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Home/ Questions/Q 8528385
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T08:47:23+00:00 2026-06-11T08:47:23+00:00

Sample code: class Sample{ private int v; public void setV(){ Lock a=new Lock(); a.lock();

  • 0

Sample code:

class Sample{
    private int v;
    public void setV(){
        Lock a=new Lock();
        a.lock();
        try{
            v=1;
        }finally{
            a.unlock();
        }
    }
    public int getV(){
        return v;
    }
}

If I have a thread constantly invoke getV and I just do setV once in another thread, Is that reading thread guaranteed to see the new value right after writing? Or do I need to make “V” volatile or AtomicReference?

If the answer is no, then should I change it into:

class Sample{
    private int v;
    private Lock a=new Lock();
    public void setV(){
        a.lock();
        try{
            v=1;
        }finally{
            a.unlock();
        }
    }
    public int getV(){
        a.lock();
        try{
            int r=v;
        }finally{
            a.unlock();
        }
        return r;
    }
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T08:47:24+00:00Added an answer on June 11, 2026 at 8:47 am

    From the documentation:

    All Lock implementations must enforce the same memory synchronization semantics as provided by the built-in monitor lock:

    • A successful lock operation acts like a successful monitorEnter action
    • A successful unlock operation acts like a successful monitorExit action

    If you use Lock in both threads (i.e. the reading and the writing ones), the reading thread will see the new value, because monitorEnter flushes the cache. Otherwise, you need to declare the variable volatile to force a read from memory in the reading thread.

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