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Home/ Questions/Q 3218418
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T15:31:32+00:00 2026-05-17T15:31:32+00:00

This question was previously asked in scala-user mailing-list without a confirmed answer. scala> val

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This question was previously asked in scala-user mailing-list without a confirmed answer.

scala> val T = new Pair(1, 2){
override def equals(obj:Any) = obj.isInstanceOf[(Int, Int)] && obj.asInstanceOf[(Int, Int)]._1 == this._1}
           }

T: (Int, Int) = (1,2)

scala> T match {
        case (1, 1) => println("matched")
        case _ => println("not matched")
   }

not matched


scala> (1, 1) match {
              case T => println("matched")
              case _ => println("not matched")
          }

not matched

scala> T == (1, 1)
res15: Boolean = true

I thought a constant(val) pattern matching result depends on the return value of “equals”, but the results show that it is not the case, then what is the criteria?

Someone had suggested that case (1, 1) => is an extractor pattern and uses Tuple2.unapply instead. so I tried these:

scala> Pair.unapply(T)
res1: Option[(Int, Int)] = Some((1,2))

scala> Pair.unapply(T).get == (1, 1)
res2: Boolean = true

Can anyone please explain why == get true but I cannot make them match?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-17T15:31:32+00:00Added an answer on May 17, 2026 at 3:31 pm

    With the resolution of #3888, I can give a final answer to this question.

    1. T match { case (1, 1) =>

      No, it has nothing to do with unapply. As extempore mentioned, case (1,1) => is a ‘Tuple Pattern’, an alias for ‘Constructor Pattern” of case class Tuple2, It only matches value constructed as Tuple2(1, 1) or Pair(1, 1).

      What really concerns unapply is the ‘Extractor Pattern’:

      object Pair {
          val T = new Pair(1,1){
              def unapply(p:(Int, Int)) :Boolean = this._1 == p._1
          }
          def main(args: Array[String]) = {
              (1, 2) match {
                  case T() => println("matched")
                  case _ => println("not matched")
              }
          } 
      }
      

      You get “matched”. Notice the () in the case clause

    2. (1, 1) match { case T => ...

      according to scala spec, section 8.1.5, this is a ‘Stable Identifier Pattern’, case T matches any value v such that T == v. So we SHOULD get “matched”. The “not matched” results
      are just caused by the bug in current compiler implementation.

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