傻傻”的JAVA编译器

2024-01-08 05:05:33

故事是从一个问题开始的:为什么 Java 中 2 * (i * i) 比 2 * i * i更快?

猛地一看,我还以为有人在钓鱼,这俩玩意不应该是一模一样吗?第二反应是计算结果溢出了int值所以导致了这个差异,于是我掏出JMH这个利器准备开始一轮验证,为了避免干扰,构造了不同的测试用例集用于纵向和横向的比较。

@BenchmarkMode(Mode.AverageTime)                  // 测试方法平均执行时间
@OutputTimeUnit(TimeUnit.NANOSECONDS)             // 输出结果的时间粒度为纳秒
@State(Scope.Thread)                              // 运行相同测试的所有线程将共享实例。可以用来测试状态对象的多线程性能(或者仅标记该范围的基准)。
@Warmup(iterations = 2, time = 1)                 // 执行5遍预热
@Measurement(iterations = 10, time = 1)           // 执行5遍测试
@Fork(1)
public class CompileBenchMarkDemo {

  @Param({"1477", "1000000000"})
  private int size;

  public static void main(String[] args) throws RunnerException {
    Options opt = new OptionsBuilder().include(CompileBenchMarkDemo.class.getSimpleName())
        .build();
    new Runner(opt).run();
  }

  @Benchmark
  public int twoSquare() {
    int n = 0;
    for (int i = 0; i < size; i++) {
      n += 2 * (i * i);
    }
    return n;
  }

  @Benchmark
  public int twoNoSquare() {
    int n = 0;
    for (int i = 0; i < size; i++) {
      n += 2 * i * i;
    }
    return n;
  }

  @Benchmark
  public long twoSquareWithLong() {
    long n = 0;
    for (long i = 0; i < size; i++) {
      n += 2 * (i * i);
    }
    return n;
  }

  @Benchmark
  public long twoNoSquareWithLong() {
    long n = 0;
    for (long i = 0; i < size; i++) {
      n += 2 * i * i;
    }
    return n;
  }

}

可以看到为了避免int溢出的干扰我使用了long来做累加,还有将i的范围限制到1477保证int不溢出

openjdk version "1.8.0_382-internal"
OpenJDK Runtime Environment (build 1.8.0_382-internal-b05)
OpenJDK 64-Bit Server VM (build 25.382-b05, mixed mode)

在上述jdk版本下最终结果如下:

Benchmark                                     (size)  Mode  Cnt          Score          Error  Units
CompileBenchMarkDemo.twoSquare            1000000000  avgt   10  461970038.267 ± 13350368.542  ns/op
CompileBenchMarkDemo.twoNoSquare          1000000000  avgt   10  744365243.050 ± 37840802.126  ns/op
CompileBenchMarkDemo.twoSquareWithLong    1000000000  avgt   10  799079820.550 ± 12835540.327  ns/op
CompileBenchMarkDemo.twoNoSquareWithLong  1000000000  avgt   10  906846479.450 ± 41247832.592  ns/op
CompileBenchMarkDemo.twoSquare                  1477  avgt   10        717.674 ±       19.992  ns/op
CompileBenchMarkDemo.twoNoSquare                1477  avgt   10       1144.875 ±      100.242  ns/op
CompileBenchMarkDemo.twoSquareWithLong          1477  avgt   10        979.668 ±       58.368  ns/op
CompileBenchMarkDemo.twoNoSquareWithLong        1477  avgt   10       1208.143 ±       97.710  ns/op

结果出乎我的意料,没想到在溢出和不溢出的情况下2 * ( i * i ) 始终优于 2 * i * i

于是我打算从字节码一探究竟

 0 iconst_0
 1 istore_1
 2 iconst_0
 3 istore_2
 4 iload_2
 5 ldc #11 <1000000000>
 7 if_icmpge 24 (+17)
10 iload_1
11 iconst_2
12 iload_2
13 iload_2
14 imul
15 imul
16 iadd
17 istore_1
18 iinc 2 by 1
21 goto 4 (-17)
24 iload_1
25 ireturn
0 iconst_0
1 istore_1
2 iconst_0
3 istore_2
4 iload_2
5 ldc #11 <1000000000>
7 if_icmpge 24 (+17)
10 iload_1
11 iconst_2
12 iload_2
13 imul
14 iload_2
15 imul
16 iadd
17 istore_1
18 iinc 2 by 1
21 goto 4 (-17)
24 iload_1
25 ireturn

可以看到字节码上除了iload_2imul的顺序不一致所有字节码都是相同的,那么这个顺序为什么会有如此大的区别呢?字节码不行那就更下一层,让我们看看汇编代码的区别。

JIT 倾向于非常积极地展开小循环,我们观察到该2 * (i * i)案例展开了 16 倍:

