3 回答

TA貢獻(xiàn)2021條經(jīng)驗(yàn) 獲得超8個(gè)贊
讓我們嘗試使用JMH復(fù)制。
@Benchmark
@Measurement(timeUnit = TimeUnit.NANOSECONDS)
@BenchmarkMode(Mode.AverageTime)
public int first() throws IOException {
return i % 2;
}
@Benchmark
@Measurement(timeUnit = TimeUnit.NANOSECONDS)
@BenchmarkMode(Mode.AverageTime)
public int second() throws IOException {
return i & 0x1;
}
好的,它是可復(fù)制的。的first速度略慢于second。現(xiàn)在讓我們找出原因。使用以下命令運(yùn)行它-prof perfnorm:
Benchmark Mode Cnt Score Error Units
MyBenchmark.first avgt 50 2.674 ± 0.028 ns/op
MyBenchmark.first:CPI avgt 10 0.301 ± 0.002 #/op
MyBenchmark.first:L1-dcache-load-misses avgt 10 0.001 ± 0.001 #/op
MyBenchmark.first:L1-dcache-loads avgt 10 11.011 ± 0.146 #/op
MyBenchmark.first:L1-dcache-stores avgt 10 3.011 ± 0.034 #/op
MyBenchmark.first:L1-icache-load-misses avgt 10 ≈ 10?3 #/op
MyBenchmark.first:LLC-load-misses avgt 10 ≈ 10?? #/op
MyBenchmark.first:LLC-loads avgt 10 ≈ 10?? #/op
MyBenchmark.first:LLC-store-misses avgt 10 ≈ 10?? #/op
MyBenchmark.first:LLC-stores avgt 10 ≈ 10?? #/op
MyBenchmark.first:branch-misses avgt 10 ≈ 10?? #/op
MyBenchmark.first:branches avgt 10 4.006 ± 0.054 #/op
MyBenchmark.first:cycles avgt 10 9.322 ± 0.113 #/op
MyBenchmark.first:dTLB-load-misses avgt 10 ≈ 10?? #/op
MyBenchmark.first:dTLB-loads avgt 10 10.939 ± 0.175 #/op
MyBenchmark.first:dTLB-store-misses avgt 10 ≈ 10?? #/op
MyBenchmark.first:dTLB-stores avgt 10 2.991 ± 0.045 #/op
MyBenchmark.first:iTLB-load-misses avgt 10 ≈ 10?? #/op
MyBenchmark.first:iTLB-loads avgt 10 ≈ 10?? #/op
MyBenchmark.first:instructions avgt 10 30.991 ± 0.427 #/op
MyBenchmark.second avgt 50 2.263 ± 0.015 ns/op
MyBenchmark.second:CPI avgt 10 0.320 ± 0.001 #/op
MyBenchmark.second:L1-dcache-load-misses avgt 10 0.001 ± 0.001 #/op
MyBenchmark.second:L1-dcache-loads avgt 10 11.045 ± 0.152 #/op
MyBenchmark.second:L1-dcache-stores avgt 10 3.014 ± 0.032 #/op
MyBenchmark.second:L1-icache-load-misses avgt 10 ≈ 10?3 #/op
MyBenchmark.second:LLC-load-misses avgt 10 ≈ 10?? #/op
MyBenchmark.second:LLC-loads avgt 10 ≈ 10?? #/op
MyBenchmark.second:LLC-store-misses avgt 10 ≈ 10?? #/op
MyBenchmark.second:LLC-stores avgt 10 ≈ 10?? #/op
MyBenchmark.second:branch-misses avgt 10 ≈ 10?? #/op
MyBenchmark.second:branches avgt 10 4.014 ± 0.045 #/op
MyBenchmark.second:cycles avgt 10 8.024 ± 0.098 #/op
MyBenchmark.second:dTLB-load-misses avgt 10 ≈ 10?? #/op
MyBenchmark.second:dTLB-loads avgt 10 10.989 ± 0.161 #/op
MyBenchmark.second:dTLB-store-misses avgt 10 ≈ 10?? #/op
MyBenchmark.second:dTLB-stores avgt 10 3.004 ± 0.042 #/op
MyBenchmark.second:iTLB-load-misses avgt 10 ≈ 10?? #/op
MyBenchmark.second:iTLB-loads avgt 10 ≈ 10?? #/op
MyBenchmark.second:instructions avgt 10 25.076 ± 0.296 #/op
注意周期和說(shuō)明上的差異?,F(xiàn)在,這很明顯。該first很關(guān)心的跡象,但second不(只按位與)。為了確保這是原因,請(qǐng)看一下程序集片段:
第一的:
0x00007f91111f8355: mov 0xc(%r10),%r11d ;*getfield i
0x00007f91111f8359: mov %r11d,%edx
0x00007f91111f835c: and $0x1,%edx
0x00007f91111f835f: mov %edx,%r10d
0x00007f6bd120a6e2: neg %r10d
0x00007f6bd120a6e5: test %r11d,%r11d
0x00007f6bd120a6e8: cmovl %r10d,%edx
第二:
0x00007ff36cbda580: mov $0x1,%edx
0x00007ff36cbda585: mov 0x40(%rsp),%r10
0x00007ff36cbda58a: and 0xc(%r10),%edx

TA貢獻(xiàn)2065條經(jīng)驗(yàn) 獲得超14個(gè)贊
這兩個(gè)操作對(duì)應(yīng)于不同的JVM處理器指令:
irem // int remainder (%) iand // bitwise and (&)
我讀過(guò)的某個(gè)地方irem
通常由JVM實(shí)現(xiàn),而iand
在硬件上可用。Oracle解釋了以下兩條指令:
iand
通過(guò)取值1和值2的按位與(連接)來(lái)計(jì)算int結(jié)果。
irem
整數(shù)結(jié)果為值1-(值1 /值2)*值2。
在我看來(lái),假設(shè)這樣做iand
會(huì)減少CPU周期,這似乎是合理的。
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