Question

所以我想对一些基本的java功能进行基准测试，以便为这个问题添加一些信息：What is the gain from declaring a method as static。

我知道写作基准有时并不容易，但这里发生的事情我无法解释。

请注意，我并没有考虑如何解决这个问题，而是为什么会发生这种情况*

测试类：

public class TestPerformanceOfStaticVsDynamicCalls {

    private static final long RUNS = 1_000_000_000L;

    public static void main( String [] args ){

        new TestPerformanceOfStaticVsDynamicCalls().run();
    }

    private void run(){

        long r=0;
        long start, end;

        for( int loop = 0; loop<10; loop++ ){

            // Benchmark

            start = System.currentTimeMillis();
            for( long i = 0; i < RUNS; i++ ) {
                r += addStatic( 1, i );
            }
            end = System.currentTimeMillis();
            System.out.println( "Static: " + ( end - start ) + " ms" );

            start = System.currentTimeMillis();
            for( long i = 0; i < RUNS; i++ ) {
                r += addDynamic( 1, i );
            }
            end = System.currentTimeMillis();
            System.out.println( "Dynamic: " + ( end - start ) + " ms" );

            // Do something with r to keep compiler happy
            System.out.println( r );

        }

    }


    private long addDynamic( long a, long b ){

        return a+b;
    }

    private static long addStatic( long a, long b ){

        return a+b;
    }

}

我期待第一个循环是热身，以下循环更快。

在Eclipse中运行它会产生奇怪的结果：

Static: 621 ms
Dynamic: 631 ms
1000000001000000000
Static: 2257 ms
Dynamic: 2501 ms
2000000002000000000
Static: 2258 ms
Dynamic: 2469 ms
3000000003000000000
Static: 2231 ms
Dynamic: 2464 ms
4000000004000000000

那么wtf？它变慢了。要进行交叉检查，我使用java / c 7运行相同的代码：

Static: 620 ms
Dynamic: 627 ms
1000000001000000000
Static: 897 ms
Dynamic: 617 ms
2000000002000000000
Static: 901 ms
Dynamic: 615 ms
3000000003000000000
Static: 888 ms
Dynamic: 616 ms
4000000004000000000

所以这里只有静态调用对于以下循环变慢了。更奇怪的是，如果我重新排列代码只在最终循环后打印r我在Eclipse中得到它：

Static: 620 ms
Dynamic: 635 ms
Static: 2285 ms
Dynamic: 893 ms
Static: 2258 ms
Dynamic: 900 ms
Static: 2280 ms
Dynamic: 905 ms
4000000004000000000

这与java / c 7：

Static: 620 ms
Dynamic: 623 ms
Static: 890 ms
Dynamic: 614 ms
Static: 890 ms
Dynamic: 616 ms
Static: 886 ms
Dynamic: 614 ms
4000000004000000000

在改变eclipse中动态/静态基准测试的顺序时：

Dynamic: 618 ms
Static: 626 ms
1000000001000000000
Dynamic: 632 ms
Static: 2524 ms
2000000002000000000
Dynamic: 617 ms
Static: 2528 ms
3000000003000000000
Dynamic: 622 ms
Static: 2506 ms
4000000004000000000

和java / c 7：

Dynamic: 625 ms
Static: 646 ms
1000000001000000000
Dynamic: 2470 ms
Static: 633 ms
2000000002000000000
Dynamic: 2459 ms
Static: 635 ms
3000000003000000000
Dynamic: 2464 ms
Static: 645 ms
4000000004000000000

那么这里发生了什么？

编辑：一些系统信息：

Java version "1.7.0_55"
OpenJDK Runtime Environment (IcedTea 2.4.7) (7u55-2.4.7-1ubuntu1)
OpenJDK 64-Bit Server VM (build 24.51-b03, mixed mode)

Intel(R) Core(TM) i7-2720QM CPU @ 2.20GHz

EDIT2：

使用Java8：

Static: 620 ms
Dynamic: 624 ms
1000000001000000000
Static: 890 ms
Dynamic: 618 ms
2000000002000000000
Static: 891 ms
Dynamic: 616 ms
3000000003000000000
Static: 892 ms
Dynamic: 617 ms
4000000004000000000

