Pattern match for variable in scope (Scala)

Question

In the following code

val x = 5
val y = 4 match {
  case x => true
  case _ => false
}

the value y is true. Scala interprets x to be a free variable in the pattern match instead of binding it to the variable with the same name in the scope.

How to solve this problem?

Why did you duplicate an existing question and then answer it yourself? http://stackoverflow.com/questions/6172557/problem-with-scala-matching-scope http://stackoverflow.com/questions/5153590/why-does-scala-complain-when-given-this-pattern-match-on-an-integral-value — dhg, Jul 19 '11 at 23:08
@dhg: I could not find that question on SO, that's why. I answer my own question since I prefer using SO as a knowledge repository instead of taking a note on my tiny hidden blog or making a mental post. Thanks for the link, I agree with closing with duplicate. — ron, Jul 21 '11 at 07:41

score 12 · Answer 1 · answered Jul 19 '11 at 21:51

12

Backticking the variable indicates to bind a scoped variable:

val x = 5
val y = 4 match { case `x` => true; case _ => false }

returns false.

Alternatively, if a variable starts with an uppercase letter, it binds to a scoped variable without backticking.

answered Jul 19 '11 at 21:51

ron

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see also http://scala-programming-language.1934581.n4.nabble.com/scala-pattern-matching-proposal-td1999364.html – ron Jul 19 '11 at 21:51

score 7 · Accepted Answer · answered Jul 19 '11 at 22:09

7

Invoking the least astonishment principle, I will simply do:

val x = 5
val y = 4 match {
  case z if z == x => true
  case _ => false
}

answered Jul 19 '11 at 22:09

paradigmatic

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Ron answer is the good one, you add a condition that can be pattern matched directly. (downvote) – Nicolas Jul 20 '11 at 07:07
I think the condition is just as fast, and remember the least astonishment principle. (upvote) – Anonymous Jul 20 '11 at 09:27

Pattern match for variable in scope (Scala)

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