Question

看看下面的代码：

class Test
{
    public static void main(String abc[])
    {
        for( int N=1; N <= 1_000_000_000; N=N*10)
        {
            long t1 = System.nanoTime();

            start(N);

            long t2 = System.nanoTime() - t1;
            System.out.println("Time taken for " + N + " : " + t2);
        }
    }

    public static void start( int N )
    {
        int j=1;
        for(int i=0; i<=N; i++)
            j=j*i;
    }
}

上述问题产生的结果是：

Time taken for 1 : 7267
Time taken for 10 : 3312
Time taken for 100 : 7908
Time taken for 1000 : 51181
Time taken for 10000 : 432124
Time taken for 100000 : 4313696
Time taken for 1000000 : 9347132
Time taken for 10000000 : 858
Time taken for 100000000 : 658
Time taken for 1000000000 : 750

问题：

1。）为什么N = 1的时间异常大于N = 10？（有时甚至超过N = 100）

2.。）为什么N = 10M及以后的时间异常低？

上述问题中指出的模式是深刻的，甚至在经过多次迭代后仍然存在。这里有memoization的连接吗？

修改

感谢您的回答。我想用实际循环替换方法调用。但现在，没有JIT优化。为什么不？是否将语句放在促进优化过程的方法中？修改后的代码如下：

class test
{
    public static void main(String abc[])
    {
        for( int k=1; k<=3; k++)
        {
            for( int N=1; N<=1_000_000_000; N=N*10)
            {
                long t1 = System.nanoTime();

                int j=1;
                for(int i=0; i<=N; i++)
                    j=j*i;

                long t2 = System.nanoTime() - t1;
                System.out.println("Time taken for "+ N + " : "+ t2);
            }
        }
    }
}

编辑2： 以上修改代码的输出：

Time taken for 1 : 2160
Time taken for 10 : 1142
Time taken for 100 : 2651
Time taken for 1000 : 19453
Time taken for 10000 : 407754
Time taken for 100000 : 4648124
Time taken for 1000000 : 12859417
Time taken for 10000000 : 13706643
Time taken for 100000000 : 136928177
Time taken for 1000000000 : 1368847843
Time taken for 1 : 264
Time taken for 10 : 233
Time taken for 100 : 332
Time taken for 1000 : 1562
Time taken for 10000 : 17341
Time taken for 100000 : 136869
Time taken for 1000000 : 1366934
Time taken for 10000000 : 13689017
Time taken for 100000000 : 136887869
Time taken for 1000000000 : 1368178175
Time taken for 1 : 231
Time taken for 10 : 242
Time taken for 100 : 328
Time taken for 1000 : 1551
Time taken for 10000 : 13854
Time taken for 100000 : 136850
Time taken for 1000000 : 1366919
Time taken for 10000000 : 13692465
Time taken for 100000000 : 136833634
Time taken for 1000000000 : 1368862705

Answer 1

1。）为什么N = 1的时间异常大于N = 10

因为这是VM首次看到该代码 - 它可能决定只是解释它，或者需要花一点时间将它JIT到本机代码，但可能没有优化。这是Java基准测试的“陷阱”之一。

2.。）为什么N = 10M及以后的时间异常低？

此时，JIT更加努力地优化代码 - 将其减少到几乎没有。

特别是，如果多次运行此代码（仅在循环中），您将看到JIT编译器优化的效果：

Time taken for 1 : 3732
Time taken for 10 : 1399
Time taken for 100 : 3266
Time taken for 1000 : 26591
Time taken for 10000 : 278508
Time taken for 100000 : 2496773
Time taken for 1000000 : 4745361
Time taken for 10000000 : 933
Time taken for 100000000 : 466
Time taken for 1000000000 : 933
Time taken for 1 : 933
Time taken for 10 : 467
Time taken for 100 : 466
Time taken for 1000 : 466
Time taken for 10000 : 933
Time taken for 100000 : 466
Time taken for 1000000 : 933
Time taken for 10000000 : 467
Time taken for 100000000 : 467
Time taken for 1000000000 : 466
Time taken for 1 : 467
Time taken for 10 : 467
Time taken for 100 : 466
Time taken for 1000 : 466
Time taken for 10000 : 466
Time taken for 100000 : 467
Time taken for 1000000 : 466
Time taken for 10000000 : 466
Time taken for 100000000 : 466
Time taken for 1000000000 : 466

正如您所看到的，在第一次循环后，无论输入是什么时间（模块噪声 - 基本上总是~460ns或~933ns，不可预测），这意味着JIT优化了循环。

如果您实际返回j，和将i的初始值更改为1而不是0，您会看到你期望的那种结果。将i的初始值更改为1是因为否则JIT会发现您最终会返回0。

Answer 2

你实际上是对Java的JIT进行基准测试。如果我稍微修改你的代码：

class Test
{
    public static void main(String abc[])
    {
        for( int N=1; N <= 1_000_000_000; N=N*10)
        {
            long t1 = System.nanoTime();

            start(N);

            long t2 = System.nanoTime() - t1;
            System.out.println("Time taken for " + N + " : " + t2);
        }

        for( int N=1; N <= 1_000_000_000; N=N*10)
        {
            long t1 = System.nanoTime();

            start(N);

            long t2 = System.nanoTime() - t1;
            System.out.println("Time taken for " + N + " : " + t2);
        }
    }

    public static void start( int N )
    {
        int j=1;
        for(int i=0; i<=N; i++)
            j=j*i;
    }
}

我明白了：

Time taken for 1 : 1811
Time taken for 10 : 604
Time taken for 100 : 1510
Time taken for 1000 : 10565
Time taken for 10000 : 104439
Time taken for 100000 : 829173
Time taken for 1000000 : 604
Time taken for 10000000 : 302
Time taken for 100000000 : 0
Time taken for 1000000000 : 0
Time taken for 1 : 0
Time taken for 10 : 302
Time taken for 100 : 0
Time taken for 1000 : 302
Time taken for 10000 : 301
Time taken for 100000 : 302
Time taken for 1000000 : 0
Time taken for 10000000 : 0
Time taken for 100000000 : 0
Time taken for 1000000000 : 302

永远不会对“冷”系统进行基准测试。总是重复每次测量几次并丢弃前几次测量，因为优化尚未开始

Answer 3

原因是1）你没有返回值，2）计算结果总是0.最后，JIT将简单地编译循环。

如果将循环更改为：

，则会得到预期的行为

public static int start(int N) {
    int j = 1;
    for (int i = 1; i <= N; i++)
        j = j * i;
    return j;
}

请注意，我已将循环初始化更改为int i = 1并添加了return j。如果我只做其中一个，那么循环（最终）仍会被编译掉。

这将产生以下系列（如果执行两次）：

Time taken for 1 : 2934
Time taken for 10 : 1466
Time taken for 100 : 3422
Time taken for 1000 : 20534
Time taken for 10000 : 191644
Time taken for 100000 : 1898845
Time taken for 1000000 : 1210489
Time taken for 10000000 : 11884401
Time taken for 100000000 : 115257525
Time taken for 1000000000 : 1061254223
Time taken for 1 : 978
Time taken for 10 : 978
Time taken for 100 : 978
Time taken for 1000 : 2444
Time taken for 10000 : 11244
Time taken for 100000 : 103644
Time taken for 1000000 : 1030089
Time taken for 10000000 : 10448535
Time taken for 100000000 : 107299391
Time taken for 1000000000 : 1072580803

Java乘法奇怪的行为

3 个答案: