问题
The definition of a functional Interface in Java 8 says:
A functional interface is defined as any interface that has exactly one explicitly declared abstract method. (The qualification is necessary because an interface may have non-abstract default methods.) This is why functional interfaces used to be called Single Abstract Method (SAM) interfaces, a term that is still sometimes seen.
So how come we have this:
List<Double> temperature =
new ArrayList<Double>(Arrays.asList(new Double[] { 20.0, 22.0, 22.5 }));
temperature.sort((a, b) -> a > b ? -1 : 1);
As the sort
method in List
is:
default void sort(Comparator<? super E> c) {
Object[] a = this.toArray();
Arrays.sort(a, (Comparator) c);
ListIterator<E> i = this.listIterator();
for (Object e : a) {
i.next();
i.set((E) e);
}
}
And the lambda expression says:
Lambda Expression should be assignable to a Functional Interface
The Comparator
interface has two abstract methods which are compare
and equals
and is annotated with @FunctionalInterface
. Does not this violate the definition of a functional interface having only one abstract method?
回答1:
It is true that the Comparator interface has 2 abstract methods. But one of them is equals
, which overrides the equals
method defined in the Object
class, and this method doesn't count.
From @FunctionalInterface:
If an interface declares an abstract method overriding one of the public methods of
java.lang.Object
, that also does not count toward the interface's abstract method count since any implementation of the interface will have an implementation fromjava.lang.Object
or elsewhere.
As such, this makes the Comparator
interface a functional interface where the functional method is compare(o1, o2).
The lambda expression (a, b) -> a > b ? -1 : 1
complies with that contract: it declares 2 parameters and return an int
.
来源:https://stackoverflow.com/questions/36844387/definition-of-functional-interface-in-java-8