在下面的代码中,有可能输入类似于此问题“ Deadlocks and Synchronized methods”的死锁,现在我知道为什么两个线程都输入一个死锁了。 死锁,但是当我执行代码时,线程总是输入死锁,因此:
1-这段代码何时不可能出现死锁?
2-如何防止它发生?
我尝试使用 wait()和 notifyAll()这样:
wait()
waver.waveBack(this)
,然后在 waveBack()中调用 notifyAll(),但是我遗漏或误解了什么没用?
package mainApp;
public class Wave {
static class Friend {
private final String name;
public Friend(String name) {
this.name = name;
}
public String getName() {
return this.name;
}
public synchronized void wave(Friend waver) {
String tmpname = waver.getName();
System.out.printf("%s : %s has waved to me!%n", this.name, tmpname);
waver.waveBack(this);
}
public synchronized void waveBack(Friend waver) {
String tmpname = waver.getName();
System.out.printf("%s : %s has waved back to me!%n", this.name, tmpname);
}
}
public static void main(String[] args) {
final Friend friendA = new Friend("FriendA");
final Friend friendB = new Friend("FriendB");
new Thread(new Runnable() {
public void run() {
friendA.wave(friendB);
}
}).start();
new Thread(new Runnable() {
public void run() {
friendB.wave(friendA);
}
}).start();
}
}
答案 0 :(得分:1)
在这种情况下,仅在持有锁时不要调用可能需要该锁的其他方法。这样可以确保始终有一个方法能够获得锁定并可以取得进展的时刻。
在wait()
之前调用waver.waveBack(this)
会引起鸡和鸡蛋的问题:永远不会调用waveBack(this)
,因为线程会在wait()
语句处停止执行,因此notifyAll()
永远不会被调用以继续执行。
在示例的上下文中,有多种方法可以防止死锁,但是让我们从您所链接的问题的sarnold
到他的answer的评论之一中,提出建议。用sarnold
来解释:“通常更容易推断出数据锁定”。
让我们假设同步方法是同步的,以确保状态的一致性更新(即某些变量需要更新,但在任何给定时间只有一个线程可以修改这些变量)。例如,让我们注册发送和接收的wave数量。下面的可运行代码应对此进行演示:
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Random;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
public class Wave {
static class Waves {
final Map<Friend, Integer> send = new HashMap<>();
final Map<Friend, Integer> received = new HashMap<>();
void addSend(Friend f) {
add(f, send);
}
void addReceived(Friend f) {
add(f, received);
}
void add(Friend f, Map<Friend, Integer> m) {
m.merge(f, 1, (i, j) -> i + j);
}
}
static class Friend {
final String name;
public Friend(String name) {
this.name = name;
}
final Waves waves = new Waves();
void wave(Friend friend) {
if (friend == this) {
return; // can't wave to self.
}
synchronized(waves) {
waves.addSend(friend);
}
friend.waveBack(this); // outside of synchronized block to prevent deadlock
}
void waveBack(Friend friend) {
synchronized(waves) {
waves.addReceived(friend);
}
}
String waves(boolean send) {
synchronized(waves) {
Map<Friend, Integer> m = (send ? waves.send : waves.received);
return m.keySet().stream().map(f -> f.name + " : " + m.get(f))
.sorted().collect(Collectors.toList()).toString();
}
}
@Override
public String toString() {
return name + ": " + waves(true) + " / " + waves(false);
}
}
final static int maxThreads = 4;
final static int maxFriends = 4;
final static int maxWaves = 50_000;
public static void main(String[] args) {
try {
List<Friend> friends = IntStream.range(0, maxFriends)
.mapToObj(i -> new Friend("F_" + i)).collect(Collectors.toList());
ExecutorService executor = Executors.newFixedThreadPool(maxThreads);
Random random = new Random();
List<Future<?>> requests = IntStream.range(0, maxWaves)
.mapToObj(i -> executor.submit(() ->
friends.get(random.nextInt(maxFriends))
.wave(friends.get(random.nextInt(maxFriends)))
)
).collect(Collectors.toList());
requests.stream().forEach(f ->
{ try { f.get(); } catch (Exception e) { e.printStackTrace(); } }
);
executor.shutdownNow();
System.out.println("Friend: waves send / waves received");
friends.stream().forEachOrdered(p -> System.out.println(p));
} catch (Exception e) {
e.printStackTrace();
}
}
}