Immutability and reordering

Question

The code below (Java Concurrency in Practice listing 16.3) is not thread safe for obvious reasons:

public class UnsafeLazyInitialization {
    private static         


        

Answer 1

It is indeed safe is UnsafeLazyInitialization.resource is immutable, i.e. the field is declared as final:

private static final Resource resource = new Resource();

It might also be considered as thread-safe if the Resource class itself is immutable and does not matter which instance you are using. In that case two calls could return different instances of Resource without issue apart from an increased memory consumption depending on the number of threads calling getInstance() at the same time).

It seems far-fetched though and I believe there is a typo, real sentence should be

UnsafeLazyInitialization is actually safe if *r*esource is immutable.