CODEWITHSUNDEEP
scala
femibyte
femibyte

Reputation: 3497

Seq[AnyVal] returned as type instead of Seq[Int]

I define the following List of tuples:

var temps = Seq(("Spain", Seq(68,70,73,75)),       
                ("Trinidad",Seq(87,83,88,84,88)), 
                ("England",Seq(52,55,58,57.5)),
                ("Eritrea",Seq(90,91.3,88,91)))

and the resulting type is as follows:

temps: Seq[(String, Seq[AnyVal])] = List((Spain,List(68, 70, 73, 75)), (Trinidad,List(87, 83, 88, 84, 88)), (England,List(52.0, 55.0, 58.0, 57.5)), (Eritrea,List(90.0, 91.3, 88.0, 91.0)))

My question is why the type of temps is inferred as Seq[(String, Seq[AnyVal])] rather than Seq[(String, Seq[Int])] or Seq[(String, Seq[Double])] which is what one would expect ?

I am a Scala newbie so pardon me if there is a simple explanation.

Upvotes: 3

Views: 421

Answers (1)

yǝsʞǝla
yǝsʞǝla

Reputation: 16412

This is how type inference works. Scala can infer type in a context/declaration scope, but not across contexts as in here. For example this works:

scala> val ss = Seq(Seq(1, 1.2))
ss: Seq[Seq[Double]] = List(List(1.0, 1.2))

Although we have both Int and Double, Int gets promoted up. However, this does not work:

scala> val ss = Seq(Seq(1, 1.2), Seq(2, 2))
ss: Seq[Seq[AnyVal]] = List(List(1.0, 1.2), List(2, 2))

because first list becomes Seq[Double] and second one - Seq[Int]. To find common super type scala reverts to Seq[AnyVal].

You can explicitly annotate the declaration to help compiler resolve the ambiguity:

scala> val ss: Seq[Seq[Double]] = Seq(Seq(1, 1.2), Seq(2, 2))
ss: Seq[Seq[Double]] = List(List(1.0, 1.2), List(2.0, 2.0))

Upvotes: 5

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