[英]Multithreading returns number before it finished processing (JAVA)
我写了下面这段代码来尝试理解多线程。 然而,结果并不是我所期望的。 似乎它在搜索完成执行之前返回了该值。 我怎样才能让它等到结果准备好然后才得到返回值?
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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();
}
}
目前的结果如下:
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
我实际上希望两个线程最后都返回 1 。
你需要解决两件事首先你有你的打印语句System.out.println("array[i] lowestValue: " + lowestValue);
在您的 for 循环内但在您的 if 条件之外,因此它为数组中的每个元素lowestValue
值的当前值,如果您希望仅在变量更新时打印它,您应该将它移到您的 if 语句中。
其次,你必须移动这个表达式System.out.println("Lowest number for this thread " + i + " is " + number);
在你调用executor.execute(searchLowestNumber);
之后因为搜索实际上正在发生,否则您将要打印的数字是您在 class '0' 的构造函数中设置的数字,因为它当时尚未更新
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