我正在Java 7中使用escape analysis进行一些测试,以便更好地了解哪些对象有资格进行堆栈分配。
这是我编写的用于测试堆栈分配的代码:
import java.util.ArrayList;
import java.util.Iterator;
public class EscapeAnalysis {
private static final long TIME_TO_TEST = 10L * 1000L; // 10s
static class Timestamp {
private long millis;
public Timestamp(long millis) {
this.millis = millis;
}
public long getTime() {
return millis;
}
public void setTime(long time) {
millis = time;
}
}
public static void main(String[] args) {
long r = 0;
System.out.println("test1");
r += test1();
System.out.println("test2");
r += test2();
System.out.println("test3");
r += test3();
System.out.println("test4");
r += test4();
System.out.println("test5");
r += test5();
System.out.println("test6");
r += test6();
System.out.println(r);
}
public static long test1() {
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
r += new Timestamp(System.currentTimeMillis()).getTime();
}
return r;
}
public static long test2() {
ArrayList<Integer> l = new ArrayList<Integer>(1000);
for (int i = 0; i < 1000; ++i) {
l.add(i);
}
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
for (Iterator<Integer> it = l.iterator(); it.hasNext(); ) {
r += it.next().longValue();
}
}
return r;
}
public static long test3() {
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
Timestamp ts = new Timestamp(System.currentTimeMillis());
ts.setTime(42);
r += ts.getTime();
}
return r;
}
public static long test4() {
ArrayList<Integer> l = new ArrayList<Integer>(1000);
for (int i = 0; i < 1000; ++i) {
l.add(i);
}
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
Iterator<Integer> it = l.iterator();
r += it.next().longValue();
r += it.next().longValue();
r += it.next().longValue();
r += it.next().longValue();
}
return r;
}
public static long test5() {
ArrayList<Integer> l = new ArrayList<Integer>(1000);
for (int i = 0; i < 1000; ++i) {
l.add(i);
}
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
Iterator<Integer> it = l.iterator();
for (int i = 0; i < l.size(); ++i) {
r += it.next().longValue();
}
}
return r;
}
public static long test6() {
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
for (Timestamp ts = new Timestamp(System.currentTimeMillis());
ts.getTime() > 0;
ts.setTime(ts.getTime() + System.currentTimeMillis())) {
r += ts.getTime();
}
}
return r;
}
}
以下是Linux上用Java 7输出的内容
java -server -version
java version "1.7.0_02"
Java(TM) SE Runtime Environment (build 1.7.0_02-b13)
Java HotSpot(TM) 64-Bit Server VM (build 22.0-b10, mixed mode)
java -server -verbose:gc -XX:CompileThreshold=1 -cp bin EscapeAnalysis
test1
test2
[GC 15616K->352K(59776K), 0,0014270 secs]
[GC 15968K->288K(59776K), 0,0011790 secs]
[GC 15904K->288K(59776K), 0,0018170 secs]
[GC 15904K->288K(59776K), 0,0011100 secs]
[GC 15904K->288K(57152K), 0,0019790 secs]
[GC 15520K->320K(56896K), 0,0011670 secs]
[GC 15232K->284K(56256K), 0,0011440 secs]
test3
test4
test5
[GC 14876K->348K(55936K), 0,0005340 secs]
[GC 14620K->348K(56000K), 0,0004560 secs]
[GC 14300K->316K(55296K), 0,0004680 secs]
[GC 13948K->316K(55488K), 0,0003590 secs]
[GC 13692K->316K(54784K), 0,0004580 secs]
[GC 13436K->316K(54976K), 0,0005430 secs]
[GC 13180K->316K(54272K), 0,0004500 secs]
[GC 12924K->316K(54464K), 0,0005090 secs]
[GC 12668K->316K(53760K), 0,0004490 secs]
[GC 12412K->316K(53888K), 0,0004350 secs]
[GC 12156K->316K(53312K), 0,0005060 secs]
test6
6737499643744733086
我正在使用GC日志来了解是否在堆栈上分配了对象(来自Escape analysis in Java的想法),这可能不是100%可靠但似乎提供了很好的提示。
根据输出,堆栈分配适用于test1,test3,test4和test6,不适用于test2和test5。我不明白为什么这不能用于for循环中的迭代器,尽管它可以工作
我已经阅读了ArrayList iterator的代码,我不明白为什么它不符合测试2和5中的堆栈分配的原因,因为它既不会逃避当前的方法,也不会逃避当前的线程。
有什么想法吗?
