LockSupport. Park and unpark, wait and notify

LockSupport.park Method to block the current thread ,unpark You can resume its operation .

LockSupport And every thread that uses it has a license (permit) relation .permit amount to 1,0 The switch of , The default is 0, Call once unpark Just add 1 become 1, Call once park Will be spending permit, Also is to 1 become 0, meanwhile park Return immediately . Call again park Will become block（ because permit by 0 了 , It's going to get stuck here , until permit Turn into 1）, Call at this time unpark Will be able to permit Set as 1. Each thread has an associated permit, permit At most one , Repeated calls to unpark It won't accumulate .
park()/unpark() The underlying principle is “ Binary semaphore ”, You can think of it as if it had only one license Semaphore, But this semaphore is executing repeatedly unpark() There will be no more licenses , There is only one license at most .

Before one of the following three situations occurs , The current thread will be dormant ：

• Some other thread calls... With the current thread as the target unpark.
• Another thread interrupts the current thread .
• The call illogically ( That is, without any reason ) return .

Thread.sleep() Will not release the lock of possession ,LockSupport.park() The lock held by the current thread will not be released ;Object.wait() Will release the lock of possession .
Thread.sleep() Time must be passed in , Can't wake up from the outside , I can only wake up myself ;Object.wait() Can pass but not pass , With time , If not notify, It will wake up automatically when it's time , No transmission means it's stuck , Need another thread to use Object.notify() Wake up the ;LockSupport.park() Method can be called by another thread LockSupport.unpark() Method wake up ;
If in wait() It was done before notify() Will throw out IllegalMonitorStateException abnormal ; If in park() It was done before unpark(), Threads will not be blocked , Just skip park(), Continue with what follows ;