我正在做一些实践作业,并尝试使用一些虚拟代码,以便更好地理解线程和锁的概念。下面是一段(有时)陷入死锁的代码。
A.java
public class A {
private B b;
public void setB(B b) {
this.b = b;
}
public synchronized void foo(boolean callBar) {
System.out.println("foo");
if (callBar) {
b.bar(false);
}
}
}B.java
public class B {
private A a;
public void setA(A a) {
this.a = a;
}
public synchronized void bar(boolean callFoo) {
System.out.println("bar");
if (callFoo) {
a.foo(false);
}
}
}Demo.java
public class Demo {
public static void main(String[] args) {
A a = new A();
B b = new B();
a.setB(b);
b.setA(a);
new Thread(() -> {
a.foo(true);
}).start();
new Thread(() -> {
b.bar(true);
}).start();
}
}解决方案:我使用Lock而不是synchronized。
A.java
public class A {
private final ReentrantLock lock = new ReentrantLock();
private B b;
public void setB(B b) {
this.b = b;
}
public ReentrantLock lock() {
return lock;
}
public boolean impendingExecute() {
Boolean thisLock = false;
Boolean otherLock = false;
try {
thisLock = lock.tryLock();
otherLock = b.lock().tryLock();
} finally {
if (!(thisLock && otherLock)) {
if (thisLock) {
lock.unlock();
}
if (otherLock) {
b.lock().unlock();
}
}
}
return thisLock && otherLock;
}
public void foo(boolean callBar) {
System.out.println("foo");
if (callBar && impendingExecute()) {
try {
b.bar(false);
} finally {
lock.unlock();
b.lock().unlock();
}
}
}
}B.java
public class B {
private final ReentrantLock lock = new ReentrantLock();
private A a;
public void setA(A a) {
this.a = a;
}
public ReentrantLock lock() {
return lock;
}
public boolean impendingExecute() {
Boolean thisLock = false;
Boolean otherLock = false;
try {
thisLock = lock.tryLock();
otherLock = a.lock().tryLock();
} finally {
if (!(thisLock && otherLock)) {
if (thisLock) {
lock.unlock();
}
if (otherLock) {
a.lock().unlock();
}
}
}
return thisLock && otherLock;
}
public void bar(boolean callFoo) {
System.out.println("bar");
if (callFoo && impendingExecute()) {
try {
a.foo(false);
} finally {
lock.unlock();
a.lock().unlock();
}
}
}
}在进行上述更改后,代码不会导致死锁。这是实现这个方法的正确方法吗(基本上,我希望对impendingExecute()方法进行审查。)此外,(有点偏离审查)是否有任何现实世界的场景,我可以遇到这种情况?
注意:我在“代码评论”上发布了这个问题,但似乎对虚拟代码的评论已经过时了。
发布于 2017-05-21 08:21:09
您可以使用java.util.concurrent.locks.ReentrantLock。此设计允许该方法尝试获取两个类的锁,如果失败则释放锁,并在必要时再试一次。如果您需要尝试直到成功,那么您需要在循环中放置一个it并使用一些策略终止。
while (true) {
if (this.lock.tryLock()) {
try {
if (ba.lock.tryLock()) {
try {
//some logic
break;
} finally {
ba.lock.unlock();
}
}
} finally {
this.lock.unlock();
}
}
int n = number.nextInt(1000);
int TIME = 1000 + n; // 1 second + random delay to prevent livelock
Thread.sleep(TIME);
}或者您可以使用此解决方案来确保以相同的顺序获取和释放多个锁:
if (compareTo(ba) < 0) {
former = this;
latter = ba;
} else {
former = ba;
latter = this;
}
synchronized (former) {
synchronized (latter) {
//Some logic
}
}
}https://stackoverflow.com/questions/44094584
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