Java 好奇循环性能

Question

我在评估我的 Java 代码时遇到了一个大问题。为了简化问题，我编写了以下代码，它们产生了同样奇怪的行为。重要的是方法 run() 和给定的双倍值率。对于我的运行时测试(在 main 方法中)，我将速率一次设置为 0.5，另一次设置为 1.0。值为 1.0 时，if 语句将在每次循环迭代中执行，值为 0.5 时，if 语句将执行一半。出于这个原因，我预计第一种情况的运行时间会更长，但事实恰恰相反。谁能给我解释一下这个现象？？

主要结果:

Test mit rate = 0.5
Length: 50000000, IF executions: 25000856
Execution time was 4329 ms.
Length: 50000000, IF executions: 24999141
Execution time was 4307 ms.
Length: 50000000, IF executions: 25001582
Execution time was 4223 ms.
Length: 50000000, IF executions: 25000694
Execution time was 4328 ms.
Length: 50000000, IF executions: 25004766
Execution time was 4346 ms.
=================================
Test mit rate = 1.0
Length: 50000000, IF executions: 50000000
Execution time was 3482 ms.
Length: 50000000, IF executions: 50000000
Execution time was 3572 ms.
Length: 50000000, IF executions: 50000000
Execution time was 3529 ms.
Length: 50000000, IF executions: 50000000
Execution time was 3479 ms.
Length: 50000000, IF executions: 50000000
Execution time was 3473 ms.

代码

public ArrayList<Byte> list = new ArrayList<Byte>();
public final int LENGTH = 50000000;

public PerformanceTest(){
    byte[]arr = new byte[LENGTH];
    Random random = new Random();
    random.nextBytes(arr);
    for(byte b : arr)
        list.add(b);
}

public void run(double rate){

    byte b = 0;
    int count = 0;

    for (int i = 0; i < LENGTH; i++) {

        if(getRate(rate)){
            list.set(i, b);
            count++;
        }
    }
    System.out.println("Length: " + LENGTH + ", IF executions: " + count);
}

public boolean getRate(double rate){
    return Math.random() < rate;
}

public static void main(String[] args) throws InterruptedException {
    PerformanceTest test = new PerformanceTest();

    long start, end;
    System.out.println("Test mit rate = 0.5");
    for (int i = 0; i < 5; i++) {
        start=System.currentTimeMillis();
        test.run(0.5);
        end = System.currentTimeMillis();
        System.out.println("Execution time was "+(end-start)+" ms.");

        Thread.sleep(500);
    }       
    System.out.println("=================================");
    System.out.println("Test mit rate = 1.0");      
    for (int i = 0; i < 5; i++) {
        start=System.currentTimeMillis();
        test.run(1.0);
        end = System.currentTimeMillis();
        System.out.println("Execution time was "+(end-start)+" ms.");
        Thread.sleep(500);
    }   
}

Answer 1

在第一种情况下，分支预测错误会破坏性能。尽管第二种情况做了一些工作，但它有点简单，因此处理器可以轻松预测下一步。请看这个Wikipedia page获取更多信息。

尝试使用 0.7 进行测试。如果我是正确的，那么性能将介于 0.5 和 1.0 之间。