其他代码排序在这里产生类似的奇怪（但是更好）的结果。

Answer 1

序言：手动编写微基准测试几乎总是注定要失败有frameworks已经解决了常见的基准测试问题。

JIT编译单元是一种方法。将几个基准合并到一个方法中会导致不可预测的结果。
JIT严重依赖于执行配置文件，即运行时统计信息。如果方法长时间运行第一个场景，JIT将优化生成的代码。当该方法突然切换到另一种情况时，不要指望它以相同的速度运行。
JIT可能会跳过优化未执行的代码。它将为此代码留下一个不常见的陷阱。如果陷阱被击中，JVM将取消优化编译的方法，切换到解释器，然后用新知识重新编译代码。例如。当你的方法run第一次在第一个热循环中编译时，JIT还不知道System.out.println。一旦执行到达println，早期编译的代码很可能会被去优化。
方法越大 - 为JIT编译器优化它就越困难。例如。可能看起来没有足够的备用寄存器来保存所有局部变量。那就是你的情况。

总而言之，您的基准测试似乎通过以下方案：

第一个热循环（addStatic）触发run方法的编译。除addStatic方法外，执行配置文件不知道任何内容。
System.out.println触发去优化，之后第二个热循环（addDynamic）导致重新编译run方法。
现在执行配置文件仅包含有关addDynamic的信息，因此JIT优化了第二个循环，第一个循环似乎有额外的寄存器溢出：

优化循环：

0x0000000002d01054: add    %rbx,%r14
0x0000000002d01057: add    $0x1,%rbx          ;*ladd
                                              ; - TestPerformanceOfStaticVsDynamicCalls::addDynamic@2
                                              ; - TestPerformanceOfStaticVsDynamicCalls::run@105

0x0000000002d0105b: add    $0x1,%r14          ; OopMap{rbp=Oop off=127}
                                              ;*goto
                                              ; - TestPerformanceOfStaticVsDynamicCalls::run@116

0x0000000002d0105f: test   %eax,-0x1c91065(%rip)        # 0x0000000001070000
                                              ;*lload
                                              ; - TestPerformanceOfStaticVsDynamicCalls::run@92
                                              ;   {poll}
0x0000000002d01065: cmp    $0x3b9aca00,%rbx
0x0000000002d0106c: jl     0x0000000002d01054

使用额外的寄存器溢出循环：

0x0000000002d011d0: mov    0x28(%rsp),%r11  <---- the problem is here
0x0000000002d011d5: add    %r10,%r11
0x0000000002d011d8: add    $0x1,%r10
0x0000000002d011dc: add    $0x1,%r11
0x0000000002d011e0: mov    %r11,0x28(%rsp)    ;*ladd
                                              ; - TestPerformanceOfStaticVsDynamicCalls::addStatic@2
                                              ; - TestPerformanceOfStaticVsDynamicCalls::run@33

0x0000000002d011e5: mov    0x28(%rsp),%r11  <---- the problem is here
0x0000000002d011ea: add    $0x1,%r11          ; OopMap{[32]=Oop off=526}
                                              ;*goto
                                              ; - TestPerformanceOfStaticVsDynamicCalls::run@44

0x0000000002d011ee: test   %eax,-0x1c911f4(%rip)        # 0x0000000001070000
                                              ;*goto
                                              ; - TestPerformanceOfStaticVsDynamicCalls::run@44
                                              ;   {poll}
0x0000000002d011f4: cmp    $0x3b9aca00,%r10
0x0000000002d011fb: jl     0x0000000002d011d0  ;*ifge
                                              ; - TestPerformanceOfStaticVsDynamicCalls::run@25

P.S。以下JVM选项对于分析JIT编译非常有用：

-XX:+PrintCompilation -XX:+UnlockDiagnosticVMOptions -XX:+PrintInlining -XX:+PrintAssembly -XX:CompileOnly=TestPerformanceOfStaticVsDynamicCalls

