Nested pattern matching in Scala

Question

I'm new to Scala and I'm trying to understand how pattern matching works. So I wrote this basic code, which returned the expected result:

def test(choice: Int): String = choice match {
  case x if x > 0 && x%2 == 0 => "positive even number"
  case x if x > 0 && x%2 != 0 => "positive odd number"
  case 0 => "null"
  case x if x < 0 && x%2 == 0 => "negative even number" 
  case x if x < 0 && x%2 != 0 => "negative odd number"
}

Now I'm trying to do something a little more elaborate:

def test(choice: Int): String = choice match {

  case x if x%2 == 0 => x match {
    case y if y > 0 => "even and positive number" 
    case y if y < 0 => "even and negative number"
    }
  case 0 => "null"
  case x if x%2 != 0 => x match {
    case y if y > 0 => "odd and positive number" 
    case y if y < 0 => "odd and negative number"
    }
}

But it failed. Here's the error message on the console:

scala> def test(choice: Int): String = choice match {
     |
     |   case x if x%2 == 0 => x match {
     | case y if y > 0 => "even and positive number"
     | Display all 600 possibilities? (y or n)
     [...]

     |  if y < 0 => "even and negative number"
<console>:5: error: '(' expected but identifier found.
 if y < 0 => "even and negative number"
    ^

    [...]

Can somebody tell me what I'm doing wrong and give me some details about what I misunderstand about Scala in general and the match method in particular.

``` | Display all 600 possibilities? (y or n) ``` Definitely means you have a tab somewhere. Have you tried to use the paste mode (scala> :paste)? — Francis Toth
– Francis Toth, Commented Feb 12, 2018 at 19:15
Pattern matching can be very useful and powerful. For this computation, however, I wouldn't use it. It just complicates the underlying logic. — jwvh
– jwvh, Commented Feb 12, 2018 at 19:19
pattern matching with only guards and no pattern seems a bad use generally — cchantep
– cchantep, Commented Feb 12, 2018 at 21:06
thank you for your help. Indeed the problem was in the tab. it compiles now — Yuva
– Yuva, Commented Feb 13, 2018 at 8:46

Jakub Kozłowski · Accepted Answer · 2018-02-12 18:15:14Z

2

It compiles for me. The order of cases doesn't matter for the success of compilation (however, the case 0 branch will never match, because case x if x%2==0 matches x=0. You may want to make the case 0 branch the first one)

I believe your problem is because of using tabs instead of spaces in the terminal.

If you use this code in a file in a project, it'll compile just as well. If you want it to work in the console, you can either:

convert the tabs to spaces before pasting the code
enter paste mode in the REPL using the :paste command, paste the code and exit paste mode with Ctrl-D (or whatever the REPL tells you the combination is - that's the one on Mac).

answered Feb 12, 2018 at 18:15

Jakub Kozłowski

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Comments

user unknown · Accepted Answer · 2018-02-12 17:55:27Z

0

In your first code, you test for x>0 in the first 2 cases, then for 0 itself.

In the second code, you don't test for x>0, but x%2 == 0 which already matches for x = 0, and a second, outer match isn't considered.

If you put the explicit 0 match on top, it might work, but I didn't look for a second error and only matched for the first error I could find. :)

answered Feb 12, 2018 at 17:55

user unknown

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1 Comment

Yuva Over a year ago

Yep, i didn't focus on that. I just wanted to made it compile. The problem was in the tab. it works now. thank you

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