如题:使用三个线程交替打印ABC,直至100次代码实战
方法一:
使用notify()、wait()
方法
public class PrintAbc {
/**
* 唤醒线程的状态值 state: threadA = 0, threadB = 1, threadC =2,
*/
int state = 0;
/**
* 循环技术,初始值0
*/
int count = 0;
public void print(PrintAbc printAbc) {
Thread threadA = new Thread(() -> {
extracted(printAbc, "A", 0, 1);
});
Thread threadB = new Thread(() -> {
extracted(printAbc, "B", 1, 2);
});
Thread threadC = new Thread(() -> {
extracted(printAbc, "C", 2, 0);
});
threadC.start();
threadB.start();
try {
Thread.sleep(1000L);
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
threadA.start();
}
/**
* 交替打印abc,直至100次
*
* @param printAbc 锁对象
* @param a 打印的字母, 对应A、B、C
* @param needState 当前线程对应的state状态值
* @param nextState 唤醒下一个线程所需state状态值
*/
private void extracted(PrintAbc printAbc, String a, int needState, int nextState) {
while (true) {
synchronized (printAbc) {
if (count >= 100) {
break;
}
if (printAbc.count < 100 && printAbc.state == needState) {
System.out.println(a);
printAbc.state = nextState;
printAbc.count++;
printAbc.notifyAll();
} else {
try {
printAbc.wait();
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
}
}
}
}
public static void main(String[] args) {
PrintAbc printAbc = new PrintAbc();
printAbc.print(printAbc);
}
}
上述代码使用notify(),wait()
,进行线程间的条件唤醒,state
的初始状态是0,对应线程A,所以第一次打印字母也一定是A
方法二
使用ReentrantLock的
的Condition
条件
public class PrintAbcByCondition {
/**
* 循环计数初始值0
*/
static int count = 0;
public void print() {
ReentrantLock reentrantLock = new ReentrantLock();
Condition conditionA = reentrantLock.newCondition();
Condition conditionB = reentrantLock.newCondition();
Condition conditionC = reentrantLock.newCondition();
Thread threadA = new Thread(() -> {
while (true) {
try {
reentrantLock.lock();
// threadA进来打印A然后唤醒threadB
if (count < 100) {
System.out.println("A");
count++;
conditionB.signal();
}
conditionA.await();
} catch (InterruptedException e) {
throw new RuntimeException(e);
} finally {
reentrantLock.unlock();
}
}
});
Thread threadB = new Thread(() -> {
while (true) {
try {
reentrantLock.lock();
// threadB进来就阻塞等待threadA使用完毕
conditionB.await();
if (count < 100) {
System.out.println("B");
count++;
conditionC.signal();
}
} catch (InterruptedException e) {
throw new RuntimeException(e);
} finally {
reentrantLock.unlock();
}
}
});
Thread threadC = new Thread(() -> {
while (true) {
try {
reentrantLock.lock();
// threadC进来就阻塞等待threadB使用完毕
conditionC.await();
if (count < 100) {
System.out.println("C");
count++;
conditionA.signal();
}
} catch (InterruptedException e) {
throw new RuntimeException(e);
} finally {
reentrantLock.unlock();
}
}
});
threadC.start();
threadB.start();
try {
Thread.sleep(1000L);
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
threadA.start();
}
public static void main(String[] args) {
new PrintAbcByCondition().print();
}
}
使用ReentrantLock的
的Condition
条件,很容易能实现三个线程之间的交替打印,需要注意的一点就是线程A是需要第一个执行,可以看到代码里threadA
在等待1秒后在执行,也能确保是第一个进行打印,原因如下:
线程B和线程C中任意一个线程拿到锁都需要等待条件成立,线程C依赖线程B,而线程B依赖线程A,所以他们会一直阻塞直至线程A执行
上述两个方法中,核心问题就是如何实现线程间的条件唤醒,如方法一,我们可以自定义state
状态变量来与各个线程绑定,每个线程都有自己对应的state
状态,当state
变量当前值与线程自身期望的state
值相同才唤醒当前线程。也可以使用juc
中ReentrantLock的
提供的Condition
条件完成线程间的条件唤醒
至此,三个线程交替打印ABC100次的实现方法介绍完毕
标签:count,printAbc,Thread,打印,state,线程,new,ABC100 From: https://www.cnblogs.com/wayn111/p/16600635.html