基於ID的線程同步

我需要一種方法來只允許一個線程修改與服務票據相關的數據。不止一個線程可能試圖同時修改票證數據。基於ID的線程同步

下面是我的方法的簡化版本。有一個更好的方法嗎？也許與java.util.concurrent包？

public class SomeClass1 
{ 
    static final HashMap<Integer, Object> ticketLockMap = new HashMap<Integer, Object>(); 


    public void process(int ticketNumber) 
    { 
     synchronized (getTicketLock(ticketNumber)) 
     { 
      // only one thread may modify ticket data here 

      // ... ticket modifications here... 
     } 
    } 


    protected static Object getTicketLock(int ticketNumber) 
    { 
     Object ticketLock; 

     // allow only one thread to use map 
     synchronized (ticketLockMap) 
     { 
      ticketLock = ticketLockMap.get(ticketNumber); 

      if (ticketLock == null) 
      { 
       // first time ticket is locked 
       ticketLock = new Object(); 
       ticketLockMap.put(ticketNumber, ticketLock); 
      } 
     } 

     return ticketLock; 
    } 
}

此外，如果我不希望HashMap中填充了未使用的鎖呢，我需要像下面這樣的更復雜的方法：

public class SomeClass2 
{ 
    static final HashMap<Integer, Lock> ticketLockMap = new HashMap<Integer, Lock>(); 


    public void process(int ticketNumber) 
    { 
     synchronized (getTicketLock(ticketNumber)) 
     { 
      // only one thread may modify ticket data here 

      // ... ticket modifications here... 

      // after all modifications, release lock 
      releaseTicketLock(ticketNumber); 
     } 
    } 


    protected static Lock getTicketLock(int ticketNumber) 
    { 
     Lock ticketLock; 

     // allow only one thread to use map 
     synchronized (ticketLockMap) 
     { 
      ticketLock = ticketLockMap.get(ticketNumber); 

      if (ticketLock == null) 
      { 
       // first time ticket is locked 
       ticketLock = new Lock(); 
       ticketLockMap.put(ticketNumber, ticketLock); 
      } 
     } 

     return ticketLock; 
    } 


    protected static void releaseTicketLock(int ticketNumber) 
    { 
     // allow only one thread to use map 
     synchronized (ticketLockMap) 
     { 
      Lock ticketLock = ticketLockMap.get(ticketNumber); 

      if (ticketLock != null && --ticketLock.inUseCount == 0) 
      { 
       // lock no longer in use 
       ticketLockMap.remove(ticketLock); 
      } 
     } 
    } 
} 


class Lock 
{ 
    // constructor/getters/setters omitted for brevity 
    int inUseCount = 1; 
}

來源

2013-06-12 Jeff Miller

您應該能夠使用'ConcurrentHashMap'而不是'HashMap'來解決票證鎖定問題，並避免執行不良的'synchronized'位。 'putIfAbsent（）'應該正是你所需要的。 – manub

@Jeff檢查這個線程http://stackoverflow.com/questions/659915/synchronizing-on-an-integer-value –

你可能會尋找Lock interface。第二種情況可以通過ReentrantLock來解決，它計算它被鎖定的次數。

鎖具有.lock()方法，該方法等待鎖獲取和.unlock方法，其應該被稱爲像

Lock l = ...; 
l.lock(); 
try { 
    // access the resource protected by this lock 
} finally { 
    l.unlock(); 
}

這可以然後與HashMap<Integer, Lock>組合。您可以省略調用並削減代碼行。

來源

2016-03-30 14:47:30

基於ID的線程同步

回答

相關問題