Function.class中的下界通配符的目的是什么？

Question

In Function.class from Java8, we have:

default <V> Function<V, R> compose(Function<? super V, ? extends T> before) {
    Objects.requireNonNull(before);
    return (V v) -> apply(before.apply(v));
}

Compose accepts:

Function<? super V, ? extends T> before

Rather than:

Function<V, ? extends T> before

Is there any plausible situation in which the fact that "V" is lower bounded matters?

Answer 1

The ? super ? super allows the returned Function 's input type ( V ) to be different from the arguments input type.

For example, this compiles with the ? super ? super version but not the alternate one.

Function<Object, String> before = Object::toString;
Function<String, Integer> after = Integer::parseInt;
Function<Integer, Integer> composed = after.compose(before);