为什么快速反平方根在Java上如此奇怪和缓慢？

我正在尝试在java上实现快速反平方根，以加快矢量归一化。但是，当我在Java中实现单精度版本时，我得到的速度与最初大致相同，然后迅速下降到速度的一半。这很有趣，因为虽然Math.sqrt使用（我推测）一个本机方法，但这涉及浮点除法，我听说这真的很慢。我计算数字的代码如下：1F / (float)Math.sqrt()

public static float fastInverseSquareRoot(float x){
    float xHalf = 0.5F * x;
    int temp = Float.floatToRawIntBits(x);
    temp = 0x5F3759DF - (temp >> 1);
    float newX = Float.intBitsToFloat(temp);
    newX = newX * (1.5F - xHalf * newX * newX);
    return newX;
}

使用我编写的一个简短的程序，每个迭代1600万次，然后聚合结果，然后重复，我得到这样的结果：

1F / Math.sqrt() took 65209490 nanoseconds.
Fast Inverse Square Root took 65456128 nanoseconds.
Fast Inverse Square Root was 0.378224 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 64131293 nanoseconds.
Fast Inverse Square Root took 26214534 nanoseconds.
Fast Inverse Square Root was 59.123647 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 27312205 nanoseconds.
Fast Inverse Square Root took 56234714 nanoseconds.
Fast Inverse Square Root was 105.895914 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 26493281 nanoseconds.
Fast Inverse Square Root took 56004783 nanoseconds.
Fast Inverse Square Root was 111.392402 percent slower than 1F / Math.sqrt()

我一直得到两者速度大致相同的数字，然后是一次迭代，其中快速反平方根节省了大约60%所需的时间，然后是几次迭代，快速反平方根需要大约两倍的时间才能运行为控件。我很困惑为什么FISR会从相同 - >快60% - >慢100%，每次我运行我的程序时都会发生这种情况。1F / Math.sqrt()

编辑： 上面的数据是我在eclipse中运行它的时候。当我运行程序时，我得到完全不同的数据：javac/java

1F / Math.sqrt() took 57870498 nanoseconds.
Fast Inverse Square Root took 88206794 nanoseconds.
Fast Inverse Square Root was 52.421004 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 54982400 nanoseconds.
Fast Inverse Square Root took 83777562 nanoseconds.
Fast Inverse Square Root was 52.371599 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 21115822 nanoseconds.
Fast Inverse Square Root took 76705152 nanoseconds.
Fast Inverse Square Root was 263.259133 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 20159210 nanoseconds.
Fast Inverse Square Root took 80745616 nanoseconds.
Fast Inverse Square Root was 300.539585 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 21814675 nanoseconds.
Fast Inverse Square Root took 85261648 nanoseconds.
Fast Inverse Square Root was 290.845374 percent slower than 1F / Math.sqrt()

编辑2：经过几次响应后，速度似乎在几次迭代后稳定下来，但它稳定到的数字是高度不稳定的。有人知道为什么吗？

这是我的代码（不完全简洁，但这是整个事情）：

public class FastInverseSquareRootTest {

    public static FastInverseSquareRootTest conductTest() {
        float result = 0F;
        long startTime, endTime, midTime;
        startTime = System.nanoTime();
        for (float x = 1F; x < 4_000_000F; x += 0.25F) {
            result = 1F / (float) Math.sqrt(x);
        }
        midTime = System.nanoTime();
        for (float x = 1F; x < 4_000_000F; x += 0.25F) {
            result = fastInverseSquareRoot(x);
        }
        endTime = System.nanoTime();
        return new FastInverseSquareRootTest(midTime - startTime, endTime
                - midTime);
    }

    public static float fastInverseSquareRoot(float x) {
        float xHalf = 0.5F * x;
        int temp = Float.floatToRawIntBits(x);
        temp = 0x5F3759DF - (temp >> 1);
        float newX = Float.intBitsToFloat(temp);
        newX = newX * (1.5F - xHalf * newX * newX);
        return newX;
    }

    public static void main(String[] args) throws Exception {
        for (int i = 0; i < 7; i++) {
            System.out.println(conductTest().toString());
        }
    }

    private long controlDiff;

    private long experimentalDiff;

    private double percentError;

    public FastInverseSquareRootTest(long controlDiff, long experimentalDiff) {
        this.experimentalDiff = experimentalDiff;
        this.controlDiff = controlDiff;
        this.percentError = 100D * (experimentalDiff - controlDiff)
                / controlDiff;
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        sb.append(String.format("1F / Math.sqrt() took %d nanoseconds.%n",
                controlDiff));
        sb.append(String.format(
                "Fast Inverse Square Root took %d nanoseconds.%n",
                experimentalDiff));
        sb.append(String
                .format("Fast Inverse Square Root was %f percent %s than 1F / Math.sqrt()%n",
                        Math.abs(percentError), percentError > 0D ? "slower"
                                : "faster"));
        return sb.toString();
    }
}