How can I use reference method in a UnaryOperator java 8

Question

Currently, I have a UnaryOperator like this

UnaryOperator<Object> defaultParser = obj -> obj;

I don't know if I can use a method reference in these kinds of operation. Example:

UnaryOperator<String> defaultParser = String::toString;

But with the generic way, not just String.

`Object::toString` then? Since `toString` is defined on object, that should be what you're looking for in "the generic way". — Silvio Mayolo, Dec 05 '18 at 02:10
@SilvioMayolo but the Unary maybe receive an object, not just String, and I want that object back. — Dang Nguyen, Dec 05 '18 at 02:25
`String::toString` is not the same as `obj -> obj`, as the former will throw a `NullPointerException` for `null` input whereas the latter won’t. If that’s not a concern or even intended behavior, you can simply use `UnaryOperator defaultParser = Objects::requireNonNull;` — Holger, Dec 05 '18 at 10:49

score 4 · Accepted Answer · answered Dec 05 '18 at 02:22

If you just want to avoid the lambda expression, UnaryOperator has static identity() method:

UnaryOperator<Object> defaultParser = UnaryOperator.identity();

If you specifically want a method reference (why??), you can define a method in your class

public static <T> T identity(T t) {
    return t;
}

Then you will be able to use it as a method reference:

UnaryOperator<Object> defaultParser = MyClass::identity;

score 2 · Answer 2 · answered Dec 05 '18 at 02:24

2

Yes, you can using the UnaryOperator.identity() as:

UnaryOperator<Object> defaultParser = UnaryOperator.identity();

which is defined with a lambda expression as

static <T> UnaryOperator<T> identity() {
    return t -> t;
}

answered Dec 05 '18 at 02:24

Naman

2 Answers2