來模擬一個死鎖（互相等待）：

TestDeadLock.java

package com.zhj.www;public class TestDeadLock implements Runnable {public int flag = 1;static Object o1 = new Object();static Object o2 = new Object();public void run() {System.out.println("flag=" + flag);if(flag ==1) {synchronized (o1) {try {Thread.sleep(500);}catch (Exception e) {e.getStackTrace();}synchronized (o2) {System.out.println("1");}}}if(flag ==0) {synchronized (o2) {try {Thread.sleep(500);}catch (Exception e) {e.getStackTrace();}synchronized (o1) {System.out.println("0");}}}}public static void main(String[] args) {TestDeadLock td1 = new TestDeadLock();TestDeadLock td2 = new TestDeadLock();td1.flag = 1;td2.flag = 0;Thread t1 = new Thread(td1);Thread t2 = new Thread(td2);t1.start();t2.start();}}

運行結果：

我們發現程序處于已知等待的情況下。

哲學家吃飯問題（多個線程死鎖）：

怎么解決？把鎖的粒度加粗，不如把整個對象鎖住，而不是只是鎖幾行。

粒度：實現數據庫的產品，鎖在行一級還是表一級。

面試題：

package com.zhj.www;
public class TestInterviewQuestion1 implements Runnable {int b =100;public synchronized void m1() {try {b = 1000;Thread.sleep(5000);System.out.println("b = "+b);} catch (Exception e) {e.getStackTrace();}}public void m2() {System.out.println(b);}public void run() {try {m1();}catch (Exception e) {e.getStackTrace();}}public static void main(String[] args) throws Exception{TestInterviewQuestion1 testInterviewQuestion1  = new TestInterviewQuestion1();Thread thread = new Thread(testInterviewQuestion1);thread.start();Thread.sleep(1000);testInterviewQuestion1.m2();}
}

運行結果：
1000
b = 1000

修改一下：

package com.zhj.www;public class TestInterviewQuestion1 implements Runnable {int b =100;public synchronized void m1() {try {b = 1000;Thread.sleep(5000);System.out.println("b = "+b);} catch (Exception e) {e.getStackTrace();}}public synchronized void m2() throws Exception {Thread.sleep(2500);b = 2000;}public void run() {try {m1();}catch (Exception e) {e.getStackTrace();}}public static void main(String[] args) throws Exception{TestInterviewQuestion1 testInterviewQuestion1  = new TestInterviewQuestion1();Thread thread = new Thread(testInterviewQuestion1);thread.start();//Thread.sleep(1000);testInterviewQuestion1.m2();System.out.println(testInterviewQuestion1.b);}
}

對每個方法加不加鎖，要考慮清楚。

互斥：某個時間段，只有一個線程進入這個方法，但不保證不進入另外一個線程。

不允許多個線程同時寫，但允許多個線程讀。