030   B2: # B2 B3 <- B1 B2  Loop: B2-B2 inner main of N18 Freq: 1e+006
030     addl    R11, RBP    # int
033     movl    RBP, R13    # spill
036     addl    RBP, #14    # int
039     imull   RBP, RBP    # int
03c     movl    R9, R13 # spill
03f     addl    R9, #13 # int
043     imull   R9, R9  # int
047     sall    RBP, #1
049     sall    R9, #1
04c     movl    R8, R13 # spill
04f     addl    R8, #15 # int
053     movl    R10, R8 # spill
056     movdl   XMM1, R8    # spill
05b     imull   R10, R8 # int
05f     movl    R8, R13 # spill
062     addl    R8, #12 # int
066     imull   R8, R8  # int
06a     sall    R10, #1
06d     movl    [rsp + #32], R10    # spill
072     sall    R8, #1
075     movl    RBX, R13    # spill
078     addl    RBX, #11    # int
07b     imull   RBX, RBX    # int
07e     movl    RCX, R13    # spill
081     addl    RCX, #10    # int
084     imull   RCX, RCX    # int
087     sall    RBX, #1
089     sall    RCX, #1
08b     movl    RDX, R13    # spill
08e     addl    RDX, #8 # int
091     imull   RDX, RDX    # int
094     movl    RDI, R13    # spill
097     addl    RDI, #7 # int
09a     imull   RDI, RDI    # int
09d     sall    RDX, #1
09f     sall    RDI, #1
0a1     movl    RAX, R13    # spill
0a4     addl    RAX, #6 # int
0a7     imull   RAX, RAX    # int
0aa     movl    RSI, R13    # spill
0ad     addl    RSI, #4 # int
0b0     imull   RSI, RSI    # int
0b3     sall    RAX, #1
0b5     sall    RSI, #1
0b7     movl    R10, R13    # spill
0ba     addl    R10, #2 # int
0be     imull   R10, R10    # int
0c2     movl    R14, R13    # spill
0c5     incl    R14 # int
0c8     imull   R14, R14    # int
0cc     sall    R10, #1
0cf     sall    R14, #1
0d2     addl    R14, R11    # int
0d5     addl    R14, R10    # int
0d8     movl    R10, R13    # spill
0db     addl    R10, #3 # int
0df     imull   R10, R10    # int
0e3     movl    R11, R13    # spill
0e6     addl    R11, #5 # int
0ea     imull   R11, R11    # int
0ee     sall    R10, #1
0f1     addl    R10, R14    # int
0f4     addl    R10, RSI    # int
0f7     sall    R11, #1
0fa     addl    R11, R10    # int
0fd     addl    R11, RAX    # int
100     addl    R11, RDI    # int
103     addl    R11, RDX    # int
106     movl    R10, R13    # spill
109     addl    R10, #9 # int
10d     imull   R10, R10    # int
111     sall    R10, #1
114     addl    R10, R11    # int
117     addl    R10, RCX    # int
11a     addl    R10, RBX    # int
11d     addl    R10, R8 # int
120     addl    R9, R10 # int
123     addl    RBP, R9 # int
126     addl    RBP, [RSP + #32 (32-bit)]   # int
12a     addl    R13, #16    # int
12e     movl    R11, R13    # spill
131     imull   R11, R13    # int
135     sall    R11, #1
138     cmpl    R13, #999999985
13f     jl     B2   # loop end  P=1.000000 C=6554623.000000

我们看到有 1 个寄存器”溢出”到堆栈上。

对于2 * i * i版本:

05a   B3: # B2 B4 <- B1 B2  Loop: B3-B2 inner main of N18 Freq: 1e+006
05a     addl    RBX, R11    # int
05d     movl    [rsp + #32], RBX    # spill
061     movl    R11, R8 # spill
064     addl    R11, #15    # int
068     movl    [rsp + #36], R11    # spill
06d     movl    R11, R8 # spill
070     addl    R11, #14    # int
074     movl    R10, R9 # spill
077     addl    R10, #16    # int
07b     movdl   XMM2, R10   # spill
080     movl    RCX, R9 # spill
083     addl    RCX, #14    # int
086     movdl   XMM1, RCX   # spill
08a     movl    R10, R9 # spill
08d     addl    R10, #12    # int
091     movdl   XMM4, R10   # spill
096     movl    RCX, R9 # spill
099     addl    RCX, #10    # int
09c     movdl   XMM6, RCX   # spill
0a0     movl    RBX, R9 # spill
0a3     addl    RBX, #8 # int
0a6     movl    RCX, R9 # spill
0a9     addl    RCX, #6 # int
0ac     movl    RDX, R9 # spill
0af     addl    RDX, #4 # int
0b2     addl    R9, #2  # int
0b6     movl    R10, R14    # spill
0b9     addl    R10, #22    # int
0bd     movdl   XMM3, R10   # spill
0c2     movl    RDI, R14    # spill
0c5     addl    RDI, #20    # int
0c8     movl    RAX, R14    # spill
0cb     addl    RAX, #32    # int
0ce     movl    RSI, R14    # spill
0d1     addl    RSI, #18    # int
0d4     movl    R13, R14    # spill
0d7     addl    R13, #24    # int
0db     movl    R10, R14    # spill
0de     addl    R10, #26    # int
0e2     movl    [rsp + #40], R10    # spill
0e7     movl    RBP, R14    # spill
0ea     addl    RBP, #28    # int
0ed     imull   RBP, R11    # int
0f1     addl    R14, #30    # int
0f5     imull   R14, [RSP + #36 (32-bit)]   # int
0fb     movl    R10, R8 # spill
0fe     addl    R10, #11    # int
102     movdl   R11, XMM3   # spill
107     imull   R11, R10    # int
10b     movl    [rsp + #44], R11    # spill
110     movl    R10, R8 # spill
113     addl    R10, #10    # int
117     imull   RDI, R10    # int
11b     movl    R11, R8 # spill
11e     addl    R11, #8 # int
122     movdl   R10, XMM2   # spill
127     imull   R10, R11    # int
12b     movl    [rsp + #48], R10    # spill
130     movl    R10, R8 # spill
133     addl    R10, #7 # int
137     movdl   R11, XMM1   # spill
13c     imull   R11, R10    # int
140     movl    [rsp + #52], R11    # spill
145     movl    R11, R8 # spill
148     addl    R11, #6 # int
14c     movdl   R10, XMM4   # spill
151     imull   R10, R11    # int
155     movl    [rsp + #56], R10    # spill
15a     movl    R10, R8 # spill
15d     addl    R10, #5 # int
161     movdl   R11, XMM6   # spill
166     imull   R11, R10    # int
16a     movl    [rsp + #60], R11    # spill
16f     movl    R11, R8 # spill
172     addl    R11, #4 # int
176     imull   RBX, R11    # int
17a     movl    R11, R8 # spill
17d     addl    R11, #3 # int
181     imull   RCX, R11    # int
185     movl    R10, R8 # spill
188     addl    R10, #2 # int
18c     imull   RDX, R10    # int
190     movl    R11, R8 # spill
193     incl    R11 # int
196     imull   R9, R11 # int
19a     addl    R9, [RSP + #32 (32-bit)]    # int
19f     addl    R9, RDX # int
1a2     addl    R9, RCX # int
1a5     addl    R9, RBX # int
1a8     addl    R9, [RSP + #60 (32-bit)]    # int
1ad     addl    R9, [RSP + #56 (32-bit)]    # int
1b2     addl    R9, [RSP + #52 (32-bit)]    # int
1b7     addl    R9, [RSP + #48 (32-bit)]    # int
1bc     movl    R10, R8 # spill
1bf     addl    R10, #9 # int
1c3     imull   R10, RSI    # int
1c7     addl    R10, R9 # int
1ca     addl    R10, RDI    # int
1cd     addl    R10, [RSP + #44 (32-bit)]   # int
1d2     movl    R11, R8 # spill
1d5     addl    R11, #12    # int
1d9     imull   R13, R11    # int
1dd     addl    R13, R10    # int
1e0     movl    R10, R8 # spill
1e3     addl    R10, #13    # int
1e7     imull   R10, [RSP + #40 (32-bit)]   # int
1ed     addl    R10, R13    # int
1f0     addl    RBP, R10    # int
1f3     addl    R14, RBP    # int
1f6     movl    R10, R8 # spill
1f9     addl    R10, #16    # int
1fd     cmpl    R10, #999999985
204     jl     B2   # loop end  P=1.000000 C=7419903.000000

在这里,由于需要保留更多的中间结果,我们观察到更多的“溢出”和对堆栈的更多访问。

因此,问题的答案很简单:2 * (i * i)2 * i * i更快,因为 JIT 生成了更优化的汇编代码。

Java的JIT是个非常有价值的东西,但有的时候它也可能“犯傻”,我们在平时写代码的过程中对于这些点倒也无需刻意去记忆,这本该是编译器自己要做的事情,祝愿Java的编译器越来越好吧。

原文链接:https://pebble-skateboard-d46.notion.site/JAVA-a91ca0b1305e49918efcdd0035a7a6e6

参考资料

https://stackoverflow.com/questions/53452713/why-is-2-i-i-faster-than-2-i-i-in-java

文章来源:https://blog.csdn.net/Ares_xb/article/details/135375630
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