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TA貢獻1829條經驗 獲得超6個贊
好吧,我將很快提供CPU如何轉換并更新帖子,但與此同時,您正在查看太小的差異而無法關心。
Java中的字節碼并不表示方法的執行速度(或不執行),有兩個JIT編譯器一旦足夠熱,它們將使此方法看起來完全不同。眾所周知,一旦編譯代碼,就會進行很少的優化,真正的優化來自。javacJIT
我已經為此進行了一些測試,要么只使用編譯器,要么用替換,要么根本不使用...(下面有很多測試代碼,你可以跳過它,只看結果,這是使用btw完成的)。此代碼使用的是JMH - 在微基準測試的java世界中使用的事實上的工具(如果手動完成,則容易出錯)。JMHC1C2GraalVMJITjdk-12
@Warmup(iterations = 10)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
@Measurement(iterations = 2, time = 2, timeUnit = TimeUnit.SECONDS)
public class BooleanCompare {
public static void main(String[] args) throws Exception {
Options opt = new OptionsBuilder()
.include(BooleanCompare.class.getName())
.build();
new Runner(opt).run();
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(1)
public boolean xor(BooleanExecutionPlan plan) {
return plan.booleans()[0] ^ plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(1)
public boolean plain(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(value = 1, jvmArgsAppend = "-Xint")
public boolean xorNoJIT(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(value = 1, jvmArgsAppend = "-Xint")
public boolean plainNoJIT(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(value = 1, jvmArgsAppend = "-XX:-TieredCompilation")
public boolean xorC2Only(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(value = 1, jvmArgsAppend = "-XX:-TieredCompilation")
public boolean plainC2Only(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(value = 1, jvmArgsAppend = "-XX:TieredStopAtLevel=1")
public boolean xorC1Only(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(value = 1, jvmArgsAppend = "-XX:TieredStopAtLevel=1")
public boolean plainC1Only(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(value = 1,
jvmArgsAppend = {
"-XX:+UnlockExperimentalVMOptions",
"-XX:+EagerJVMCI",
"-Dgraal.ShowConfiguration=info",
"-XX:+UseJVMCICompiler",
"-XX:+EnableJVMCI"
})
public boolean xorGraalVM(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
@Benchmark
@BenchmarkMode(Mode.AverageTime)
@Fork(value = 1,
jvmArgsAppend = {
"-XX:+UnlockExperimentalVMOptions",
"-XX:+EagerJVMCI",
"-Dgraal.ShowConfiguration=info",
"-XX:+UseJVMCICompiler",
"-XX:+EnableJVMCI"
})
public boolean plainGraalVM(BooleanExecutionPlan plan) {
return plan.booleans()[0] != plan.booleans()[1];
}
}
結果:
BooleanCompare.plain avgt 2 3.125 ns/op
BooleanCompare.xor avgt 2 2.976 ns/op
BooleanCompare.plainC1Only avgt 2 3.400 ns/op
BooleanCompare.xorC1Only avgt 2 3.379 ns/op
BooleanCompare.plainC2Only avgt 2 2.583 ns/op
BooleanCompare.xorC2Only avgt 2 2.685 ns/op
BooleanCompare.plainGraalVM avgt 2 2.980 ns/op
BooleanCompare.xorGraalVM avgt 2 3.868 ns/op
BooleanCompare.plainNoJIT avgt 2 243.348 ns/op
BooleanCompare.xorNoJIT avgt 2 201.342 ns/op
我不是一個多才多藝的人來閱讀匯編程序,盡管我有時喜歡這樣做......這里有一些有趣的事情。如果我們這樣做:
