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我正在制作一个应该与 Math.pow() 执行相同操作的方法，但我只是不知道如何使 y 使用双精度值并获得带小数的双精度值结果......有什么想法（最好使用 for 循环）？在下面的方法中，我使用“x”作为底数，使用“y”作为指数。

public static double power(double x, double y) {
    double result = 1;
    if (y <= 0)
        return 0;
    for (int count = 0; count < (int)y; count++)
        result *= x;
    return result;
}
4

3 回答 3

3

您可以使用 Math.log 和 Math.exp 来实现这一点。

public static void main(String[] args) throws InterruptedException {


         System.out.println(power(2,2.5));

    }


  public static double power(double x, double y) {

    double val = y *  Math.log(x);

    double result = Math.exp(val);

            return result;
    }

输出是

5.65685424949238
于 2012-11-15T05:06:56.287 回答
0

创建不使用其他数学函数的 math.pow 副本需要对按位运算有基本的了解。我不使用java，但这里是c#中的代码

  public int ReturnPower(int firstNum, int secondNum)
    {

        int finalNum = 1;

        while (secondNum != 0)
        {

            //Use the bitwise operator & to check both strings
            if ((secondNum & 1) == 1)
            {
                finalNum = finalNum * firstNum;
            }

       //increment firstNum for the next iteration
            firstNum = firstNum * firstNum;

       //Bitwise operator righshift '>>' is used to reconfigure secondNum
            secondNum = secondNum >> 1;
        }



        //return((int)Math.Pow(firstNum, secondNum));
        return (finalNum);


    }

**旁注最后两行将返回相同的输出。'(int)' 放置在 Math.pow 之前以对其进行类型转换，以允许传递类型 'int' 而不是 'double'。

于 2017-01-31T18:51:06.550 回答
0

这是一个有效/详细的实现：

public static double pow(double x, double y)
{
  // Special cases first.
  if (y == 0)
    return 1;
  if (y == 1)
    return x;
  if (y == -1)
    return 1 / x;
  if (x != x || y != y)
    return Double.NaN;

  // When x < 0, yisint tells if y is not an integer (0), even(1),
  // or odd (2).
  int yisint = 0;
  if (x < 0 && floor(y) == y)
    yisint = (y % 2 == 0) ? 2 : 1;
  double ax = abs(x);
  double ay = abs(y);

  // More special cases, of y.
  if (ay == Double.POSITIVE_INFINITY)
    {
      if (ax == 1)
        return Double.NaN;
      if (ax > 1)
        return y > 0 ? y : 0;
      return y < 0 ? -y : 0;
    }
  if (y == 2)
    return x * x;
  if (y == 0.5)
    return sqrt(x);

  // More special cases, of x.
  if (x == 0 || ax == Double.POSITIVE_INFINITY || ax == 1)
    {
      if (y < 0)
        ax = 1 / ax;
      if (x < 0)
        {
          if (x == -1 && yisint == 0)
            ax = Double.NaN;
          else if (yisint == 1)
            ax = -ax;
        }
      return ax;
    }
  if (x < 0 && yisint == 0)
    return Double.NaN;

