how to convert java method return to scala?

Question

how to convert this java code to scala ? Scala doesn't prefer 'return', then scala has a better version ?

public void myMethod(String name){
  if(s==null) return;
  //other logic and many lines ....
}

There nothing wrong with that. `return` is useful for exiting early from a guard condition. If you want to avoid it, just put the rest of your method into the `else` part of that if. However, if you want to write "idiomatic" scala. Return a value, don't write void methods (`Unit` in Scala). Keep your methods and functions short as well, doing so makes them more reusable and reduces the need for early returns. — puhlen, Mar 10 '17 at 14:56
I agree with you. so many lines put into an else , it is hard to read. — user1615666, Mar 10 '17 at 15:44
Shouldn't be any harder to read than what you already have. I don't see much different between `if (condition) return; //lots of code` and `if (condition) else{ //lots of code}` from a readability perspective — puhlen, Mar 10 '17 at 15:50

score 1 · Accepted Answer · answered Mar 10 '17 at 14:56

Let's do this in two steps:

def myMethod(name: String): Unit =
  if(name == null) () else{
    //rest of code that works via side-effect
  }

The () denotes an instance of Unit.

That's ok but there's something even better! This take advantage that the Option constructor has special null handling, e.g. Option(null) == None.

def myMethod(name: Option[String]): Unit = name.foreach{
  //rest of code that works via side-effect
}

Wherein if you might have something which could be null you will make sure that the type system specifies that the variable might or might not contain a value (yes, Java now has Optional too and for similar reasons!)

score 0 · Answer 2 · answered Mar 10 '17 at 15:11

0

Or refactor the code in this way:

def myMethod(name: String) = {
  if (name != null) {
  //other logic and many lines ....
  }
}

answered Mar 10 '17 at 15:11

freedev

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score 0 · Answer 3 · edited May 23 '17 at 12:25

The short answer is that you don't need to use the return keyword. The return is Scala is pretty much explained here pretty much but I'd like to point out a few things.

Expressions vs. Statements

Most constructs in Java are statements which means they only executed, not evaluated. This is quite important because in Scala most of these constructs are expressions. Let's see how this works on an if-construct.

// Initialize i to 1 if some condition is true otherwise to 0
int i;
if (someCondition) i = 1;
else i = 0;

Assuming we'd like to initialize i in one step (maybe we'd like making it final) we need to use a ternary operator:

final int i = someCondition ? 1 : 0;

This works because the right hand side is actually an expression. Now let's write this in Scala:

val i = if (someCondition) 1 else 0

This is a conditional expression.

Functions

A good thing to keep in mind about functions is that the return value of a function is equal to the result of its last expression. This being you can define the if-expression as the latest in your function and thus its result will be returned. Applying this to your code leaves us with:

def myMethod(name: String): Int = {
  if (s == null) -1
  else { ... }
}

I've updated your method to return an Int in this case for the sake of argument. You probably need to replace it with something more meaningful. On the else branch you'd also need to return an Int in this case. Anyway I hope this clears your request.

As a closing note this is a nice guide explaining why you should avoid using return. Personally I haven't used it at all even though there are some examples there.

how to convert java method return to scala?

3 Answers3

Expressions vs. Statements

Functions