使用4个子线程求出1到100的和。且每个子线程只能计算25个数。主线程必须等待子线程的计算结果。

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原创最新推荐文章于 2026-03-27 10:12:05 发布 1.3k 阅读 AI 写同款· GEO 优化 ›

Java 通过主线程实现子线程的超时控制 package com.bill99.seashell.fsp.quartz.test; import java.util.concurrent.Callable; import java.util.concurrent.ExecutionException; import java.util.concurrent.ExecutorService; import java.util.con 阅读详情

public class FourThread {

static int sum11 = 0;
static int sum12 = 0;
static int sum13 = 0;
static int sum14 = 0;
static int sum = 0;

public static void main(String[] args) {
/**
* 使用4个子线程求出1到100的和,且每个子线程只能计算25个数,主线程必须等待子线程
*/
new Thread(){
public void run() {
int sum1=0;
for(int i=1;i<=25;i++){
sum1+=i;
}
sum11 += sum1;
System.out.println(Thread.currentThread().getName() + " " + sum11);
};
}.start();

new Thread(){
public void run() {

int sum2 = 0;
for (int i = 26; i <= 50; i++) {
sum2 += i;
}
sum12 += sum2;
System.out.println(Thread.currentThread().getName() + " " + sum12);

};
}.start();

new Thread(){

public void run() {

int sum3 = 0;
for (int i = 51; i <= 75; i++) {
sum3 += i;
}
sum13 += sum3;
System.out.println(Thread.currentThread().getName() + " " + sum13);

};
}.start();

new Thread(){
public void run() {
int sum4 = 0;
for (int i = 76; i <= 100; i++) {
sum4 += i ;
}
sum14 = sum4;
System.out.println(Thread.currentThread().getName() + " " + sum14 );

};

}.start();

try {
Thread.sleep(100);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
sum = sum11 + sum12 + sum13 + sum14;
System.out.println( "sum: " + sum);

}

}