单利模式为何要用volatile

Posted 2020-10-31 coderDu

防止指令重排：对volatile的写操作先于对volatile的读操作。

synchronized虽然保证了原子性，但却没有保证指令重排序的正确性，会出现A线程执行初始化，但可能因为构造函数里面的操作太多了，所以A线程的uniqueInstance实例还没有造出来，但已经被赋值了。而B线程这时过来了，错以为uniqueInstance已经被实例化出来，一用才发现uniqueInstance尚未被初始化.

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The volatile prevents memory writes from being re-ordered, making it impossible for other threads to read uninitialized fields of your singleton through the singleton‘s pointer.

Consider this situation: thread A discovers that uniqueInstance == null, locks, confirms that it‘s still null, and calls singleton‘s constructor. The constructor makes a write into member XYZ inside Singleton, and returns. Thread A now writes the reference to the newly created singleton into uniqueInstance, and gets ready to release its lock.

Just as thread A gets ready to release its lock, thread B comes along, and discovers that uniqueInstance is not null. Thread B accesses uniqueInstance.XYZ thinking that it has been initialized, but because the CPU has reordered writes, the data that thread A has written into XYZ has not been made visible to thread B. Therefore, thread B sees an incorrect value inside XYZ, which is wrong.

When you mark uniqueInstance volatile, a memory barrier is inserted. All writes initiated prior to that of uniqueInstance will be completed before the uniqueInstance is modified, preventing the reordering situation described above.

解释：维基百科

1. Thread A notices that the value is not initialized, so it obtains the lock and begins to initialize the value.

2. Due to the semantics of some programming languages, the code generated by the compiler is allowed to update the shared variable to point to a partially constructed object before A has finished performing the initialization. For example, in Java if a call to a constructor has been inlined then the shared variable may immediately be updated once the storage has been allocated but before the inlined constructor initializes the object.[6]

3. Thread B notices that the shared variable has been initialized (or so it appears), and returns its value. Because thread B believes the value is already initialized, it does not acquire the lock. If B uses the object before all of the initialization done by A is seen by B (either because A has not finished initializing it or because some of the initialized values in the object have not yet percolated to the memory B uses (cache coherence)), the program will likely crash.