我創建了兩個單獨的線程，一個用于寫入列表，第二個用于從列表中刪除。package com.produ.consu;public class Test { public static void main(String[] args) { Operations operations = new Operations(); Runnable r1 = new ThreadsClass(operations); Runnable r2 = new ThreadsClass(operations); Thread t1 = new Thread(r1); Thread t2 = new Thread(r2); t1.setName("READ"); t2.setName("WRITE"); t1.start(); t2.start(); }}以上是我創建線程的測試類。package com.produ.consu;public class ThreadsClass implements Runnable { Operations operations; ThreadsClass(Operations operations){ this.operations=operations; } @Override public void run() { // TODO Auto-generated method stub if(Thread.currentThread().getName().equals("WRITE")) { operations.writeList(); } else { operations.readList(); } }}以上是基于線程名稱調用同步方法的類：import java.util.ArrayList;import java.util.List;public class Operations { List<Integer> list=null; int count=0; boolean flag; Operations(){ list=new ArrayList<>(); flag=true; } public synchronized void writeList() { // TODO Auto-generated method stub while(true) { if(flag) { count++; list.add(count); System.out.println("inise if block...."+Thread.currentThread().getName()); System.out.println(list); flag=false; try { wait(); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } } else { notify(); } } }上面是我提到功能的地方。 所以寫線程必須將元素寫入列表并等待它被第二個線程刪除。一旦刪除第二個應該通知第一個并等待直到元素插入。但是得到...inise if block....寫 [1]甚至沒有被刪除，它必須是一個連續的過程。給我關于給定代碼的建議。
3 回答
HUH函數
TA貢獻1836條經驗 獲得超4個贊
您也應該wait()在else塊內，否則while如果不滿意，它將繼續運行if并且其他線程將沒有機會執行。
執行讀/寫操作后，線程應調用notify以喚醒另一個線程，并且本身應進入等待狀態。
public synchronized void writeList() {
while (true) {
if (flag) {
count++;
list.add(count);
System.out.println("inise if block...." + Thread.currentThread().getName());
System.out.println(list);
flag = false;
try {
notify(); //notify the read thread that write is complete
wait(); // go into the waiting state so that no further write is done until the current element is removed by the read thread.
} catch (InterruptedException e) {
e.printStackTrace();
}
} else {
try {
wait(); //wait in else, otherwise while will run endlessly
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
public synchronized void readList() {
while (true) {
System.out.println("in read");
if (!flag) {
Integer i = list.remove(0);
System.out.println(i + "..removed at index by" + Thread.currentThread().getName());
flag = true;
try {
notify(); //notify write thread that read is complete
wait(); //go into wait until new element is inserted
} catch (InterruptedException e) {
e.printStackTrace();
}
} else {
try {
wait(); //wait in else otherwise while runs endlessly
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
茅侃侃
TA貢獻1842條經驗 獲得超22個贊
您應該用更細粒度的同步替換方法同步：使用同步塊包裝訪問列表和更新計數的操作。
或者，使用鏈接阻塞隊列在線程之間傳遞“消息”（或者，在更高級的場景中，使用 Akka、Kafka、zeromq 等消息傳遞解決方案或其他類似解決方案）
紅糖糍粑
TA貢獻1815條經驗 獲得超6個贊
由于您一次只添加一個元素，因此您也可以使用 Exchanger?？匆幌?Java 并發包。
import java.io.IOException;
import java.util.concurrent.Exchanger;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.TimeUnit;
public class Foo {
public static void main(String[] args) throws IOException, InterruptedException {
final Exchanger<Integer> exchanger = new Exchanger<>();
ExecutorService executorService = Executors.newCachedThreadPool();
executorService.execute(() -> {
// producer
int count = 0;
while (!Thread.currentThread().isInterrupted()) {
try {
exchanger.exchange(count++);
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
}
});
executorService.execute(() -> {
// reader
while (!Thread.currentThread().isInterrupted()) {
try {
System.out.println("consume " + exchanger.exchange(null));
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
}
});
System.in.read();
executorService.shutdownNow();
executorService.awaitTermination(10, TimeUnit.SECONDS);
System.out.println("Shut down");
}
}
添加回答
舉報
0/150
提交
取消