How can an immutable object, with non-final fields, be thread unsafe?

Question

say we have this

// This is trivially immutable.
public class Foo {
    private String bar;
    public Foo(String bar) {
        this.bar = bar;
    }
    public String getBar() {
        return bar;
    }
}

What makes this thread unsafe ? Following on from this question.

Answer 1

Foo is thread safe once it has been safely published. For example, this program could print "unsafe" (it probably won't using a combination of hotspot/x86) - if you make bar final it can't happen:

public class UnsafePublication {

    static Foo foo;

    public static void main(String[] args) {
        new Thread(new Runnable() {
            @Override
            public void run() {
                while (foo == null) {}
                if (!"abc".equals(foo.getBar())) System.out.println("unsafe");
            }
        }).start();

        new Thread(new Runnable() {
            @Override
            public void run() {
                foo = new Foo("abc");
            }
        }).start();
    }
}

Answer 2

Due to JVM optimization, you can never assume that operations are executed in the order they are written, unless it matters for the same thread. So when you call the constructor and then pass a reference to the resulting object to another thread, the JVM might not actually write the value of foo.bar before it is needed within the same thread.

That means that in a multithreaded environment, the getBar method could be called before the value in the constructor was written to it.