I've heard of the "fast inverse square root", discussed here, and I wanted to put it in my Java program (just for research purposes, so ignore anything about the native libraries being faster).

I was looking at the code, and the C code directly converts the float into an int with some C pointer magic. If you try to do this in Java with casts, it doesn't work: java truncates the float (as you would expect), and you can't get the pointer of a primitive (as you can in C).

So how do you do this?

解决方案

Remember to benchmark your code before using this.

If it turns out you don't need it, or it's slower on the CPU architecture you are using, then it's better to go without having this obtuse code in your project.

The Java libraries have a way to get from the float number to the raw bits.

As seen in the Javadoc for java.lang.Float ( http://docs.oracle.com/javase/6/docs/api/java/lang/Float.html ), we have the floatToIntBits function, as well as intBitsToFloat.

This means we can write the "fast inverse square root" in Java as follows:

public static float invSqrt(float x) {

float xhalf = 0.5f * x;

int i = Float.floatToIntBits(x);

i = 0x5f3759df - (i >> 1);

x = Float.intBitsToFloat(i);

x *= (1.5f - xhalf * x * x);

return x;

}

Here is the version for doubles:

public static double invSqrt(double x) {

double xhalf = 0.5d * x;

long i = Double.doubleToLongBits(x);

i = 0x5fe6ec85e7de30daL - (i >> 1);

x = Double.longBitsToDouble(i);

x *= (1.5d - xhalf * x * x);

return x;

}