需要帮助将代码从C转换为Java

Question

从这篇文章.这是代码:

float InvSqrt(float x){ // line 0
   float xhalf = 0.5f * x;
   int i = *(int*)&x; // store floating-point bits in integer
   i = 0x5f3759d5 - (i >> 1); // initial guess for Newton's method
   x = *(float*)&i; // convert new bits into float
   x = x*(1.5f - xhalf*x*x); // One round of Newton's method
   return x;
}

......我甚至不知道那是C还是C++.[好吧显然它是C,谢谢]有人可以为我翻译成Java吗？它(只是,我希望)第2和第4行令我感到困惑.

Answer 1

您想要使用这些方法:

• Float.floatToIntBits
• Float.intBitsToFloat

并且可能存在问题strictfp等.

它大致是这样的:(注意:这是未经测试的!)

float InvSqrt(float x){ // line 0
   float xhalf = 0.5f * x;
   int i = Float.floatToIntBits(x); // store floating-point bits in integer
   i = 0x5f3759d5 - (i >> 1); // initial guess for Newton's method
   x = Float.intBitsToFloat(i); // convert new bits into float
   x = x*(1.5f - xhalf*x*x); // One round of Newton's method
   return x;
}

Answer 2

这些线用于之间进行转换float,并int作为位模式.Java有静态方法java.lang.Float- 其他一切都是相同的.

static float InvSqrt(float x) { // line 0
    float xhalf = 0.5f * x;
    int i = Float.floatToIntBits(x); // store floating-point bits in integer
    i = 0x5f3759d5 - (i >> 1); // initial guess for Newton's method
    x = Float.intBitsToFloat(i); // convert new bits into float
    x = x * (1.5f - xhalf * x * x); // One round of Newton's method
    return x;
}