current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Code Review Stack Exchange is a question and answer site for peer programmer code reviews. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
up vote 2 down vote favorite

Java 8 streams are my new golden hammer that I try to use as often as possible with often enormous gains in brevity and readability. Getting multiple return values, such as a maximum value and its associated element, is a bit cumbersome though (and requires an additional "pair" class).

Should I go back to "normal Java" for this task or is this syntax preferable? Is there a more concise way? 

List<String> names = Arrays.asList("John","Paul","Ringo"); 
Pair<String,Integer> longestName = names.stream()
 .map(n->new Pair<>(n,n.length())) // pretend n.length() to be a lengthy operation
 .max(Comparator.comparing(Pair::getB)).get();
System.out.println(longestName.getA()+" "+longestName.getB());

P.S.: With "lengthy operation" I mean long running, as in I don't want to calculate it twice.

share|improve this question
up vote 2 down vote

Your concept is fine. My only criticism is in code style, and perhaps you should have a specific/custom container instead of the Pair. getA() and getB() should be getName() and getLength(). This also allows you to use primitive values and not Integer, and removes the confusing generic types.

Also:

  • don't get from Optionals at the stream end.
  • use white-space, it's your friend.
  • use meaningful names for lambda parameters (name is better than n )

The code I would write would look like:

Optional<Operation> maxOp = names.stream()
                  .map (name -> new Operation(name, name.length()))
                  .max (Comparator.comparingInt(Operation::getLength()));
Operation longest = maxOp.get();

System.out.println(longest.getName() + " " + longest.getLength());
share|improve this answer
up vote 2 down vote

You don't need the Pair. You can get the String with the maximum length by doing

names.stream().max(Comparator.comparingInt(String::length))
share|improve this answer
    
Agreed. Unlike C strings, where strlen() is O(n), String::length should be fast. – 200_success♦ Mar 25 at 17:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.