同步块中的volatile变量是否会阻止锁定省略？

Question

例如，

public class Counter {
    private volatile long count;

    public long vaule() {
         //Since count is volatile and this is read only,
         // We don't need to use synchronized here 
        return count;
    }

    // This method will only be accessed from a single thread.
    public void increment() {
        synchronized (this) {
            count++;
        }
    }

}

假设上述增量只能从一个线程访问，那么由于为同步引入的锁定消除（Elision）优化，JIT编译器将优化该方法中的synchronized ？