Java 8 Lambdas: Mapping a Stream to type Integer and then calling sum() won't compile

Question

I was playing around with Java8 lambda expressions. As an example I then tried to sum up the ages persons contained in a list:

import java.util.Arrays;
import java.util.List;

public class Person {
    public static void main(String[] args) {
        List<Person> persons = Arrays.asList(new Person("FooBar", 12), new Person("BarFoo", 16));
        Integer sumOfAges = persons.stream().map(Person::getAge).sum();
        System.out.println("summedUpAges: " + sumOfAges);
    }

    private final String name;
    private final Integer age;

    public Person(String name, Integer age) {
        this.name = name;
        this.age = age;
    }

    public String getName() {
        return name;
    }

    public Integer getAge() {
        return age;
    }
}

When I try to compile this snippet with the following java compiler:

openjdk version "1.8.0-ea"
OpenJDK Runtime Environment (build 1.8.0-ea-lambda-night
OpenJDK 64-Bit Server VM (build 25.0-b21, mixed mode)

I get the following compile error:

java: cannot find symbol
  symbol:   method sum()
  location: interface java.util.stream.Stream<java.lang.Integer>

But if I change the return value of the getAge() method from Integer to int, I get the expected result. But sometimes it is not possible or desired to change signatures on the fly. Is there any way to make this working when getAge() returns a Integer type?

Thanks in advance

Answer 1

I think they are changing the method

IntStream map(ToIntFunction<? super T> mapper)

to

IntStream mapToInt(ToIntFunction<? super T> mapper)

then your

persons.stream().mapToInt(Person::getAge).sum()

will always work, whether getAge() returns int or Integer.

See this thread: http://mail.openjdk.java.net/pipermail/lambda-libs-spec-experts/2013-March/001480.html

BTW you can post your questions to lambda-dev mailing list, they'd like to hear real world experiences.

Answer 2

They are not changing the methods, both of them are available where as map() is a generalized method that returns Stream, and they have separated the int, long, double Stream into separate method such as mapToInt(), mapToLong() and mapToDouble().

Yes, you can use reduce() method

reduce(T identity, BinaryOperator< T > accumulator)

Performs a reduction on the elements of this stream, using an associative accumulation function, and returns an Optional describing the reduced value, if any.

How reduce() works: suppose we have stream of int array and we apply the reduce function

reduce(0, (l, r) -> l + r)

In a list of integers such as 1, 2, 3, 4, and 5, the seed 0 is added to 1 and the result (1) is stored as the accumulated value, which then serves as the left-hand value in addition to serving as the next number in the stream (1+2). The result (3) is stored as the accumulated value and used in the next addition (3+3). The result (6) is stored and used in the next addition (6+4), and the result is used in the final addition (10+5), yielding the final result 15.

So you example would look like:

import java.util.Arrays;
import java.util.List;
import java.util.stream.Stream;

public class Person {

    public static void main(String[] args) {
        List<Person> persons = Arrays.asList(new Person("FooBar", 12), new Person("BarFoo", 16));
        Stream<Integer> stream = persons.stream().map(x -> x.getAge());
        int sum = stream.reduce(0, (l, r) -> l + r);
        System.out.println(sum);

    }

    private final String name;
    private final Integer age;

    public Person(String name, Integer age) {
        this.name = name;
        this.age = age;
    }

    public String getName() {
        return name;
    }

    public Integer getAge() {
        return age;
    }
}