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` According to the REPL, this has type `shapeless.::[Int with shapeless.record.KeyTag[String("foo"),Int],shapeless.HNil]` I am trying to define a simple function: ``` def fun(x: ::[Int with KeyTag[String("foo"), Int], HNil]) = x("foo") ``` but it does not even compile. I cannot use a String("foo") in the type declaration, and get an error. I have two questions: - How can I specify the type of the extensible record in my code? - When working with records with more fields, the length and complexity of the type declaration will be unmanageable. Is there a way to create an alias for the type, given a particular instance of a record, or some other workaround? EDIT I have found that: ``` val rec1 = ("foo" ->> 42) :: HNil val rec2 = ("foo" ->> 43) :: HNil var x = rec1 x = rec2 ``` works well. I conclude rec1, rec2, and x are of the same type. I just don't k…

Accepted Answer

Here's something a little more general that I think might answer your question. Suppose we want to write a method that will work on any record with a `"foo"` key. We can use a combination of a witness and a selector:

```
import shapeless._, record._, syntax.singleton._

val w = Witness("foo")

def fun[L <: HList](xs: L)(implicit sel: ops.record.Selector[L, w.T]) = xs("foo")
```

And then:

```
scala> fun(("foo" ->> 42) :: HNil)
res0: Int = 42
```

Or:

```
scala> fun(("bar" ->> 'a) :: ("foo" ->> 42) :: HNil)
res1: Int = 42
```

If we really wanted to only allow records with no other fields, we could write the following:

```
def fun(l: Int with KeyTag[w.T, Int] :: HNil) = l("foo")
```

But that's somewhat at odds with the way records are generally used.

We have to define the witness precisely because Scala 2.10 doesn't provide any way to refer to a singleton type directly—see for example my fork of Alois Cochard's Shona project for some discussion.

I will add as a final disclaimer that I'm only just now getting familiar with Shapeless 2.0 myself, but I don't think even Miles is magical enough to get around this limitation.

Passing a Shapeless Extensible Record to a Function

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