J260713/授课代码/c260810/src/homework0808/Homework_03.java

package homework0808;

import java.util.concurrent.Callable;
import java.util.concurrent.FutureTask;

/**
 * @author WanJl
 * @version 1.0
 * @title Homework_03
 * @description
 * @create 2026/8/10
 */
public class Homework_03 {
    // ========== ① 实现 Callable 接口，泛型指定返回类型 ==========
    static class SumCallable implements Callable<Integer> {
        private int n;   // 求 1~n 的和

        public SumCallable(int n) {
            this.n = n;
        }

        @Override
        public Integer call() throws Exception {
            // TODO 1: int sum = 0; 计算 1~n 的累加和
            int sum = 0;
            // TODO 2: 打印 Thread.currentThread().getName() + "求和结果：" + sum
            for (int i = 1; i <=n; i++) {
                sum+=i;
            }
            System.out.println(Thread.currentThread().getName() + "求和结果：" + sum);
            // TODO 3: return sum;   // 返回计算结果（get() 会拿到这个值）
            return sum;
        }
    }

    public static void main(String[] args) throws Exception {
        // TODO 4: 创建 3 个 SumCallable 对象（求 1~100 / 1~1000 / 1~10000 的和）
        SumCallable sc1=new SumCallable(100);
        SumCallable sc2=new SumCallable(1000);
        SumCallable sc3=new SumCallable(10000);
        // TODO 5: 用 FutureTask<Integer> 包装每个 Callable 对象
        FutureTask<Integer> ft1=new FutureTask<>(sc1);
        FutureTask<Integer> ft2=new FutureTask<>(sc2);
        FutureTask<Integer> ft3=new FutureTask<>(sc3);
        // TODO 6: new Thread(futureTask, "线程名") 创建 3 个线程并 start()
        Thread t1=new Thread(ft1,"线程1");
        Thread t2=new Thread(ft2,"线程2");
        Thread t3=new Thread(ft3,"线程3");
        t1.start();
        t2.start();
        t3.start();
        // TODO 7: 调用 ft.get() 获取每个线程的返回值，打印并在 main 中汇总总结果
        System.out.println(ft1.get());
        System.out.println(ft2.get());
        System.out.println(ft3.get());

        // TODO 8: （思考）对比三种实现方式：run() 无返回值 vs call() 有返回值，各自适合什么场景
    }

}