让2个线程按顺序打印

zzkang0206

本帖最后由吴光新于 2014-2-15 03:16 编辑

public class Test1 {
private A a;
private B b;

public static void main(String[] args){
  Test1 t = new Test1();
  t.show();
}

public void show(){
  a = new A();
  b = new B();
  a.start();
  b.start();
}

class A extends Thread{
  @Override
  public void run() {
for (int i = 1; i < 100; i+=2) {
System.out.println("A=" + i);
}
  }
}
class B extends Thread{
  @Override
  public void run() {
for (int i = 0; i < 100; i+=2) {
System.out.println("B=" + i);
}
  }
}
}

打印结果是
A=1
A=3
A=5
B=0
A=7
A=9
A=11
A=13
....

请修改程序，让两个线程同时打印
A=1
B=2
A=3
B=4
A=5
B=6
......

回答正确，前三名各奖励2技术分

小Zeor……… · 小Zeor………

本帖最后由小Zeor……… 于 2014-2-15 10:27 编辑

抢沙发，应该是线程加锁的问题。占个位置一会贴出来解答，有技术分，哇咔咔

public class Test1 {
private A a;
private B b;
private R r;
public static void main(String[] args) {
Test1 t = new Test1();
t.show();
}
public void show() {
r = new R();
a = new A();
b = new B();
Thread t1 = new Thread(a);
Thread t2 = new Thread(b);
t1.start();
t2.start();
}
class R {
boolean flag = true;
}
class A extends Thread {
@Override
public void run() {
int i = 0;
while (i < 100) {
{
if (r.flag) {
synchronized(r){System.out.println("A=" + i);
i++;
i++;
r.flag = false;
}
}
}
}
}
}
class B extends Thread {
@Override
public void run() {
int i = 1;
while (i < 100) {
if (!r.flag) {
synchronized (r) {
System.out.println("B=" + i);
i++;
i++;
r.flag = true;
}
}
}
}
}
}

复制代码

分解式 · 分解式

求技术分

奋斗的小胖子 · 奋斗的小胖子

本帖最后由奋斗的小胖子于 2014-2-15 09:44 编辑

public class Test1
{
public static void main(String[] args)
{
show s1=new show();
A a1=new A(s1);
B b1=new B(s1);
Thread t1=new Thread(a1);
Thread t2=new Thread(b1);
t1.start();
t2.start();
}
}
class show
{
private show s;
int num=0;
private boolean flag=false;
show(){}
Lock lock=new ReentrantLock();
Condition A_lock=lock.newCondition();
Condition B_lock=lock.newCondition();
public void showA()
{
lock.lock();
try {
while(flag==true)
{
A_lock.await();
if(num<100)
{
num++;
System.out.println("A="+num);
}
}
flag=true;
B_lock.signal();
}
catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
finally
{
lock.unlock();
}
}
public void showB()
{
lock.lock();
try {
while(flag==false)
{
B_lock.await();
if(num<100)
{
num++;
System.out.println("B="+num);
}
}
flag=false;
A_lock.signal();
}
catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
finally
{
lock.unlock();
}
}
}
class A implements Runnable
{
show s;
public A(show s)
{
this.s=s;
}
public void run()
{
while(true)
{
s.showA();
}
}
}
class B implements Runnable
{
show s;
public B(show s)
{
this.s=s;
}
public void run()
{
while(true)
{
s.showB();
}
}
}

复制代码

我是参考生产消费的例子写的，为了抢楼没写注释，就先发了

LSCZ3633 · LSCZ3633

package com.itheima;
/*
* 使两个线程一个一个有序进行
*
* 思路：定义一个标签boolean 判断哪个线程应该执行
* 定义一个公共变量i
* 加入一个同步代码块 synchronized(r)
*
*
*
* */
public class Test1 {
private A a;
private B b;
private R r;
public static void main(String[] args){
Test1 t = new Test1();
t.show();
}
public void show(){
r=new R();
a = new A(r);
b = new B(r);
a.start();
b.start();
}
}
class R{
int i=1;
boolean flag=true;
R(){}
}
class A extends Thread{
private R r;
A(R r){
this.r=r;
}
@Override
public void run() {
for (; r.i < 100; ) {
if(r.flag){
synchronized(r){
System.out.println("A=" + r.i++);
r.flag=false;
}
}
}
}
}
class B extends Thread{
private R r;
B(R r){
this.r=r;
}
@Override
public void run() {
for (; r.i < 100; ) {
if(!r.flag){
synchronized(r){
System.out.println("B=" + r.i++);
r.flag=true;
}
}
}
}
}

