十个线程分别计算1-10.11-20......91-100,最后计算总和是多少?
分析:首先应该有十个线程,然后每个线程计算一部分,将十个线程的结果相加即可。
给出几种写法:
第一种:sync关键字
public class 十个线程 {
private int shu = 0;
private int count = 0;
public static void main(String[] args) {
// for (int k = 0; k < 10; k++) {
十个线程 十个线程 = new 十个线程();
for (int i = 0; i < 10; i++) {
Thread t = new Thread(new Runnable() {
@Override
public void run() {
十个线程.print(十个线程.shu);
}
});
t.start();
try {
t.join();//按顺序执行
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println(十个线程.count);
// }
}
public void print(int start) {
for (int i = 1; i <= 10; i++) {
// System.out.println(start + i);
shu = start + i;
count += shu;
}
}
}
后面两种几乎没有区别
第二种:executor框架
public class 十个线程基于executor {
private volatile int shu=0;
private int j;
public static void main(String[] args) {
十个线程基于executor 十个线程 = new 十个线程基于executor();
ExecutorService executorService = Executors.newFixedThreadPool(10);
for (int i = 0; i < 10; i++) {
FutureTask<Integer> future = new FutureTask<Integer>(new Callable<Integer>() {
@Override
public Integer call() throws Exception {
return 十个线程.print(十个线程.j);
}
});
executorService.submit(future);
try {
System.out.println(future.get());
} catch (InterruptedException e) {
e.printStackTrace();
} catch (ExecutionException e) {
e.printStackTrace();
}
}
executorService.shutdown();
System.out.println(executorService.isShutdown());
}
public Integer print(int start){
for (int i = 1; i <= 10; i++) {
System.out.println(start+i);
j=start+i;
shu += start+i;
}
System.out.println(Thread.currentThread().getName());
return shu;
}
}
第三种:faturetask
public class 十个线程基于fature {
private int shu = 0;
private int j;
public static void main(String[] args) {
// for (int k = 0; k < 10; k++) {
十个线程基于fature 十个线程 = new 十个线程基于fature();
for (int i = 0; i < 10; i++) {
FutureTask<Integer> integerFutureTask = new FutureTask<Integer>(new Callable<Integer>() {
@Override
public Integer call() throws Exception {
return 十个线程.print(十个线程.j);
}
});
Thread thread = new Thread(integerFutureTask);
thread.start();
try {
// thread.join(); 无需join
System.out.println(integerFutureTask.get() + "=======");//已经帮忙计算好了所有的和
} catch (InterruptedException e) {
e.printStackTrace();
} catch (ExecutionException e) {
e.printStackTrace();
}
}
// }
}
public Integer print(int start) {
for (int i = 1; i <= 10; i++) {
System.out.println(start + i);
j = start + i;
shu += start + i;
}
return shu;
}
}
本文转自 jimolvxing.blog.csdn.net/article/det…，如有侵权，请联系删除。
