Convert Java method to Scala method

Question

I need to convert the following java method to scala and am having difficulty because scala doesn't allow you to return values in the middle of methods. Could someone give me a hand or at least a start on how to convert this:

public boolean isAllowed(String method, String path, Map<String,String> apiUrlMap) {

    if (apiUrlMap != null) {
        Set<Entry<String, String>> apiSet = apiUrlMap.entrySet();

        for (Entry<String, String> apiUrl : apiSet) {
            String aUrl = apiUrl.getKey();
            String aMeth = apiUrl.getValue();
            if (aUrl.equals("#")) {
                if (aMeth.contains(method)) {
                    return true;
                }
            }

            if (aUrl.endsWith("#")) {
                String testUrl = aUrl.replaceFirst("/#", "");
                if (path.startsWith(testUrl)) {
                    if (aMeth.contains(method)) {
                        return true;
                    }
                }
            }

            if (aUrl.equals(path) || path.equals(aUrl +"/")) {
                if (aMeth.contains(method)) {
                    return true;
                }
            }

        }
    }

    return false;
}

"... because scala doesn't allow you to return values in the middle of methods" - Why do you think that? The return keyword also exists in Scala. — Jesper
– Jesper, Commented Mar 26, 2012 at 17:14
I guess it does, just started learning scala and didn't see any examples with actual return statements. Granted I bet there is a better way to do this besides this very imperative java way? — chiappone
– chiappone, Commented Mar 26, 2012 at 17:18
If you use IntelliJ, you can just paste the Java code in and it'll convert it to Scala. Unidiomatic Scala, but it should work and you can use it as a basis. — Luigi Plinge
– Luigi Plinge, Commented Mar 27, 2012 at 2:52

dhg · Accepted Answer · 2012-03-26 17:24:08Z

7

Scala does let you return, it's just not recommended style.

For example, the following works fine:

def f(x: Int): String = {
  if(x > 5)
    return "greater"
  else
    return "less"
}

One catch is that Scala requires you to have an explicit return type specified if you use the return keyword.

The better style would be to simply have the if-expression be the last expression of the method:

def f(x: Int): String = {
  if(x > 5)
    "greater"
  else
    "less"
}

edited Mar 26, 2012 at 17:24

answered Mar 26, 2012 at 17:15

dhg

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2 Comments

chiappone Over a year ago

Thanks do you have an example of a scala recommended way to do this?

dhg Over a year ago

@chiappone edited. you should try rewriting your method and just post questions if you have trouble. try to make your questions specific and avoid posting lots of code since that makes it harder to tell exactly where your issue is.

hbatista · Accepted Answer · 2012-03-26 17:43:50Z

4

Having a quick go at it (not tested)

def isAllowed(method:String, path:String, apiUrlMap:Map[String,String]) = 
  apiUrlMap exists { case (aUrl, aMeth) => {
        aUrl.equals("#") ||  
        aUrl.endsWith("#") && path.startsWith(aUrl.replaceFirst("/#", "")) || 
        (aUrl.equals(path) || path.equals(aUrl +"/"))
      } && aMeth.contains(method)
    }

answered Mar 26, 2012 at 17:43

hbatista

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2 Comments

Ed Staub Over a year ago

Nice. Is the case used just to get meaningful names on Pair._1 and Pair._2, or does it perform some other function also?

hbatista Over a year ago

@Ed the case is just used to deconstruct (pattern match) the pair, no other function

Nicolas · Accepted Answer · 2012-03-26 18:18:20Z

The idea in scala is to use function call instead of control structure. Here is a (quickly written) example:

def isAllowed(
    method: String,
    path: String,
    apiUrlMap: Option[Map[String, String]]) =

  apiUrlMap flatMap {m =>
    m collectFirst {
      case ("#", aMeth) 
          if aMeth contains `method` =>
        true 
      case (aUrl, aMeth)
          if (aUrl endsWith "#") && 
             (path startsWith aUrl.replaceFirst("/#", "")) &&
             (aMeth contains `method`) =>
        true 
      case (`path`, aMeth)
          if (aMeth contains `method`) =>
        true 
      case (aUrl, aMeth)
          if (path == aUrl + "/") && 
             (aMeth contains `method`) =>
        true 
    }
  } getOrElse false

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