复制代码

LuckyQS · LuckyQS

public class Test1 {
private A a;
private B b;
private R r;
public static void main(String[] args){
Test1 t = new Test1();
t.show();
}
public void show(){
r=new R();
a = new A(r);
b = new B(r);
a.start();
b.start();
}
}
class R{
int i=1;
boolean flag=true;
R(){}
}
class A extends Thread{
private R r;
A(R r){
this.r=r;
}
@Override
public void run() {
while(r.i<100) {
if(r.flag){
synchronized(r){
System.out.println("A="+r.i++);
r.flag=false;
}
}
}
}
}
class B extends Thread{
private R r;
B(R r){
this.r=r;
}
@Override
public void run() {
while(r.i<100) {
if(!r.flag){
synchronized(r){
System.out.println("B="+r.i++);
r.flag=true;
}
}
}
}
}
只要让两个线程控制同一份元素,定义一个boolean变量来改变线程的执行权

倉促、尋找你 · 倉促、尋找你

手机回复，就不贴代码了，线程同步建议使用实现Runnable接口的方法去做这个题目不难，你搜一下资料应该就搞定了

孤独的天奇 · 孤独的天奇

奋斗的小胖子发表于 2014-2-15 09:17
我是参考生产消费的例子写的，为了抢楼没写注释，就先发了

嗯，我觉得，就应该是这样子。

孤独的天奇 · 孤独的天奇

package com.itheima;

import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

public class Resource {

Lock lock = new ReentrantLock();
boolean flag = false;
Condition conditionA = lock.newCondition();
Condition conditionB = lock.newCondition();

public void printA(){
lock.lock();
try {
for (int i = 1; i <= 100; i+=2) {
while(flag)
conditionA.await();
      System.out.println("A=" + i);
      flag = true;
      conditionB.signal();
   }
} catch (Exception e) {
// TODO: handle exception
}finally{
lock.unlock();
}

}
package com.itheima;

public class A implements Runnable{

private Resource res;

public A(Resource res){
this.res = res;
}

@Override
public void run() {
res.printA();

}
}

package com.itheima;

public class B implements Runnable {

private Resource res;

public B(Resource res){
this.res = res;
}

@Override
public void run() {
res.printB();
}

}

package com.itheima;

public class Test {

/**
   * @param args
   */
public static void main(String[] args) {
Resource res = new Resource();
new Thread(new A(res)).start();
new Thread(new B(res)).start();

}

}

孤独的天奇 · 孤独的天奇

package com.itheima;

import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

public class Resource {

Lock lock = new ReentrantLock();
boolean flag = false;
Condition conditionA = lock.newCondition();
Condition conditionB = lock.newCondition();

public void printA(){
lock.lock();
try {
for (int i = 1; i <= 100; i+=2) {
while(flag)
conditionA.await();
      System.out.println("A=" + i);
      flag = true;
      conditionB.signal();
   }
} catch (Exception e) {
// TODO: handle exception
}finally{
lock.unlock();
}

}

public void printB(){
lock.lock();
try {
for (int i = 2; i <= 100; i += 2) {
while(!flag)
conditionB.await();
System.out.println("B=" + i);
flag = false;
      conditionA.signal();
}
} catch (Exception e) {
lock.unlock();
}
}
}

Resource 类没有发全，补上

小Zeor……… · 小Zeor………

@吴光新求加分。。。。要下一步考试。。

奋斗的小胖子 · 奋斗的小胖子

QQ多少啊？什么奖励啊？

奋斗的小胖子 · 奋斗的小胖子

没有权限啊

奋斗的小胖子 · 奋斗的小胖子

请修改我的权限

wh963572516 · wh963572516

自己写的，不知道对不对，我自己验证的是和你要求的一样

//公共资源
class Resource
{
public int i = 1;
}
//线程 1
class Thread1 implements Runnable {
private Resource r;
public Thread1(Resource r) {
this.r = r;
}
public void run() {
synchronized (r) {
while(r.i<100){
System.out.println("A="+r.i);
r.i++;
r.notify();//唤醒
try{
r.wait();//等待
}catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
//线程2
class Thread2 implements Runnable {
private Resource r;
public Thread2(Resource r) {
this.r = r;
}
public void run() {
synchronized (r) {
while(r.i <100){
System.out.println("B="+r.i);
r.i++;
r.notify();//唤醒
try {
r.wait();//等待
}catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
//主线程
public class ThreadDemo6{
public static void main(String[] args){
Resource r =new Resource();
Thread t1=new Thread(new Thread1(r));
Thread t2=new Thread(new Thread2(r));
t1.start();
t2.start();
}
}

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让2个线程按顺序打印

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