How can I match case with shapeless variable?

Question

How can I match case with shapeless variable ?

Let's say I have variable of following type shapeless.::[String,shapeless.::[String,shapeless.HNil]]

Currently I have to do this

authHeaders.hrequire(shape_value => {
    val (client_id, client_secret) = value.tupled
    isAuthorized(client_id, client_secret)
  }
  )

Can I somehow unwind String :: String :: HNil to String pair so that I don't have to do it in separate statement ?

Answer 1

There is method unapply in object shapeless.:::

def unapply[H, T <: HList](x: H :: T): Option[(H, T)]

So you could just match on HList like this:

scala> val ::(a, ::(b, HNil)) = "1" :: "x" :: HNil
a: String = 1
b: String = x

Or with alternative syntax for unapply method with Tuple2 result: a :: b instead of ::(a, b):

scala> val a :: b :: HNil = "1" :: "x" :: HNil
a: String = 1
b: String = x

scala> "1" :: "x" :: HNil match {
     |   case a :: b :: HNil => s"$a :: $b :: HNil"
     | }
res0: String = 1 :: x :: HNil

In your case:

authHeaders.hrequire{
  case client_id :: client_secret :: HNil => isAuthorized(client_id, client_secret)
}

Alternative

You could use tupled method to convert function of N arguments to function of single TupleN argument.

For function:

val isAuthorized: (String, String) => Boolean = ???
authHeaders.hrequire{ isAuthorized tupled _.tupled }

For method:

def isAuthorized(s1: String, s2: String): Boolean = ???
authHeaders.hrequire{ (isAuthorized _) tupled _.tupled }