Scala method type parameterization

Question

I am trying to get a better handle on understanding methods parameterized by types and have this piece of code -

 def inferType[T, U](x: T, y: U): Unit = y match {
    case _: T => println("T")
    case _    => println("Not T")
  }
 inferType("0", 11) // call 1
 inferType(0, 11)   // call 2

I would have expected call 1 to have printed "Not T" and call 2 to have printed "T". However, its printing "T" in both cases. Obviously I am missing something here. Can someone help me with why pattern matching is not matching the generic type T here?

https://stackoverflow.com/questions/12218641/scala-what-is-a-typetag-and-how-do-i-use-it/ — Dmytro Mitin, Jan 14 '21 at 22:52
https://stackoverflow.com/questions/38570948/type-erasure-in-scala — Dmytro Mitin, Jan 14 '21 at 22:53
https://blog.bruchez.name/2015/11/generalized-type-constraints-in-scala.html — Luis Miguel Mejía Suárez, Jan 14 '21 at 22:54

score 2 · Answer 1 · answered Jan 15 '21 at 09:27

2

The JVM is performing type-erasure, so at runtime all generic types are actually Object, and your match is checking whether y is of type T = Object, which it is.

You can get around this using TypeTag, but this is an anti-pattern in most cases.

answered Jan 15 '21 at 09:27

francoisr

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Scala method type parameterization

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