答案 0 :(得分:8)
EA是C2编译器基于它生成的IR分析的东西,因此您需要它才能在享受优势之前编译该方法。每个测试只调用一次,因此没有机会进行编译。关于热点内部维基(https://wikis.oracle.com/display/HotSpotInternals/Overview+of+Ideal,+C2的+高+级+中级+表示和https://wikis.oracle.com/display/HotSpotInternals/EscapeAnalysis)中的EA和C2 IR的详细信息
这是一个试图展示影响的版本
import com.sun.management.ThreadMXBean;
import java.lang.management.ManagementFactory;
import java.util.ArrayList;
import java.util.Iterator;
public class EscapeAnalysisTest {
private static final long TIME_TO_TEST = 10L * 1000L; // 10s
static class Timestamp {
private long millis;
public Timestamp(long millis) {
this.millis = millis;
}
public long getTime() {
return millis;
}
public void setTime(long time) {
millis = time;
}
}
public static void main(String[] args) {
System.out.println("****");
doIt();
System.out.println("****");
doIt();
System.out.println("****");
doIt();
System.out.println("****");
doIt();
System.out.println("****");
}
private static void doIt() {
final ThreadMXBean mxbean = (ThreadMXBean) ManagementFactory.getThreadMXBean();
final long tid = Thread.currentThread().getId();
long r = 0;
final long allocPre = mxbean.getThreadAllocatedBytes(tid);
r += test1();
long alloc1 = mxbean.getThreadAllocatedBytes(tid);
System.out.println("test1 - " + (alloc1 - allocPre));
r += test2();
final long alloc2 = mxbean.getThreadAllocatedBytes(tid);
System.out.println("test2 - " + (alloc2 - alloc1));
r += test3();
final long alloc3 = mxbean.getThreadAllocatedBytes(tid);
System.out.println("test3 - " + (alloc3 - alloc2));
r += test4();
final long alloc4 = mxbean.getThreadAllocatedBytes(tid);
System.out.println("test4 - " + (alloc4 - alloc3));
r += test5();
final long alloc5 = mxbean.getThreadAllocatedBytes(tid);
System.out.println("test5 - " + (alloc5 - alloc4));
r += test6();
final long alloc6 = mxbean.getThreadAllocatedBytes(tid);
System.out.println("test6 - " + (alloc6 - alloc5));
System.out.println(r);
}
public static long test1() {
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
r += new Timestamp(System.currentTimeMillis()).getTime();
}
return r;
}
public static long test2() {
ArrayList<Integer> l = new ArrayList<Integer>(1000);
for (int i = 0; i < 1000; ++i) {
l.add(i);
}
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
for (Iterator<Integer> it = l.iterator(); it.hasNext(); ) {
r += it.next().longValue();
}
}
return r;
}
public static long test3() {
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
Timestamp ts = new Timestamp(System.currentTimeMillis());
ts.setTime(42);
r += ts.getTime();
}
return r;
}
public static long test4() {
ArrayList<Integer> l = new ArrayList<Integer>(1000);
for (int i = 0; i < 1000; ++i) {
l.add(i);
}
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
Iterator<Integer> it = l.iterator();
r += it.next().longValue();
r += it.next().longValue();
r += it.next().longValue();
r += it.next().longValue();
}
return r;
}
public static long test5() {
ArrayList<Integer> l = new ArrayList<Integer>(1000);
for (int i = 0; i < 1000; ++i) {
l.add(i);
}
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
Iterator<Integer> it = l.iterator();
for (int i = 0; i < l.size(); ++i) {
r += it.next().longValue();
}
}
return r;
}
public static long test6() {
long r = 0;
long start = System.currentTimeMillis();
while (System.currentTimeMillis() - start < TIME_TO_TEST) {
for (Timestamp ts = new Timestamp(System.currentTi());
ts.getTime() > 0;
ts.setTime(ts.getTime() + System.currentTimeMillis())) {
r += ts.getTime();
}
}
return r;
}
}
在使用-server -XX:CompileThreshold=1
****
test1 - 109048
test2 - 89243416
test3 - 16664
test4 - 42840
test5 - 71982168
test6 - 1400
-5351026995119026839
****
test1 - 16432
test2 - 85921464
test3 - 16664
test4 - 42840
test5 - 66777600
test6 - 1368
7844020592566674506
****
test1 - 48
test2 - 18256
test3 - 272
test4 - 18264
test5 - 18264
test6 - 272
-2137858376905291730
****
test1 - 48
test2 - 18256
test3 - 272
test4 - 18264
test5 - 18264
test6 - 272
3273987624143297143
****
这里的一个危险是这个方法的编译从根本上改变了它,我没有试图防止这种情况,因此可能需要使用LogCompilation
或PrintCompilation
进行检查。 / p>
答案 1 :(得分:2)
Escape Analysis严重依赖于函数调用的内联。
与任何其他微基准测试一样 - 特别是在服务器VM上 - 需要预热。如果删除-XX:CompileThreshold=1
并在循环中执行主测试,您将注意到在1-2次迭代后它将停止收集垃圾,因为编译器收集了足够的分析信息以内联方法,然后执行转义分析。
答案 2 :(得分:2)
我刚刚调查了同样的事情,但对于Java 8.我把答案放在一个重复的问题中,因为我没有及时找到这个问题。
首先,它依赖于实现。这个答案适用于OpenJDK 1.8,也可能适用于Oracle JVM 1.8。
其次,正如其他人所说,堆栈分配仅在C2编译器编译方法时才会发生,只有在方法被调用足够次数后才会发生。
如果是这样,如果
,则可以对象进行堆栈分配如果你不了解Hotspot的一些特定怪癖,那么内联尤其是不可预测的。有关详细信息,请参阅链接的答案。
编辑:我尝试在java 8(OpenJDK)上运行测试,所有内容都在那里内联。因此,java 7和8之间的堆栈分配存在差异。