java - 从内部类引用的局部变量必须是 final 或有效的 final

Question

这个程序是我类(class)的最终作业，我无法弄清楚为什么我收到错误“从内部类引用的局部变量必须是最终的或有效的最终”。该程序正在运行并发线程来对# 数组进行排序，然后找到该数组的高值和低值。当我在没有并发的情况下创建它时，我没有出现这个错误。我正在为最终确定高变量和低变量的位置而苦苦挣扎。

public void HiLo(int[] numbers){

    int high = numbers[0];
    int low = numbers[0];

    Runnable r2 = new Runnable(){
        @Override
        public void run() {
            System.out.println("The highest value is: ");
            for (int index = 1; index < numbers.length; index++){
                if (numbers[index] > high)
                    high = numbers[index];
                System.out.println(high);
                }
            System.out.println();
            System.out.println("The lowest value is: ");
            for (int ind = 1; ind < numbers.length; ind++){
                if (numbers[ind] < low)
                    low = numbers[ind];
                System.out.println(low);
            }
        }
    };
    pool.execute(r2);
}

这是产生错误的代码块。如果我让 int high = numbers[0];或 int low = numbers[0]; final 然后我得到一个错误，我不能使该值最终并且相反变量的错误消失。

这是程序的其余部分。感谢您的帮助。

package concurrentthread;

import java.util.Arrays;
import java.util.Scanner;
import java.util.concurrent.Executor;
import java.util.concurrent.Executors;


public class ConcurrentThread {

    static Executor pool = Executors.newFixedThreadPool(2);

public static void main(String[] args) {
    int size;

    Scanner keyboard = new Scanner(System.in);

    ConcurrentThread sort = new ConcurrentThread();
    ConcurrentThread hilo = new ConcurrentThread();

    System.out.println("This program will calculate the highest and lowest "
                + "numbers entered by the user \nand also sort them in "
                + "ascending order");
    System.out.println();
    System.out.print("How many numbers would you like in the array? ");
        size = keyboard.nextInt();

    final int[] numbers = new int[size];

    for (int index = 0; index < numbers.length; index++){
        System.out.print("Please enter a number between 1 and 100: ");
        numbers[index] = keyboard.nextInt(); 
    }

    System.out.println();
    sort.Sort(numbers);
    hilo.HiLo(numbers);

    //System.exit(0);
}

public void Sort(int[] numbers){
    int sort = numbers[0];

    Runnable r1 = () -> {
        Arrays.sort(numbers);
        System.out.println("The sorted values are: ");
        for (int index = 0; index < numbers.length; index++)
            System.out.print(numbers[index] + " ");

        System.out.println();
    };
    pool.execute(r1);
}

public void HiLo(int[] numbers){

    final int high = numbers[0];
    int low = numbers[0];

    Runnable r2 = new Runnable(){
        @Override
        public void run() {
            System.out.println("The highest value is: ");
            for (int index = 1; index < numbers.length; index++){
                if (numbers[index] > high)
                    high = numbers[index];
                System.out.println(high);
                }
            System.out.println();
            System.out.println("The lowest value is: ");
            for (int ind = 1; ind < numbers.length; ind++){
                if (numbers[ind] < low)
                    low = numbers[ind];
                System.out.println(low);
            }
        }
    };
    pool.execute(r2);
}

Answer 1

您不断更新 run() 方法中的 high 和 low，根据定义使它们不是有效的 final。

因为无论如何在 run() 方法之外都不需要它们，所以只需将这两行移到里面。

public void HiLo(int[] numbers){

    Runnable r2 = new Runnable(){
        @Override
        public void run() {
            int high = numbers[0];
            int low = numbers[0];
            System.out.println("The highest value is: ");