Lambda expression to add object into list

Question

there is a List and i want to create books object and add all the book object into this List. here I did with normal approach like this .

public static List<Books> listOfBooks = new ArrayList<>();
public static void addProduct() {
        Books book1 = new Books("java", "games", 200.56);
        Books book2 = new Books("c", "dennis", 300.56);
        Books book3 = new Books("c++", "Bjarne", 250.56);
        Books book4 = new Books("javaScript", "Brendan", 209.56);
        Books book5 = new Books("Sql", "Donald", 249.56);
        listOfBooks.add(book1);
        listOfBooks.add(book2);
        listOfBooks.add(book3);
        listOfBooks.add(book4);
        listOfBooks.add(book5);
    }

I want the same operation by using lambda expression but i do not know how to do it. please help me out?

yeah @flown i know the basic of lamda and while learning i got these doubt . — chand mohd
– chand mohd, Commented Nov 2, 2017 at 6:48
Basically lambda expressions are single abstract method interfaces. So, which interface would you implement to do your task? Or are you trying to use the Stream API? — Flown
– Flown, Commented Nov 2, 2017 at 6:51
@Flown i'm not dealing with interface right now yes i'm trying to use stream API. and i'm trying to re factor above piece of code by lamda expression but i'm not getting how to do so i asked — chand mohd
– chand mohd, Commented Nov 2, 2017 at 7:02

janos · Accepted Answer · 2017-11-02 06:57:30Z

1

You could write:

Arrays.asList(book1, book2, ...).forEach(b -> listOfBooks.add(b));

Or shorter using a method reference instead of a lambda:

Arrays.asList(book1, book2, ...).forEach(listOfBooks::add);

But that's not really better than writing:

listOfBooks.addAll(Arrays.asList(book1, book2, ...));

edited Nov 2, 2017 at 6:57

answered Nov 2, 2017 at 6:48

janos

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4 Comments

chand mohd Over a year ago

thank you @janos but if i want to just write this piece of code by using lamda expression then how would i do that?

janos Over a year ago

@chandmohd I changed the first variation to use a lambda expression. But that's missing the point. This is not a typical or recommended use case for lambdas.

chand mohd Over a year ago

i got the point but as you said that this is not recommended use .why ? are you saying in term of performance? if yes then how would i see which method taking how much time to execute ?

janos Over a year ago

@chandmohd If you want to add all elements of list X to list Y, it's best to use Y.addAll(X) because that's the API designed for this, and you can expect implementations to be more optimal than other hand-crafted solutions. If you have individual items and you want to add them to a list, then the natural way is to add them one by one, rather than creating temporary objects to wrap them just so that you can use lambdas. A better example using lambdas would be something that requires passing a function as parameter.

Flown · Accepted Answer · 2017-11-02 07:08:12Z

0

The Stream API is using lambda expression, but are not part of it.

public static List<Books> listOfBooks = new ArrayList<>();
public static void addProduct() {
  String[] languages = {"java", "c", "c++", "javascript", "Sql"};
  String[] authors = {"games", "dennis", "Bjarne", "Brendan", "Donald"};
  double[] prices = {200.56d, 300.56d, 250.56d, 209.56d, 249.56d};
  listOfBooks = IntStream.range(0, languages.length)
                        .mapToObj(i -> new Books(languages[i], authors[i], prices[i]))
                        .collect(Collectors.toList());
  // Alternatively:
  /* IntStream.range(0, languages.length)
          .mapToObj(i -> new Books(languages[i], authors[i], prices[i]))
          .forEach(listOfBooks::add);
  */
}

answered Nov 2, 2017 at 7:08

Flown

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1 Comment

chand mohd Over a year ago

Thank you @Flown appreciated!!

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