Answer 2

看起来Java正在向变量r添加值。我做了一些更改，添加了方法run2()：

public class TestPerformanceOfStaticVsDynamicCalls {
    private static final long RUNS = 1_000_000_000L;
    public static void main(String[] args) {
        System.out.println("Test run 1 =================================");
        new TestPerformanceOfStaticVsDynamicCalls().run();
        System.out.println("Test run 2 =================================");
        new TestPerformanceOfStaticVsDynamicCalls().run2();
    }
    private void run2() {
        long r = 0;
        long start, end;
        for (int loop = 0; loop < 10; loop++) {
            // Benchmark
            long stat = 0;
            start = System.currentTimeMillis();
            for (long i = 0; i < RUNS; i++) {
                stat += addStatic(1, i);
            }
            end = System.currentTimeMillis();
            System.out.println("Static: " + (end - start) + " ms");
            long dyna = 0;
            start = System.currentTimeMillis();
            for (long i = 0; i < RUNS; i++) {
                dyna += addDynamic(1, i);
            }
            end = System.currentTimeMillis();
            System.out.println("Dynamic: " + (end - start) + " ms");
            // If you really want to have values in "r" then...
            r += stat + dyna;
            // Do something with r to keep compiler happy
            System.out.println(r);
        }
    }
    private void run() {
        long r = 0;
        long start, end;
        for (int loop = 0; loop < 10; loop++) {
            // Benchmark
            start = System.currentTimeMillis();
            for (long i = 0; i < RUNS; i++) {
                r += addStatic(1, i);
            }
            end = System.currentTimeMillis();
            System.out.println("Static: " + (end - start) + " ms");
            start = System.currentTimeMillis();
            for (long i = 0; i < RUNS; i++) {
                r += addDynamic(1, i);
            }
            end = System.currentTimeMillis();
            System.out.println("Dynamic: " + (end - start) + " ms");
            // If you really want to have values in "r" then...
            // Do something with r to keep compiler happy
            System.out.println(r);
        }
    }
    private long addDynamic(long a, long b) {
        return a + b;
    }
    private static long addStatic(long a, long b) {
        return a + b;
    }
}

结果是：

Test run 1 =================================
Static: 582 ms
Dynamic: 579 ms
1000000001000000000
Static: 2065 ms
Dynamic: 2352 ms
2000000002000000000
Static: 2084 ms
Dynamic: 2345 ms
3000000003000000000
Static: 2095 ms
Dynamic: 2347 ms
4000000004000000000
Static: 2102 ms
Dynamic: 2338 ms
5000000005000000000
Static: 2073 ms
Dynamic: 2345 ms
6000000006000000000
Static: 2074 ms
Dynamic: 2341 ms
7000000007000000000
Static: 2102 ms
Dynamic: 2355 ms
8000000008000000000
Static: 2062 ms
Dynamic: 2354 ms
9000000009000000000
Static: 2057 ms
Dynamic: 2350 ms
-8446744063709551616
Test run 2 =================================
Static: 584 ms
Dynamic: 582 ms
1000000001000000000
Static: 587 ms
Dynamic: 577 ms
2000000002000000000
Static: 577 ms
Dynamic: 579 ms
3000000003000000000
Static: 577 ms
Dynamic: 577 ms
4000000004000000000
Static: 578 ms
Dynamic: 579 ms
5000000005000000000
Static: 578 ms
Dynamic: 580 ms
6000000006000000000
Static: 577 ms
Dynamic: 579 ms
7000000007000000000
Static: 578 ms
Dynamic: 577 ms
8000000008000000000
Static: 580 ms
Dynamic: 578 ms
9000000009000000000
Static: 576 ms
Dynamic: 579 ms
-8446744063709551616

至于为什么直接添加到r，我没有任何线索。也许有人可以提供更多见解，了解为什么在r内访问loop block会让事情变得更慢。

Answer 3

只需一个注释。如果我对long和r使用i，我只能观察到这种奇怪的行为。如果我将它们转换为int，那么我会得到这些时间：

Static: 352 ms
Dynamic: 353 ms
Static: 348 ms
Dynamic: 349 ms
Static: 349 ms
Dynamic: 348 ms
Static: 349 ms
Dynamic: 344 ms

所以一个可能的结论是在这些情况下避免long。至少在Linux / Amd64 Java7中，性能很重要。

一些运行/ JIT的故障后Java循环变慢了？

3 个答案: