否是使用cos和sin函数引起的问题，是否有一个体面的工作来避免这种情况？

我必须制作大量的迭代，因为我正在为大学的项目工作，所以它必须准确。我查了几篇关于如何避免使用sin和cos的文章虽然我需要严格遵循一些公式，否则我生成的结果可能不准确，所以我不确定是否妥协。

    struct Particle // Need to define what qualities our particle has
{
   double dPosition;
   double dAngle;

};

Particle Subject;

void M1(double &x, double &y) //Defines movement if particle doesn't touch inner   boundary
{
    x = x + 2*y;
}

double d = 0.25; //This can and will be changed when I need to find a distance between
                // the two cricles at a later stage


void M2(double &x,double &y, double d) //Defines movement of a particle if it impacts the inner boundary
{
    double z = asin(-(sin(y)+d*cos(x + y))/0.35);
    double y1 = y;
    y = asin(-0.35*sin(z) + d*cos(x + y + 2*z));
    x = y + y1 + x + 2*z;
}

int main()
{
    cout << "Please tell me where you want this particle to start positions-wise? (Between 0 and 2PI" << endl;
    cin >> Subject.dPosition;
    cout << "Please tell me the angle that you would like it to make with the normal? (Between 0 and PI/2)" << endl;
    cin >> Subject.dAngle;
    cout << "How far would you like the distances of the two middle circles to be?" << endl;
    double d;
    cin >> d;

    // These two functions are to understand where the experiment begins from.
    // I may add a function to change where the circle starts however I will use radius = 0.35 throughout

    cout << "So position is: " << Subject.dPosition << endl;
    cout << "And angle with the normal is: " << Subject.dAngle <<endl;

    int n=0;
    while (n <= 100) //This is used to iterate the process and create an array of Particle data points
    {               // in order to use this data to build up Poincare diagrams.

    {
        while (Subject.dPosition > 2*M_PI)
            Subject.dPosition = Subject.dPosition - 2*M_PI;
    }
    {
        if (0.35 >= abs(0.35*cos(Subject.dPosition + Subject.dAngle)+sin(Subject.dAngle))) //This is the condition of hitting the inner boundary
            M2(Subject.dPosition, Subject.dAngle, d); //Inner boundary collision
        else
            M1(Subject.dPosition, Subject.dAngle); // Outer boundary collision
    };
    cout << "So position is: " << Subject.dPosition << endl;
    cout << "And angle with the normal is: " << Subject.dAngle <<endl;
    n++;
}
    return 0;
}

Answer 1

Nan在c ++中显示为无限，零偏差的指示，以及其他一些不可表示数字的变体。

修改：

正如Matteo Itallia指出的，inf用于无限/零分割。我找到了这些方法：

template<typename T> inline bool isnan(T value) { return value != value; } // requires #include <limits> template<typename T> inline bool isinf(T value) { return std::numeric_limits<T>::has_infinity && value == std::numeric_limits<T>::infinity(); }

参考：http://bytes.com/topic/c/answers/588254-how-check-double-inf-nan

Answer 2

如果值超出[-1，+ 1]并传递给asin（），则结果为nan

如果您需要检查Nan，请尝试以下

if( value != value ){
    printf("value is nan\n");
}

使用sin和cos函数迭代时得到Nan

2 个答案: