Lambda expression java 8 map method

Question

The syntax of map method in java 8 is :

<R> Stream<R> map(Function<? super T,? extends R> mapper)

but i can use a lambda expression :

personList.stream().filter(p -> p.getPersonType().equals("student"))
            .map(p -> new Student(p.getId(), p.getName()))
            .collect(Collectors.toList());

How does the argument in map method equates to a Function datatype.Please help me understand this .

Thanks

Solution 1

The function Function<? super T,? extends R> mapper of the map method basically represents any function taking one parameter and returning a value so in this specific case the lambda p -> new Student(p.getId(), p.getName()) is a function taking a Person p and returning a Student hence fits into that perfectly.

To look at it another way, the lambda is equivalent to:

.map(new Function<Person, Student>() {
     @Override
      public Student apply(Person p) {
            return new Student(p.getId(), p.getName());
      }
})

Solution 2

You can write like this Function<Person,Student> function = p -> new Student(p.getId(), p.getName())

so as you see it is represent a function.

personList.stream().filter(p -> p.getPersonType().equals("student"))
        .map(function::apply) // p -> function.apply(p)
        .collect(Collectors.toList());

Solution 3

In the context of your Stream<Person> (assuming personList is a List<Person>), the lambda expression p -> new Student(p.getId(), p.getName()) implements a functional interface that accepts a Person instance (an element of your Stream) and returns a Student instance.

Hence it matches the Function<Person,Student> functional interface, which is the interface required by the map() method.

Solution 4

At runtime, the following lambda expression:

p -> new Student(p.getId(), p.getName())

will be represented by the class that implements Function<T, R> interface.

An instance of this functional interface can be passed as a parameter to Stream.map(...) method:

<R> Stream<R> map(Function<? super T, ? extends R> mapper);

In order to get a better understanding of how this works, one could replace a lambda with a good old anonymous class that implements the corresponding interface.

.map(p -> new Student(p.getId(), p.getName())) is equivalent to:

.map(new Function<Person, Student>() {
    @Override
    public Student apply(Person p) {
        return new Student(p.getId(), p.getName());
    }
})

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Comments

[jayendra bhatt]

while i got stuck here i found this explaination infoq.com/articles/Java-8-Lambdas-A-Peek-Under-the-Hood quite helpfull too.