Java中的Streams，无法想出这段代码

Question

I've found the following code snippet:

Function<Integer, Predicate<Integer>> smallerThan = x -> y -> y < x;
List<Integer> l = Arrays.asList(5, 6, 7, 23, 4, 5645,
    6, 1223, 44453, 60182, 2836, 23993, 1);

List<Integer> list2 = l.stream()
    .filter(smallerThan.apply(l.get(0)))
    .collect(Collectors.toList());

System.out.println(list2);

As output I receive:

[4, 1]

How does the smallerThan function in this example work, considering that we only pass one parameter smallerThan.apply(l.get(0)) ?

Answer 1

smallerThan is a Function that accepts a single Integer and returns a Predicate<Integer> (a Predicate<Integer> is a function that accepts a single Integer and returns a boolean ).

smallerThan.apply(l.get(0)) returns a Predicate<Integer> that looks like this:

y -> y < l.get(0)

ie it returns true if the input passed to it is smaller than l.get(0) .

When you pass that Predicate to filter , your Stream pipeline keeps only the elements smaller than l.get(0) .

Your pipeline can be re-written as:

List<Integer> list2 = l.stream()
    .filter(y -> y < l.get(0))
    .collect(Collectors.toList());

Since l.get(0) is 5 , your pipeline returns all the elements of the original list smaller than 5 .

Answer 2

That is called "currying" and it was possible before Java 8 too via anonymous classes for example, but it was much more verbose. Ultimately, it's a Function that returns a Function and while not that spread in Java (yet), in other functional languages it is used quite heavily.

Answer 3

The function smallerThan accepts a number and returns a Predicate object, in which case we apply this predicate to every element of the stream.

So, l.get(0) will retrieve the first value of the list l which is (5) , we then pass it to the smallerThan function and this function returns a predicate with the criteria y -> y < x; which reads as "given a number, return true if it is less than 5" hence output is [4, 1]