C1 編譯器僅包含 !=
/*
* run many iterations of this with :
* java -XX:+UnlockDiagnosticVMOptions
* -XX:TieredStopAtLevel=1
* "-XX:CompileCommand=print,com/so/BooleanCompare.compare"
* com.so.BooleanCompare
*/
public static boolean compare(boolean left, boolean right) {
return left != right;
}
我們得到:
0x000000010d1b2bc7: push %rbp
0x000000010d1b2bc8: sub $0x30,%rsp ;*iload_0 {reexecute=0 rethrow=0 return_oop=0}
; - com.so.BooleanCompare::compare@0 (line 22)
0x000000010d1b2bcc: cmp %edx,%esi
0x000000010d1b2bce: mov $0x0,%eax
0x000000010d1b2bd3: je 0x000000010d1b2bde
0x000000010d1b2bd9: mov $0x1,%eax
0x000000010d1b2bde: and $0x1,%eax
0x000000010d1b2be1: add $0x30,%rsp
0x000000010d1b2be5: pop %rbp
對我來說,這個代碼有點明顯:把0放進去,->如果不等于的話,把1放進去。返回。eaxcompare (edx, esi)eaxeax & 1
帶有 ^的 C1 編譯器:
public static boolean compare(boolean left, boolean right) {
return left ^ right;
}
# parm0: rsi = boolean
# parm1: rdx = boolean
# [sp+0x40] (sp of caller)
0x000000011326e5c0: mov %eax,-0x14000(%rsp)
0x000000011326e5c7: push %rbp
0x000000011326e5c8: sub $0x30,%rsp ;*iload_0 {reexecute=0 rethrow=0 return_oop=0}
; - com.so.BooleanCompare::compare@0 (line 22)
0x000000011326e5cc: xor %rdx,%rsi
0x000000011326e5cf: and $0x1,%esi
0x000000011326e5d2: mov %rsi,%rax
0x000000011326e5d5: add $0x30,%rsp
0x000000011326e5d9: pop %rbp
我真的不知道為什么這里需要,否則我想這也相當簡單。and $0x1,%esi
但是如果我啟用C2編譯器,事情會更有趣。
/**
* run with java
* -XX:+UnlockDiagnosticVMOptions
* -XX:CICompilerCount=2
* -XX:-TieredCompilation
* "-XX:CompileCommand=print,com/so/BooleanCompare.compare"
* com.so.BooleanCompare
*/
public static boolean compare(boolean left, boolean right) {
return left != right;
}
# parm0: rsi = boolean
# parm1: rdx = boolean
# [sp+0x20] (sp of caller)
0x000000011a2bbfa0: sub $0x18,%rsp
0x000000011a2bbfa7: mov %rbp,0x10(%rsp)
0x000000011a2bbfac: xor %r10d,%r10d
0x000000011a2bbfaf: mov $0x1,%eax
0x000000011a2bbfb4: cmp %edx,%esi
0x000000011a2bbfb6: cmove %r10d,%eax
0x000000011a2bbfba: add $0x10,%rsp
0x000000011a2bbfbe: pop %rbp
我甚至沒有看到經典的后記,而是通過以下方式看到一些非常不尋常的東西(至少對我來說):push ebp; mov ebp, esp; sub esp, x
sub $0x18,%rsp
mov %rbp,0x10(%rsp)
....
add $0x10,%rsp
pop %rbp
再一次,比我更全能的人可以解釋。否則,它就像一個更好的版本:C1
xor %r10d,%r10d // put zero into r10d
mov $0x1,%eax // put 1 into eax
cmp %edx,%esi // compare edx and esi
cmove %r10d,%eax // conditionally move the contents of r10d into eax
AFAIK比因為分支預測更好 - 這至少是我讀過的......cmp/cmovecmp/je
使用 C2 編譯器的異或:
public static boolean compare(boolean left, boolean right) {
return left ^ right;
}
0x000000010e6c9a20: sub $0x18,%rsp
0x000000010e6c9a27: mov %rbp,0x10(%rsp)
0x000000010e6c9a2c: xor %edx,%esi
0x000000010e6c9a2e: mov %esi,%eax
0x000000010e6c9a30: and $0x1,%eax
0x000000010e6c9a33: add $0x10,%rsp
0x000000010e6c9a37: pop %rbp
它看起來確實與編譯器生成的幾乎相同。C1
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