I've seen a lot of methods where a new class is instantiated in a lambda method reference but can't seem to understand why. When is the new keyword needed in a method reference?
For example, the following passes compilation:
UnaryOperator<String>stringToUpperCase = String::toUpperCase;
But this doesn't:
UnaryOperator<String>stringToUpperCase = new String()::toUpperCase;
String::toUpperCase is a method reference that can be applied to any String instance.
new String()::toUpperCase is a method reference that can be applied to a specific String instance (the instance created by new String()).
Since UnaryOperator<String> expects a method that takes a String and returns a String, String::toUpperCase fits (since you can apply it on a String and get the upper case version of that String).
On the other hand, new String()::toUpperCase doesn't fit UnaryOperator<String>, since it is executed on an already specified String, so you can't pass another String instance to it.
It can, however, by assigned to a Supplier<String>, since it simply supplies an empty String instance:
Supplier<String> emptyStringToUpperCase = new String()::toUpperCase;
This is similar to:
Supplier<String> emptyStringToUpperCase = () -> new String().toUpperCase();
while this:
UnaryOperator<String> stringToUpperCase = String::toUpperCase;
is similar to:
UnaryOperator<String> stringToUpperCase = s -> s.toUpperCase();
There are four kinds of method references as shown below and your type falls in the second category, but UnaryOperator<String> essentially needs to represent a method which accepts any String argument and returns a String. However, the non-working method reference that you have used is actually working on a particular String object (i.e. not any String object)
Refer: https://docs.oracle.com/javase/tutorial/java/javaOO/methodreferences.html
new String()::toUpperCase is indeed a method reference, of the third kind (new String() is an object which has the toUpperCase method). It doesn't take arguments, but returns a String. It could be used as a Supplier<String>. But it is a very complicated way to say () -> "".
s -> ""will do the same thing