并发王者课 - 青铜 6:借花献佛 - 如何格式化 Java 内存工具 JOL 输出
发布于: 2021 年 05 月 27 日
在前面的文章《一探究竟-如何从 synchronized 理解 Java 对象头中的锁》中,我们介绍并体验了 JOL 工具。虽然 JOL 很赞,但它的输出对我们不是很友好,如果不借助工具,我们很难直观理解其中的含义。
下面这段代码是对 JOL 输出的翻译,建议你收藏。代码非我所写,文末已经注明出处。
import org.openjdk.jol.info.ClassLayout;
import java.nio.ByteOrder;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class PrintObjectHeader {
/**
* Get binary data
*
* @param o
* @return
*/
public static String getObjectHeader(Object o) {
ByteOrder order = ByteOrder.nativeOrder();//Byte order
String table = ClassLayout.parseInstance(o).toPrintable();
Pattern p = Pattern.compile("(0|1){8}");
Matcher matcher = p.matcher(table);
List<String> header = new ArrayList<String>();
while (matcher.find()) {
header.add(matcher.group());
}
//Little-endian machines, need to traverse in reverse
StringBuilder sb = new StringBuilder();
if (order.equals(ByteOrder.LITTLE_ENDIAN)) {
Collections.reverse(header);
}
for (String s : header) {
sb.append(s).append(" ");
}
return sb.toString().trim();
}
/**
* Parsing object header function for 64bit jvm
* In 64bit jvm, the object header has two parts: Mark Word and Class Pointer, Mark Word takes 8 bytes, Class Pointer takes 4 bytes
*
* @param s Binary string of object header (each 8 bits, separated by a space)
*/
public static void parseObjectHeader(String s) {
String[] tmp = s.split(" ");
System.out.print("Class Pointer: ");
for (int i = 0; i < 4; ++i) {
System.out.print(tmp[i] + " ");
}
System.out.println("\nMark Word:");
if (tmp[11].charAt(5) == '0' && tmp[11].substring(6).equals("01")) {//0 01 lock-free state, regardless of GC mark
//notice: Mark word structure without lock: unused(25bit) + hashcode(31bit) + unused(1bit) + age(4bit) + biased_lock_flag(1bit) + lock_type(2bit)
// The reason why hashcode only needs 31bit is: hashcode can only be greater than or equal to 0, eliminating the negative range, so you can use 31bit to store
System.out.print("\thashcode (31bit): ");
System.out.print(tmp[7].substring(1) + " ");
for (int i = 8; i < 11; ++i) System.out.print(tmp[i] + " ");
System.out.println();
} else if (tmp[11].charAt(5) == '1' && tmp[11].substring(6).equals("01")) {//1 01, which is the case of biased lock
//notice: The object is in a biased lock, its structure is: ThreadID(54bit) + epoch(2bit) + unused(1bit) + age(4bit) + biased_lock_flag(1bit) + lock_type(2bit)
// ThreadID here is the thread ID holding the biased lock, epoch: a timestamp of the biased lock, used for optimization of the biased lock
System.out.print("\tThreadID(54bit): ");
for (int i = 4; i < 10; ++i) System.out.print(tmp[i] + " ");
System.out.println(tmp[10].substring(0, 6));
System.out.println("\tepoch: " + tmp[10].substring(6));
} else {//In the case of lightweight locks or heavyweight locks, regardless of the GC mark
//notice: JavaThread*(62bit,include zero padding) + lock_type(2bit)
// At this point, JavaThread* points to the monitor of the lock record/heavyweight lock in the stack
System.out.print("\tjavaThread*(62bit,include zero padding): ");
for (int i = 4; i < 11; ++i) System.out.print(tmp[i] + " ");
System.out.println(tmp[11].substring(0, 6));
System.out.println("\tLockFlag (2bit): " + tmp[11].substring(6));
System.out.println();
return;
}
System.out.println("\tage (4bit): " + tmp[11].substring(1, 5));
System.out.println("\tbiasedLockFlag (1bit): " + tmp[11].charAt(5));
System.out.println("\tLockFlag (2bit): " + tmp[11].substring(6));
System.out.println();
}
public static void printObjectHeader(Object o) {
if (o == null) {
System.out.println("null object.");
return;
}
parseObjectHeader(getObjectHeader(o));
}
}
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参考资料
https://www.programmersought.com/article/21094532407/
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发布于: 2021 年 05 月 27 日阅读数: 69
版权声明: 本文为 InfoQ 作者【秦二爷】的原创文章。
原文链接:【http://xie.infoq.cn/article/cdcd380ede561ea0fd836a79f】。文章转载请联系作者。
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