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Home/ Questions/Q 5999815
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T00:36:20+00:00 2026-05-23T00:36:20+00:00

I am reading this book called Java Concurrency in Practice and the author gives

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I am reading this book called “Java Concurrency in Practice” and the author gives an example of an unsafe object publication. Here is the example.

public Holder holder;

public void initialize(){
   holder = new Holder(42);
}

and

public class Holder {
    private int n;
    public Holder(int n) { this.n = n; }
    public void assertSanity() {
        if (n != n)
            throw new AssertionError("This statement is false.");
    }
}

So does this mean that other thread has access to an object when it is not even fully constructed? I guess that when a thread A calls holder.initialize(); and thread B calls holder.assertSanity(); the condition n != n will not be met if thread A has not yet executed this.n = n;

Does this also mean that if I have a simpler code like

int n;

System.out.println(n == n); //false?
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  1. Editorial Team
    Editorial Team
    2026-05-23T00:36:21+00:00Added an answer on May 23, 2026 at 12:36 am

    A problem can occur if the assertSanity method is pre-empted between the first and second load of n (the first load would see 0 and the second load would see the value set by the constructor). The problem is that the basic operations are:

    1. Allocate space for the object
    2. Call the constructor
    3. Set holder to the new instance

    The compiler/JVM/CPU is allowed to reorder steps #2 and #3 since there are no memory barriers (final, volatile, synchronized, etc.)

    From your second example, it’s not clear if “n” is a local variable or a member variable or how another thread might be simultaneously mutating it.

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