I wanted to ask - what is the difference between

Set<CurrencyType> set1 = new HashSet<>() {{
    add("a");
    add("b");
    add("c");
}}

and

Set<CurrencyType> set2 = Set.of(
    "a",
    "b",
    "c"
}}

In debug mode in a @Test named MySillyTest (this name will come back very soon) i can see that set1 is an instance of MySillyTest$1, but I assume it is just a HashSet. set2 on the other hand is an instance of ImmutableCollections$SetN. What is the real difference between those two? What implementation of java.util.Set is Set.of() using? Is there a performance/memory usage/cpu usage difference between those two?

CodePudding user response：

Don’t know, don’t care

The concrete class used by Set.of (and List.of, Map.of) is not documented. All we know is that the object returned (a) implements the interface, and (b) is unmodifiable.

That is all we need to know. We should not care about the particular concrete class used under the covers.

Being of unknown concrete class gives freedom to the implementors of the of methods.

• Those programmers are free to optimize according to the nature of your arguments. For example, if you are passing enum arguments, the highly optimized EnumSet class might be used behind the scenes.
• Those programmers are free to change their concrete class between versions of Java. For example, Java 17 implementation might return one concrete class while Java 18 returns another.

So you should never depend on a particular concrete class being utilized by the of/copyOf methods.

You asked:

What is the real difference between those two?

In your first one, we know the concrete class. And the resulting set is modifiable.

In your second one, we do not know the concrete class, nor do we care about the concrete class. And the resulting set is unmodifiable.

Avoid double-brace

As others said, it’s generally best to avoid double-brace initialization.

If you want the convenience of compact literals-style initialization of your modifiable collection, combine with the of methods. You can pass an existing collection to the constructor.

Set< String > set =
    new HashSet<>(
        Set.of( "a", "b", "c" )  // Pass a collection to constructor. 
    )
;

CodePudding user response：

The implementation class resulting from Set.of(...) is not guaranteed. It could change depending on the runtime implementation or in future versions. However, some of its characteristics—chiefly immutability—are guaranteed.

When you use "double-brace initialization", you are defining a new anonymous class that derives from the specified type. So MySillyTest$1 extends HashSet because that's what you specified. Note that double-brace initialization has problems; I don't allow it, and I discourage others from using it.

The important difference between the two is the immutability resulting from Set.of(...). If you need a mutable set, it's not an option. But if you can use an immutable set, it provides superior readability and performance.

Even if you need a mutable set, however, don't look at double-brace initialization as an alternative to Set.of(...); just use a HashSet in the conventional way.