Collections.synchronizedList 是如何让一个集合变成线程安全的呢?为什么线程安全的集合效率非常低下?
那一定是你没有仔细查看实现源码。懒得找源码请直接看下文
// Synch Wrappers
/**
* Returns a synchronized (thread-safe) collection backed by the specified
* collection. In order to guarantee serial access, it is critical that
* <strong>all</strong> access to the backing collection is accomplished
* through the returned collection.<p>
*
* It is imperative that the user manually synchronize on the returned
* collection when iterating over it:
* <pre>
* Collection c = Collections.synchronizedCollection(myCollection);
* ...
* synchronized (c) {
* Iterator i = c.iterator(); // Must be in the synchronized block
* while (i.hasNext())
* foo(i.next());
* }
* </pre>
* Failure to follow this advice may result in non-deterministic behavior.
*
* <p>The returned collection does <i>not</i> pass the <tt>hashCode</tt>
* and <tt>equals</tt> operations through to the backing collection, but
* relies on <tt>Object</tt>'s equals and hashCode methods. This is
* necessary to preserve the contracts of these operations in the case
* that the backing collection is a set or a list.<p>
*
* The returned collection will be serializable if the specified collection
* is serializable.
*
* @param c the collection to be "wrapped" in a synchronized collection.
* @return a synchronized view of the specified collection.
*/
public static <T> Collection<T> synchronizedCollection(Collection<T> c) {
return new SynchronizedCollection<>(c);
}
重点来了~
/**
* @serial include
*/
static class SynchronizedCollection<E> implements Collection<E>, Serializable {
private static final long serialVersionUID = 3053995032091335093L;
final Collection<E> c; // Backing Collection
final Object mutex; // Object on which to synchronize
SynchronizedCollection(Collection<E> c) {
if (c==null)
throw new NullPointerException();
this.c = c;
mutex = this;
}
SynchronizedCollection(Collection<E> c, Object mutex) {
this.c = c;
this.mutex = mutex;
}
public int size() {
synchronized (mutex) {return c.size();}
}
public boolean isEmpty() {
synchronized (mutex) {return c.isEmpty();}
}
public boolean contains(Object o) {
synchronized (mutex) {return c.contains(o);}
}
public Object[] toArray() {
synchronized (mutex) {return c.toArray();}
}
public <T> T[] toArray(T[] a) {
synchronized (mutex) {return c.toArray(a);}
}
public Iterator<E> iterator() {
return c.iterator(); // Must be manually synched by user!
}
public boolean add(E e) {
synchronized (mutex) {return c.add(e);}
}
public boolean remove(Object o) {
synchronized (mutex) {return c.remove(o);}
}
public boolean containsAll(Collection<?> coll) {
synchronized (mutex) {return c.containsAll(coll);}
}
public boolean addAll(Collection<? extends E> coll) {
synchronized (mutex) {return c.addAll(coll);}
}
public boolean removeAll(Collection<?> coll) {
synchronized (mutex) {return c.removeAll(coll);}
}
public boolean retainAll(Collection<?> coll) {
synchronized (mutex) {return c.retainAll(coll);}
}
public void clear() {
synchronized (mutex) {c.clear();}
}
public String toString() {
synchronized (mutex) {return c.toString();}
}
private void writeObject(ObjectOutputStream s) throws IOException {
synchronized (mutex) {s.defaultWriteObject();}
}
}
通过以上源码可以知道,实现线程安全的方法就是暴力加锁,这效率能不低下吗?
当然,也不能说这代码有有啥问题,毕竟搞开发的,需求至上~