Passing a Shapeless Extensible Record to a Function

Question

I am trying to learn Shapeless (using version 2.10.2). I have created a very simple extensible record:

val rec1 = (\"foo\" ->> 42) :: HNil

Answer 1

As of shapeless 2.1.0 there is a new syntax to express record types:

scala> :paste
// Entering paste mode (ctrl-D to finish)

import shapeless._
import shapeless.record._
import shapeless.syntax.singleton._

def fun(x: Record.`"foo" -> Int`.T) = x("foo")

// Exiting paste mode, now interpreting.

import shapeless._
import shapeless.record._
import shapeless.syntax.singleton._
fun: (x: shapeless.::[Int with shapeless.labelled.KeyTag[String("foo"),Int],shapeless.HNil])Int

scala> fun( ("foo" ->> 42) :: HNil )
res2: Int = 42

scala> fun( ("foo" ->> 42) :: ("bar" ->> 43) :: HNil )
:30: error: type mismatch;
 found   : shapeless.::[Int with shapeless.labelled.KeyTag[String("foo"),Int],shapeless.::[Int with shapeless.labelled.KeyTag[String("bar"),Int],shapeless.HNil]]
 required: shapeless.::[Int with shapeless.labelled.KeyTag[String("foo"),Int],shapeless.HNil]
       fun( ("foo" ->> 42) :: ("bar" ->> 43) :: HNil )

But the Selector is probably the best approach for OP's use-case.