  // Now we can start!
  double t;
  double t1;
  double t2;
  double u;
  double v;
  double w;
  if (ay > TWO_31)
    {
      if (ay > TWO_64) // Automatic over/underflow.
        return ((ax < 1) ? y < 0 : y > 0) ? Double.POSITIVE_INFINITY : 0;
      // Over/underflow if x is not close to one.
      if (ax < 0.9999995231628418)
        return y < 0 ? Double.POSITIVE_INFINITY : 0;
      if (ax >= 1.0000009536743164)
        return y > 0 ? Double.POSITIVE_INFINITY : 0;
      // Now |1-x| is <= 2**-20, sufficient to compute
      // log(x) by x-x^2/2+x^3/3-x^4/4.
      t = x - 1;
      w = t * t * (0.5 - t * (1 / 3.0 - t * 0.25));
      u = INV_LN2_H * t;
      v = t * INV_LN2_L - w * INV_LN2;
      t1 = (float) (u + v);
      t2 = v - (t1 - u);
    }
  else
  {
    long bits = Double.doubleToLongBits(ax);
    int exp = (int) (bits >> 52);
    if (exp == 0) // Subnormal x.
      {
        ax *= TWO_54;
        bits = Double.doubleToLongBits(ax);
        exp = (int) (bits >> 52) - 54;
      }
    exp -= 1023; // Unbias exponent.
    ax = Double.longBitsToDouble((bits & 0x000fffffffffffffL)
                                 | 0x3ff0000000000000L);
    boolean k;
    if (ax < SQRT_1_5)  // |x|<sqrt(3/2).
      k = false;
    else if (ax < SQRT_3) // |x|<sqrt(3).
      k = true;
    else
      {
        k = false;
        ax *= 0.5;
        exp++;
      }

    // Compute s = s_h+s_l = (x-1)/(x+1) or (x-1.5)/(x+1.5).
    u = ax - (k ? 1.5 : 1);
    v = 1 / (ax + (k ? 1.5 : 1));
    double s = u * v;
    double s_h = (float) s;
    double t_h = (float) (ax + (k ? 1.5 : 1));
    double t_l = ax - (t_h - (k ? 1.5 : 1));
    double s_l = v * ((u - s_h * t_h) - s_h * t_l);
    // Compute log(ax).
    double s2 = s * s;
    double r = s_l * (s_h + s) + s2 * s2
      * (L1 + s2 * (L2 + s2 * (L3 + s2 * (L4 + s2 * (L5 + s2 * L6)))));
    s2 = s_h * s_h;
    t_h = (float) (3.0 + s2 + r);
    t_l = r - (t_h - 3.0 - s2);
    // u+v = s*(1+...).
    u = s_h * t_h;
    v = s_l * t_h + t_l * s;
    // 2/(3log2)*(s+...).
    double p_h = (float) (u + v);
    double p_l = v - (p_h - u);
    double z_h = CP_H * p_h;
    double z_l = CP_L * p_h + p_l * CP + (k ? DP_L : 0);
    // log2(ax) = (s+..)*2/(3*log2) = exp + dp_h + z_h + z_l.
    t = exp;
    t1 = (float) (z_h + z_l + (k ? DP_H : 0) + t);
    t2 = z_l - (t1 - t - (k ? DP_H : 0) - z_h);
  }

  // Split up y into y1+y2 and compute (y1+y2)*(t1+t2).
  boolean negative = x < 0 && yisint == 1;
  double y1 = (float) y;
  double p_l = (y - y1) * t1 + y * t2;
  double p_h = y1 * t1;
  double z = p_l + p_h;
  if (z >= 1024) // Detect overflow.
    {
      if (z > 1024 || p_l + OVT > z - p_h)
        return negative ? Double.NEGATIVE_INFINITY
          : Double.POSITIVE_INFINITY;
    }
  else if (z <= -1075) // Detect underflow.
    {
      if (z < -1075 || p_l <= z - p_h)
        return negative ? -0.0 : 0;
    }

  // Compute 2**(p_h+p_l).
  int n = round((float) z);
  p_h -= n;
  t = (float) (p_l + p_h);
  u = t * LN2_H;
  v = (p_l - (t - p_h)) * LN2 + t * LN2_L;
  z = u + v;
  w = v - (z - u);
  t = z * z;
  t1 = z - t * (P1 + t * (P2 + t * (P3 + t * (P4 + t * P5))));
  double r = (z * t1) / (t1 - 2) - (w + z * w);
  z = scale(1 - (r - z), n);
  return negative ? -z : z;
}

从这里复制http://developer.classpath.org/doc/java/lang/StrictMath-source.html

于 2018-10-23T13:06:49.470 回答