目前,我正在开发一个项目,我必须使用 Threads
和 BlockingQueue
制作一个简单的程序:
该程序是凯撒编码,为什么 LinkedBlockingQueue
很好,因为它是一个赋值:)。
程序应从控制台读取字母(包括回车符和换行符 char(10)
char(13)
),这些字母将充当标志识别未编码字母的结束位置!
所以队列看起来像这样(假设移位因子为 3):
['a'->'a'->'a'->'13'->'d'->'d'->'d']
- 然后在编码时,正在读取的字母将被写入编码队列并从字母队列中删除。
所以我所做的如下:
代码:
import java.io.*;
import java.util.concurrent.*;
public class Caesar {
private final static int CR = 13,LF=10, SHIFT_FACTOR=3, OFFSET=23;
public static void main(String[] args) throws InterruptedException {
var br = new BufferedReader(new InputStreamReader(System.in));
var letters = new LinkedBlockingQueue<Character>();
var encoded = new LinkedBlockingQueue<Character>();
var done = new LinkedBlockingQueue<Boolean>();
Thread t1=new Thread(()-> takeInput(letters,br));
Thread t2=new Thread(()->encode(letters,encoded));
Thread t3=new Thread(()-> send(encoded,done));
t1.start();
t2.start();
t3.start();
t1.join();
t2.join();
t3.join();
System.out.println(done.take());
}
private static void takeInput(BlockingQueue<Character>letters, BufferedReader br ) {
System.out.println("Enter your input!.");
int temp;
try {
while((temp=br.read())!=-1) {
letters.put((char) temp);
System.out.println("added "+ temp+ " and letters queue is "+letters);
if (temp == CR || temp == LF){
break;
}
}
}catch (IOException | InterruptedException e){ e.printStackTrace(); }
finally {
try {
br.close();
} catch (IOException e) { e.printStackTrace(); }
}
}
public static void encode(BlockingQueue<Character>letters, BlockingQueue<Character>encoded) {
try {
char tempLetter;
while(!letters.isEmpty()){//dont know why its not going in here!!!! BLOCKED HERE debugger dont want to get in here!
System.out.println("inside encode");
tempLetter=letters.take();
if (tempLetter == CR ||tempLetter ==LF){
encoded.put(tempLetter);
break;
}else if (tempLetter == ' ' || tempLetter == '.') {
encoded.put(tempLetter);
}else if (cap(tempLetter) < 'X') {
encoded.put((char)(tempLetter+SHIFT_FACTOR));/
}else{
encoded.put((char)(tempLetter - OFFSET));
}
}
}catch (InterruptedException e) { e.printStackTrace(); }
}
private static Character cap(Character temp) {
return temp >= 'a' ? Character.toUpperCase(temp):temp;/
}
private static void send( BlockingQueue<Character> encoded, BlockingQueue<Boolean> done){
try {
char temp=encoded.take();
while (!encoded.isEmpty()){
System.out.println("stuck in encode!");
if(temp==CR || temp==LF)
break;
System.out.print(temp);
temp=encoded.take();
}
done.put(true);
}catch (InterruptedException e) {
e.printStackTrace();
}
}
}
所需输出
Enter your input!.
aaaa
started encoding
the encoded elements are : -> dddd
我得到的输出:!
Enter your input!.
aaa
added 97 and letters queue is [a]
added 97 and letters queue is [a, a]
added 97 and letters queue is [a, a, a]
added 10 and letters queue is [a, a, a,
]
(the program is on hold without and doesn't want to enter the while loop in the encrypt!!!!)
教授给了我们一个GO代码,作为实现功能的引用:
这里是 pastebin 上的 go 代码的链接
随意提出任何建议:)
最佳答案
t1.start();
t1.join();
t2.start();
t2.join();
t3.start();
t3.join();
您启动一个线程,然后等待它完成,然后再启动下一个线程。
将所有连接放在开始之后:
t1.start();
t2.start();
t3.start();
t1.join();
t2.join();
t3.join();
关于java - 使用 ListBlockingQueue 线程的并发运行由于某种原因被阻塞,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62373946/