英文:
Will jvm evaluate a obvious return value on each execution?
问题
我确信几乎每个人都熟悉在SO上标记为Java的最低票数问题。为了完整起见，复制粘贴片段如下：
k = (j = (i = 0) + 2) + 1;
return i|= j|= k|= (j+= i) - - (k+++k) - - (i =+j);
无论如何，上述片段始终返回11。因此，我的问题是：JVM是否会在每次调用时计算这种/类似的荒谬表达式？
英文:
I am sure almost everybody is familiar with the most downvoted question(java tagged) on SO. Copy pasting the snippet for completeness:
k = (j = (i = 0) + 2) + 1;
return i|= j|= k|= (j+= i) - - (k+++k) - - (i =+j);
Above snippet always returns 11 no matter what. So my question is: will jvm evaluate this/similar madness on each invocation?
答案1
得分: 2
我不知道这是否算作一个答案，但似乎 JVM 可以证明不需要进行评估：
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
@Warmup(iterations = 20, time = 20)
@Measurement(iterations = 20, time = 20)
public class MostDownVoted {
public static void main(String[] args) throws Exception {
Options opt = new OptionsBuilder()
.include(MostDownVoted.class.getSimpleName())
.build();
new Runner(opt).run();
}
@State(Scope.Benchmark)
public static class Holder {
int k;
int i;
int j;
@Setup(Level.Iteration)
public void setUp() {
k = ThreadLocalRandom.current().nextInt();
i = ThreadLocalRandom.current().nextInt();
j = ThreadLocalRandom.current().nextInt();
}
}
@Fork(1)
@Benchmark
public int test1(Holder h) {
h.k = (h.j = (h.i = 0) + 2) + 1;
return h.i |= h.j |= h.k |= (h.j += h.i) - -(h.k++ + h.k) - -(h.i = +h.j);
}
@Fork(1)
@Benchmark
public int test2(Holder h) {
return 11;
}
@Benchmark
@Fork(value = 1, jvmArgsAppend = "-XX:TieredStopAtLevel=1")
public int test3(Holder h) {
h.k = (h.j = (h.i = 0) + 2) + 1;
return h.i |= h.j |= h.k |= (h.j += h.i) - -(h.k++ + h.k) - -(h.i = +h.j);
}
}
结果显示，一旦 C2 编译器介入，与 return 11 的结果相当：
MostDownVoted.test1 avgt 20 2.816 ± 0.003 ns/op
MostDownVoted.test2 avgt 20 2.122 ± 0.016 ns/op
MostDownVoted.test3 avgt 20 3.979 ± 0.758 ns/op
英文:
I don't know if this counts as an answer, but it seems that the JVM can prove that no evaluation is needed:
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
@Warmup(iterations = 20, time = 20)
@Measurement(iterations = 20, time = 20)
public class MostDownVoted {
public static void main(String[] args) throws Exception {
Options opt = new OptionsBuilder()
.include(MostDownVoted.class.getSimpleName())
.build();
new Runner(opt).run();
}
@State(Scope.Benchmark)
public static class Holder {
int k;
int i;
int j;
@Setup(Level.Iteration)
public void setUp() {
k = ThreadLocalRandom.current().nextInt();
i = ThreadLocalRandom.current().nextInt();
j = ThreadLocalRandom.current().nextInt();
}
}
@Fork(1)
@Benchmark
public int test1(Holder h) {
h.k = (h.j = (h.i = 0) + 2) + 1;
return h.i |= h.j |= h.k |= (h.j += h.i) - -(h.k++ + h.k) - -(h.i = +h.j);
}
@Fork(1)
@Benchmark
public int test2(Holder h) {
return 11;
}
@Benchmark
@Fork(value = 1, jvmArgsAppend = "-XX:TieredStopAtLevel=1")
public int test3(Holder h) {
h.k = (h.j = (h.i = 0) + 2) + 1;
return h.i |= h.j |= h.k |= (h.j += h.i) - -(h.k++ + h.k) - -(h.i = +h.j);
}
}
The results show that once C2 compiler kicks in, the results with return 11 are on par with what you have :
MostDownVoted.test1 avgt 20 2.816 ± 0.003 ns/op
MostDownVoted.test2 avgt 20 2.122 ± 0.016 ns/op
MostDownVoted.test3 avgt 20 3.979 ± 0.758 ns/op
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