Convert a Scala list to a tuple?

Shapeless 2.0 changed some syntax. Here's the updated solution using shapeless.

import shapeless._
import HList._
import syntax.std.traversable._

val x = List(1, 2, 3)
val y = x.toHList[Int::Int::Int::HNil]
val z = y.get.tupled

The main issue being that the type for .toHList has to be specified ahead of time. More generally, since tuples are limited in their arity, the design of your software might be better served by a different solution.

Still, if you are creating a list statically, consider a solution like this one, also using shapeless. Here, we create an HList directly and the type is available at compile time. Remember that an HList has features from both List and Tuple types. i.e. it can have elements with different types like a Tuple and can be mapped over among other operations like standard collections. HLists take a little while to get used to though so tread slowly if you are new.

scala> import shapeless._
import shapeless._

scala> import HList._
import HList._

scala>   val hlist = "z" :: 6 :: "b" :: true :: HNil
hlist: shapeless.::[String,shapeless.::[Int,shapeless.::[String,shapeless.::[Boolean,shapeless.HNil]]]] = z :: 6 :: b :: true :: HNil

scala>   val tup = hlist.tupled
tup: (String, Int, String, Boolean) = (z,6,b,true)

scala> tup
res0: (String, Int, String, Boolean) = (z,6,b,true)

an example using shapeless :

import shapeless._
import syntax.std.traversable._
val x = List(1, 2, 3)
val xHList = x.toHList[Int::Int::Int::HNil]
val t = xHList.get.tupled

Note: the compiler need some type informations to convert the List in the HList that the reason why you need to pass type informations to the toHList method

You can't do this in a typesafe way. Why? Because in general we can't know the length of a list until runtime. But the "length" of a tuple must be encoded in its type, and hence known at compile time. For example, (1,'a',true) has the type (Int, Char, Boolean), which is sugar for Tuple3[Int, Char, Boolean]. The reason tuples have this restriction is that they need to be able to handle a non-homogeneous types.

You can do it using scala extractors and pattern matching (link):

val x = List(1, 2, 3)

val t = x match {
  case List(a, b, c) => (a, b, c)
}

Which returns a tuple

t: (Int, Int, Int) = (1,2,3)

Also, you can use a wildcard operator if not sure about a size of the List

val t = x match {
  case List(a, b, c, _*) => (a, b, c)
}