我正在尝试在 java 上实现快速平方根逆以加快向量归一化。但是，当我在 Java 中实现单精度版本时，我的速度与开始时大致相同1F / (float)Math.sqrt()，然后迅速下降到一半的速度。这很有趣，因为虽然 Math.sqrt 使用（我认为）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：
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迭代每 1600 万次，然后汇总结果并重复，我得到如下结果：
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()
我始终得到两个速度大致相同的数字，然后进行迭代，其中快速平方根可节省大约 60% 所需的时间1F / Math.sqrt()，然后进行几次迭代，快速平方根运行所需的时间大约是其两倍控制。我很困惑为什么 FISR 会从相同 -> 快 60% -> 慢 100%，并且每次运行我的程序时都会发生这种情况。
编辑：上面的数据是我在 Eclipse 中运行它时的数据。当我运行程序时，javac/java我得到完全不同的数据：
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()
EDIT2：经过几次响应后，似乎速度在几次迭代后稳定下来，但它稳定到的数字非常不稳定。有人知道为什么吗？
这是我的代码（不完全简洁，但这是全部内容）：
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();
}
}