在 Intel Core i5
处理器和 Ubuntu 操作系统下运行基准测试。
java version "1.8.0_144"
Java(TM) SE Runtime Environment (build 1.8.0_144-b01)
Java HotSpot(TM) 64-Bit Server VM (build 25.144-b01, mixed mode)
我正在比较Collectors.counting
和Collectors.summingLong(x -> 1L)
的性能。以下是基准测试:
public List<Integer> ints = new ArrayList<>();
Collector<Integer, ?, Long> counting = Collectors.counting();
Collector<Integer, ?, Long> summingLong = Collectors.summingLong(x -> 1L);
@Setup
public void setup() throws Exception{
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
ints.add(new Random().nextInt(1000));
}
@Benchmark
@OutputTimeUnit(TimeUnit.NANOSECONDS)
@BenchmarkMode(Mode.AverageTime)
public Long counting() {
return ints.stream().collect(counting);
}
@Benchmark
@OutputTimeUnit(TimeUnit.NANOSECONDS)
@BenchmarkMode(Mode.AverageTime)
public Long summingLong() {
return ints.stream().collect(summingLong);
}
我得到的结果是使用Collectors.counting
比Collectors.summingLong
慢了3倍。
于是我使用-prof perfnorm
并进行了25次分支测试。以下是结果:
Benchmark Mode Cnt Score Error Units
MyBenchmark.counting avgt 125 87.115 ± 3.787 ns/op
MyBenchmark.counting:CPI avgt 25 0.310 ± 0.011 #/op
MyBenchmark.counting:L1-dcache-load-misses avgt 25 1.963 ± 0.171 #/op
MyBenchmark.counting:L1-dcache-loads avgt 25 258.716 ± 41.458 #/op
MyBenchmark.counting:L1-dcache-store-misses avgt 25 1.890 ± 0.168 #/op
MyBenchmark.counting:L1-dcache-stores avgt 25 131.344 ± 16.433 #/op
MyBenchmark.counting:L1-icache-load-misses avgt 25 0.035 ± 0.007 #/op
MyBenchmark.counting:LLC-loads avgt 25 0.411 ± 0.143 #/op
MyBenchmark.counting:LLC-stores avgt 25 0.007 ± 0.001 #/op
MyBenchmark.counting:branch-misses avgt 25 0.029 ± 0.017 #/op
MyBenchmark.counting:branches avgt 25 139.669 ± 21.901 #/op
MyBenchmark.counting:cycles avgt 25 261.967 ± 29.392 #/op
MyBenchmark.counting:dTLB-load-misses avgt 25 0.036 ± 0.003 #/op
MyBenchmark.counting:dTLB-loads avgt 25 258.111 ± 41.008 #/op
MyBenchmark.counting:dTLB-store-misses avgt 25 0.001 ± 0.001 #/op
MyBenchmark.counting:dTLB-stores avgt 25 131.394 ± 16.166 #/op
MyBenchmark.counting:iTLB-load-misses avgt 25 0.001 ± 0.001 #/op
MyBenchmark.counting:iTLB-loads avgt 25 0.001 ± 0.001 #/op
MyBenchmark.counting:instructions avgt 25 850.262 ± 113.228 #/op
MyBenchmark.counting:stalled-cycles-frontend avgt 25 48.493 ± 8.968 #/op
MyBenchmark.summingLong avgt 125 37.238 ± 0.194 ns/op
MyBenchmark.summingLong:CPI avgt 25 0.311 ± 0.002 #/op
MyBenchmark.summingLong:L1-dcache-load-misses avgt 25 1.793 ± 0.013 #/op
MyBenchmark.summingLong:L1-dcache-loads avgt 25 93.785 ± 0.640 #/op
MyBenchmark.summingLong:L1-dcache-store-misses avgt 25 1.727 ± 0.013 #/op
MyBenchmark.summingLong:L1-dcache-stores avgt 25 56.249 ± 0.408 #/op
MyBenchmark.summingLong:L1-icache-load-misses avgt 25 0.020 ± 0.003 #/op
MyBenchmark.summingLong:LLC-loads avgt 25 0.843 ± 0.117 #/op
MyBenchmark.summingLong:LLC-stores avgt 25 0.004 ± 0.001 #/op
MyBenchmark.summingLong:branch-misses avgt 25 0.008 ± 0.002 #/op
MyBenchmark.summingLong:branches avgt 25 61.472 ± 0.260 #/op
MyBenchmark.summingLong:cycles avgt 25 110.949 ± 0.784 #/op
MyBenchmark.summingLong:dTLB-load-misses avgt 25 0.031 ± 0.001 #/op
MyBenchmark.summingLong:dTLB-loads avgt 25 93.662 ± 0.616 #/op
MyBenchmark.summingLong:dTLB-store-misses avgt 25 ≈ 10⁻³ #/op
MyBenchmark.summingLong:dTLB-stores avgt 25 56.302 ± 0.351 #/op
MyBenchmark.summingLong:iTLB-load-misses avgt 25 0.001 ± 0.001 #/op
MyBenchmark.summingLong:iTLB-loads avgt 25 ≈ 10⁻³ #/op
MyBenchmark.summingLong:instructions avgt 25 357.029 ± 1.712 #/op
MyBenchmark.summingLong:stalled-cycles-frontend avgt 25 10.074 ± 1.096 #/op
我注意到的是:
分支、指令和周期差别几乎达到了3倍。缓存操作也是如此。分支似乎预测得很好,而且缓存未命中也不太多(这只是我的看法)。
因此,问题可能出在编译代码上。使用
-prof perfasm
运行它(太长了,无法放在这里)。在编译代码中,我注意到了以下内容: I.
