Swapping values in an array using Pattern Matching in Scala

Question

I am trying to solve the following problem from Scala for the impatient. The question is as follows:

Using pattern matching, write a function swap that swaps the first two elements of an array provided its length is at least two.

My solution is:

def swap(sArr:Array[Int]) = sArr.splitAt(2) match { 
                               case (Array(x,y),Array(z)) => Array(y,x,z)
                               case (Array(x,y),Array()) => Array(y,x)
                               case _ => sArr
                            }

My problem is with the first case statement. I think it would pattern-match something like (Array(1,2),Array(3)) whereas I intend it to pattern-match (Array(1,2),Array(3,4,5.....))

Can somebody point out how that would be possible.

Thanks

dhg · Accepted Answer · 2012-05-01 21:17:29Z

9

The problem with your code is that Array(z) means "match a one-element array". What you want is for z to be the whole array, no matter how many elements:

def swap(sArr: Array[Int]) = 
  sArr.splitAt(2) match { 
    case (Array(x, y), z) => Array(y, x) ++ z
    case _ => sArr
}

However, I would write it with the sequence-matching syntax _* so that you don't have to manually split the array:

def f(a: Array[Int]) = 
  a match {
    case Array(x, y, z @ _*) => Array(y, x) ++ z
    case _ => a
}

edited May 1, 2012 at 21:17

answered May 1, 2012 at 20:27

dhg

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1 Comment

sc_ray Over a year ago

Thanks. I guess the @ _* denotes the zero or more here for c?

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