Avoid mutable variable in scala

Question

I have a piece of code like this

def filter(t: String) : Boolean = {
    var found = false;
    for(s <- listofStrings) {
      if ( t.contains(s)) { found = true}
    }
    found
  }

The compiler gives a warning that its not good practise to use a mutable variable. How do I avoid this ?

Disclaimer: I used a variant of this code in an assignment and the submission is done. I would like to know what the right thing to do is

Eduardo · Accepted Answer · 2012-10-12 07:27:13Z

8

You could do:

def filter(t:String) = listofStrings.exists(t.contains(_))

answered Oct 12, 2012 at 7:27

Eduardo

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6 Comments

Matthew Farwell Over a year ago

This is wrong. As defined, filter is equivalent to contains(), you don't need the exists.

Eduardo Over a year ago

@Matthew Farwell: No. He wants to check if the String t contains any of the strings in the list of strings, not if the String t is contained in the list of strings.

Malte Schwerhoff Over a year ago

@MatthewFarwell I disagree (or don't understand your objection). The OP's filter(t) method is true iff one of the elements in listofStrings is a substring of t, NOT if t is contained in listofStrings.

Matthew Farwell Over a year ago

Eduardo, you're right, I misread the original algorithm. Sorry.

Eduardo Over a year ago

@Matthew: say the list of strings was "a"::"b""::Nil and t was "a1": this should return true because t contains "a". Doing list.contains(t) would yield false

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Malte Schwerhoff · Accepted Answer · 2012-10-12 08:01:53Z

3

If you what to use as few built-in collection functions as possible, use recursion:

def filter(t: String, xs: List[String]): Boolean = xs match {
  case Nil => false
  case x :: ys => t.contains(x) || filter(t, ys)
}

println(filter("Brave New World", List("few", "screw", "ew"))) // true

println(filter("Fahrenheit 451", List("20", "30", "80"))) // false

answered Oct 12, 2012 at 8:01

Malte Schwerhoff

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