Collectors.summingLong
assembly
我们有3个循环迭代数组并计数。第一个只计算一个元素。0x00007f9abd226dfd: mov %edi,%ebp ;contains the iteration index
incl %ebp
;...
0x00007f9abd226e27: incl %edi
0x00007f9abd226e29: cmp %ebp,%edi
0x00007f9abd226e2b: jnl 0x7f9abd226e34 ;exit after the first iteration
每次迭代需要4个元素(这是循环展开吗?),并且在第一个元素后退出。
0x00007f9abd226ea6: add $0x1,%rsi
;...
0x00007f9abd226ed0: add $0x2,%rsi
;...
0x00007f9abd226efa: add $0x3,%rsi
;...
0x00007f9abd226f1c: add $0x4,%rbx
;...
0x00007f9abd226f20: mov %rbx,0x10(%r14)
第三项计算其余元素。
II。 Collectors.counting
汇编
我们仅有一个循环以逐个计算所有元素(未展开)。此外,我们在计数结果的循环内联了装箱转换。另外,在循环中似乎没有内联装箱转换。
0x00007f80dd22dfb5: mov $0x1,%esi
0x00007f80dd22dfba: nop
0x00007f80dd22dfbb: callq 0x7f80dd046420
看起来这段代码对lambda表达式中提供的1L
进行了装箱操作,但是原因并不清楚。实际执行加法时,我们有以下代码:
0x00007f80dd22dfec: mov $0x1,%r10d
0x00007f80dd22dff2: add 0x10(%r11),%r10
此外,我们将计数结果存储在栈内
mov %r10d,0x10(%rsp)
,而不是像第一种情况那样存储在堆内。如果我正确理解了正在发生的事情,我的问题是:循环展开和装箱转换导致了3倍的性能下降吗?如果是这样,为什么在counting
情况下运行时没有展开循环?请注意,Collectors.summingLong
的GC压力比Collectors.counting
高2.5倍。这并不是很清楚(我只能猜测我们在Collectors.counting
中将中间值存储在栈中)。MyBenchmark.counting avgt 5 96.956 ± 4.412 ns/op
MyBenchmark.counting:·gc.alloc.rate avgt 5 734.538 ± 33.083 MB/sec
MyBenchmark.counting:·gc.alloc.rate.norm avgt 5 112.000 ± 0.001 B/op
MyBenchmark.counting:·gc.churn.PS_Eden_Space avgt 5 731.423 ± 340.767 MB/sec
MyBenchmark.counting:·gc.churn.PS_Eden_Space.norm avgt 5 111.451 ± 48.411 B/op
MyBenchmark.counting:·gc.churn.PS_Survivor_Space avgt 5 0.017 ± 0.067 MB/sec
MyBenchmark.counting:·gc.churn.PS_Survivor_Space.norm avgt 5 0.003 ± 0.010 B/op
MyBenchmark.counting:·gc.count avgt 5 16.000 counts
MyBenchmark.counting:·gc.time avgt 5 12.000 ms
MyBenchmark.summingLong avgt 5 38.371 ± 1.733 ns/op
MyBenchmark.summingLong:·gc.alloc.rate avgt 5 1856.581 ± 81.706 MB/sec
MyBenchmark.summingLong:·gc.alloc.rate.norm avgt 5 112.000 ± 0.001 B/op
MyBenchmark.summingLong:·gc.churn.PS_Eden_Space avgt 5 1876.736 ± 192.503 MB/sec
MyBenchmark.summingLong:·gc.churn.PS_Eden_Space.norm avgt 5 113.213 ± 9.916 B/op
MyBenchmark.summingLong:·gc.churn.PS_Survivor_Space avgt 5 0.033 ± 0.072 MB/sec
MyBenchmark.summingLong:·gc.churn.PS_Survivor_Space.norm avgt 5 0.002 ± 0.004 B/op
MyBenchmark.summingLong:·gc.count avgt 5 62.000 counts
MyBenchmark.summingLong:·gc.time avgt 5 48.000 ms
final Long one = Long.valueOf(1L); Collector<Integer, ?, Long> counting = Collectors.reducing(0L, e -> one, Long::sum);
并查看是否有所不同?这段代码是Collectors.counting(...)
的实现,它事先装箱了1L
。请看它是否有所区别。 - Sergey KalinichenkoCollectors.counting
的实现,并用e -> one
替换了e -> 1L
,其中one
是预装箱的值1L
。 - Sergey Kalinichenkocounting
方法返回的收集器仅供作为下游收集器使用。通过Stream
的count
方法直接获得相同的功能,因此**没有理由使用collect(counting())
**。”换句话说,你不应该这样做 :) 可能是因为你在这里所描述的原因。 - Andy TurnerjvmArgs
来运行每个基准测试,并尽可能地分配更多堆内存。至于gdb
,我从未试过,所以这超出了我的能力范围 :( - EugeneCollectors.counting
的情况下运行时没有展开循环? - St.Antario