synchronized似乎没有像我预期的那样工作。它不应该使括号内的代码相对于由同一对象进行synchronized的单独代码块是原子的吗?
我发现它在我的代码中根本不同步。
private Object movementMutex_ = new Object();
// Thread
public void run()
{
while (run_)
{
synchronized(movementMutex_)
{
if (timeToMove_)
{
Log.v("meh", "timeToMove_ was true, moving");
makeMove();
Log.v("meh", "Move Complete. Setting timeToMove_ to false");
timeToMove_ = false;
Log.v("meh", "timeToMove_ is now false");
}
}
}
}
// Called by a different thread so that this thread knows when to make a move
public void move()
{
Log.v("meh", "awaiting movementMutex in move()");
// Synchronizing so that timeToMove_ doesn't get set true while in the middle of moving and thus setting it back false prematurely
synchronized(movementMutex_)
{
Log.v("meh", "move called, setting timeToMove_");
timeToMove_ = true;
Log.v("meh", "timeToMove_ is now true");
}
}查看Log打印输出,我发现它们的打印顺序出乎意料。根据我对synchronized的理解,粗体的语句应该无法被非粗体的打印输出解释,但事实并非如此,我试图避免的事情正在发生:我错过了下一步,因为我在它仍然为真的时候将其设置为真,并立即将其转换为假。
08-12 10:47:19.860: V/meh(27639): awaiting movementMutex in move()
08-12 10:47:19.985: V/meh(27639): move called, setting timeToMove_
08-12 10:47:19.985: V/meh(27639): timeToMove_ is now true
08-12 10:47:19.985: V/meh(27639): **timeToMove_ was true, moving**
08-12 10:47:20.352: V/meh(27639): awaiting movementMutex in move() 下一行应该是不可能的。它不是在等待movementMutex!
08-12 10:47:20.352: V/meh(27639): move called, setting timeToMove_
08-12 10:47:20.360: V/meh(27639): timeToMove_ is now true
08-12 10:47:20.360: V/meh(27639): **Move Complete. Setting timeToMove_ to false**
08-12 10:47:20.360: V/meh(27639): **timeToMove_ is now false** 发布于 2012-08-15 00:35:25
正如对该问题的注释中所提到的,线程能够重新进入它自己的同步锁。"makeMove()“函数调用实际上是调用"move()”的发起者,这意味着它在同一线程下执行,因此不会被锁定。简而言之,下面的代码不会导致死锁;它会毫无问题地执行,这就是我所经历的。
synchronized(movementMutex_)
{
synchronized(movementMutex_)
{
doSomething();
}
}https://stackoverflow.com/questions/11923296
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