Question

我写了下面的代码来尝试理解多线程。但是，结果不是我所期望的。看起来它在搜索完成执行之前返回了值。我怎样才能让它等到结果准备好然后才获得返回值？

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import java.util.concurrent.Executor;  
import java.util.concurrent.Executors;  
import java.util.concurrent.ThreadPoolExecutor;  
import java.util.concurrent.TimeUnit;  
import java.util.ArrayList;

//Finding the smallest number in an array
public class Main
{
    public static class Search implements Runnable {
        private int[] array;
        private int lowestNumber; 
        private int taskNumber;

        public Search(int[] array, int taskNumber){
            this.lowestNumber = 0;
            this.taskNumber = taskNumber;
            this.array = array;
        }

        public int getLowestNumber(){
            return lowestNumber;
        }

        protected void setLowestNumber(int lowestNumber){
            this.lowestNumber = lowestNumber;
        }

        protected void searchArrayLowestNumber(){
            int lowestValue = 0; 
            int arrayLength = array.length;

            for(int i = 0; i < arrayLength; i++){
                if( i == 0 ){
                    lowestValue = array[i];
                }
                if(array[i] < lowestValue){
                    lowestValue = array[i];
                } 
                System.out.println("array[i] lowestValue: " + lowestValue);
            }
            setLowestNumber(lowestValue);

        }

        public void run(){
                System.out.println("Accessing search...task number: " + taskNumber);
                searchArrayLowestNumber();

        }

    }

    public static void main(String[] args) throws InterruptedException {

        ThreadPoolExecutor executor = (ThreadPoolExecutor) Executors.newFixedThreadPool(2);
        int[][] arrayA = {{12, 13, 1}, {10, 34, 1}};

        for (int i = 0; i <= 1; i++) 
        {
            int[] tempArray = new int[3];
            for(int j = 0; j < 3; j++){
                tempArray[j] = arrayA[i][j];
            }
            Search searchLowestNumber = new Search(tempArray, i);
            int number = searchLowestNumber.getLowestNumber();
             try{
                Long duration = (long) (Math.random() * 10);
                System.out.println("Lowest number for this thread " + i + " is " + number);
                TimeUnit.SECONDS.sleep(duration); 
             }catch (InterruptedException e) {
                e.printStackTrace();
             }

            executor.execute(searchLowestNumber);
        }
        executor.shutdown();
    }
}

当前结果如下：

Lowest number for this thread 0 is 0                                                                                                                                               
Lowest number for this thread 1 is 0                                                                                                                                               
Accessing search...task number: 0                                                                                                                                                  
array[i] lowestValue: 12                                                                                                                                                           
array[i] lowestValue: 12                                                                                                                                                           
array[i] lowestValue: 1                                                                                                                                                            
Accessing search...task number: 1                                                                                                                                                  
array[i] lowestValue: 10                                                                                                                                                           
array[i] lowestValue: 10                                                                                                                                                           
array[i] lowestValue: 1

我实际上期望两个线程最后都返回1。

Answer 1

您需要修复两件事首先，您需要在for循环内拥有打印语句System.out.println("array[i] lowestValue: " + lowestValue);，但在if条件语句之外，因此，如果您希望仅将lowestValue的当前值对数组中的每个元素打印一次，在变量更新时打印出来，则应将其移至if语句中。

第二，您必须在调用System.out.println("Lowest number for this thread " + i + " is " + number);之后将表达式executor.execute(searchLowestNumber);移动到，因为那里实际上正在进行搜索，否则，将要打印的数字就是您要设置的数字在您的类“ 0”的构造函数中，因为此时尚未对其进行更新

多线程在完成处理之前返回数字（JAVA）

1 个答案: