这是我第一次在StackOverflow上提问。我遇到的问题如下：
我有一个生产者和消费者类。在Producer类中，我逐行读取一个文件，并将这些文本行放入字符串列表中。当列表中的行数为x时。此列表被添加到ArrayBlockingQueue。我有一个在主线程中启动的生产者线程。除此之外，我还启动了一些Consumer线程。消费者线程从队列中获取一个项目，该项目应该是一个列表，并遍历此行列表以查找特定单词。找到单词后，它会增加一个计数变量。
发生的事情是，当消费者从队列中取出一个项目时，它说它是空的。我不知道为什么，因为我的制作人当然应该将其添加到队列中。
我的代码如下：
消费类：
public static class Consumer implements Callable<Integer> {
int count = 0;
@Override
public Integer call() throws Exception {
List<String> list = new ArrayList<>();
list = arrayBlockingQueueInput.take();
do {
if (!list.isEmpty()){
for (int i = 0; i < arrayBlockingQueueInput.take().size(); i++) {
for (String element : list.get(i).split(" ")) {
if (element.equalsIgnoreCase(findWord)) {
count++;
}
}
}
} else {
arrayBlockingQueueInput.put(list);
}
} while (list.get(0) != "HALT");
return count;
}
}
生产者类别：
public static class Producer implements Runnable {
@Override
public void run() {
try {
FileReader file = new FileReader("src/testText.txt");
BufferedReader br = new BufferedReader(file);
while ((textLine = br.readLine()) != null) {
if (textLine.isEmpty()) {
continue;
}
/* Remove punctuation from the text, except of punctuation that is useful for certain words.
* Examples of these words are don't or re-enter */
textLine = textLine.replaceAll("[[\\W]&&[^']&&[^-]]", " ");
/* Replace all double whitespaces with single whitespaces.
* We will split the text on these whitespaces later */
textLine = textLine.replaceAll("\\s\\s+", " ");
textLine = textLine.replaceAll("\\n", "").replaceAll("\\r", "");
if (results.isEmpty()) {
results.add(textLine);
continue;
}
if (results.size() <= SIZE) {
results.add(textLine);
if (results.size() == SIZE) {
if (arrayBlockingQueueInput.size() == 14){
List<String> list = new ArrayList<String>();
list.add(HALT);
arrayBlockingQueueInput.put(list);
} else{
arrayBlockingQueueInput.put(results);
results.clear();
}
}
}
}
/* Count the remaining words in the list
* (last lines of the file do perhaps not fill up until the given SIZE, therefore need to be counted here)
* Fill the list with empty items if the size of the list does not match with the given SIZE */
while (results.size() != SIZE) {
results.add("");
}
arrayBlockingQueueInput.put(results);
List<String> list = new ArrayList<String>();
list.add(HALT);
arrayBlockingQueueInput.put(list);
results.clear();
} catch (InterruptedException e) {
producerIsRunning = false;
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
主类：
public void main() throws IOException, InterruptedException {
System.out.println("Enter the word you want to find: ");
Scanner scan = new Scanner(System.in);
findWord = scan.nextLine();
System.out.println("Starting...");
long startTime = System.currentTimeMillis();
Thread producer = new Thread(new Producer());
producer.start();
ExecutorService executorService = Executors.newFixedThreadPool(CORE);
List<Future<Integer>> futureResults = new ArrayList<Future<Integer>>();
for (int i = 0; i < CORE; i++) {
futureResults.add(executorService.submit(new Consumer()));
}
executorService.shutdown();
for (Future<Integer> result : futureResults) {
try {
wordsInText += result.get();
} catch (ExecutionException | InterruptedException e) {
e.printStackTrace();
}
}
producer.join();
long stopTime = System.currentTimeMillis();
System.out.println("The word " + findWord + " appears " + wordsInText + " times in the given text");
System.out.println("Elapsed time was " + (stopTime - startTime) + " milliseconds.");
}
有人可以解释为什么会这样吗？我还要补充一点，我们尝试使用毒丸告知消费者生产者处于暂停状态。
要回答这个问题，为什么我们要这样做？对于学校，我们尝试并行处理某个编程问题。我们选择的问题是字符串匹配。我们首先提出了串行解决方案和并行解决方案。对于下一个作业，我们必须改进并行解决方案，老师告诉我们这是一种实现方法。
谢谢！
尼克
答案 0 :(得分：2)
您将列表添加到队列中并清除它：
$sftp = new Net_SFTP('sftp.sftpurl.co.uk', 2222);
if (!$sftp->login('username', 'password')) {
exit('Login Failed');
}
您需要执行以下操作将列表副本添加到队列中，以使arrayBlockingQueueInput.put(results);
results.clear();
不会清除队列中的列表：
clear()答案 1 :(得分：0)
在老师的帮助下，他帮助我们找到了问题。有两个错误。其中一个在生产者阶层内。我有一些代码可以在主while循环内向生产者发出暂停信号。不应这样做。
除此之外，我应该在do-while循环中完成Do-While之前在Consumer类中执行.take（）。
正确的代码如下：
消费类：
public static class Consumer implements Callable<Integer> {
int count = 0;
@Override
public Integer call() throws Exception {
List<String> list = new ArrayList<>();
do {
list = arrayBlockingQueueInput.take();
if (!list.get(0).equals(HALT)){
for (int i = 0; i < list.size(); i++) {
for (String element : list.get(i).split(" ")) {
if (element.equalsIgnoreCase(findWord)) {
count++;
}
}
}
} else {
arrayBlockingQueueInput.put(list);
}
} while (!list.get(0).equals(HALT));
return count;
}
}
生产者类别：
public static class Producer implements Runnable {
@Override
public void run() {
try {
FileReader file = new FileReader("src/testText.txt");
BufferedReader br = new BufferedReader(file);
while ((textLine = br.readLine()) != null) {
if (textLine.isEmpty()) {
continue;
}
/* Remove punctuation from the text, except of punctuation that is useful for certain words.
* Examples of these words are don't or re-enter */
textLine = textLine.replaceAll("[[\\W]&&[^']&&[^-]]", " ");
/* Replace all double whitespaces with single whitespaces.
* We will split the text on these whitespaces later */
textLine = textLine.replaceAll("\\s\\s+", " ");
textLine = textLine.replaceAll("\\n", "").replaceAll("\\r", "");
if (results.isEmpty()) {
results.add(textLine);
continue;
}
if (results.size() <= SIZE) {
results.add(textLine);
if (results.size() == SIZE) {
arrayBlockingQueueInput.put(new ArrayList<String>(results));
results.clear();
}
}
}
/* Count the remaining words in the list
* (last lines of the file do perhaps not fill up until the given SIZE, therefore need to be counted here)
* Fill the list with empty items if the size of the list does not match with the given SIZE */
while (results.size() != SIZE) {
results.add("");
}
arrayBlockingQueueInput.put(new ArrayList<String>(results));
List<String> list = new ArrayList<String>();
list.add(HALT);
arrayBlockingQueueInput.put(list);
results.clear();
} catch (InterruptedException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
主类：
public void main() throws IOException, InterruptedException {
System.out.println("Enter the word you want to find: ");
Scanner scan = new Scanner(System.in);
findWord = scan.nextLine();
System.out.println("Starting...");
long startTime = System.currentTimeMillis();
Thread producer = new Thread(new Producer());
producer.start();
ExecutorService executorService = Executors.newFixedThreadPool(CORE);
List<Future<Integer>> futureResults = new ArrayList<Future<Integer>>();
for (int i = 0; i < CORE; i++) {
futureResults.add(executorService.submit(new Consumer()));
}
executorService.shutdown();
for (Future<Integer> result : futureResults) {
try {
wordsInText += result.get();
} catch (ExecutionException | InterruptedException e) {
e.printStackTrace();
}
}
producer.join();
long stopTime = System.currentTimeMillis();
System.out.println("The word " + findWord + " appears " + wordsInText + " times in the given text");
System.out.println("Elapsed time was " + (stopTime - startTime) + " milliseconds.");
}
感谢@Ivan帮助我解决了调用结果的.clear方法。没有这个，代码解决方案将无法工作。