How to use the @as decorator?

1

I am experimenting with the @as decorator with the following code:

type status = [
  @as("GOOD") #Good |
  @as("BAD") #Bad
]

@val external setStatus: status => unit = "setStatus"

setStatus(#Good)

This compiles to:

setStatus("Good");

But I was expecting it would compile to:

setStatus("GOOD");

I suspect I am not using the decorator correctly. Any suggestions on how to correct this?

View in playground

Thanks

2

After some more investigation I’ve discovered this:

@val external setStatus: @string[
  @as("GOOD") #Good |
  @as("BAD") #Bad
] => unit = "setStatus"

setStatus(#Good)

Which produces:

setStatus("GOOD");

So it seems that:

  1. It’s intended for use with externals, and
  2. the @string decorator is also required.

How does that sound?

Update:

Some more investigation, I see it can be used with record types as well.

type action = {@as("type") type_: string}

let one: action = {type_: "One"}

which produces:

var one = {
  type: "One"
};

So it seems, that the @as decorator:

  1. may be used with record types directly, and
  2. may be used with polymorphic variants on externals with the @string or @int decorators.

How does that sound?

3

Your two statements are correct.

Poly vars are more flexible with the identifier names for its constructors, so you can easily do it without the as decorator as well, e.g. #GOOD.

Docs on poly vars are currently in the last review phase, so hopefully this will be more clear in the future!

4 Likes
4

@ryyppy can I not use @as in a defined poly variant type?

type typeopt = @string[
  | @as("required") #Required
  | @as("maxLength") #LengthMax
  | @as("minLength") #LengthMin
  | @as("max") #Max
  | @as("min") #Min
  | @as("validate") #Validate
]

This is at a foreign boundary but the type is used in numerous functions, for example, so writing per function is rough.

Thanks
A

5

You don’t need @as. Polyvariant values compile to the exact strings as their names in the code. E.g. #required will compile to the string "required" and so on.

6

The strings I’m trying to capture are long and have whitespace and capitalization so using them as the variant symbols is really annoying.

7

You can assign them to variables and use those instead.

let foo = #"foo foo"
let bar = #"bar bar"

Of course, in pattern matches you will still need to use the original poly variant literals. But usually pattern matches are only required in a few places.

8

Or i could use @as? What do you have against @as?

9

Only the fact that it doesn’t work :wink:

10

@as will only be reserved for nominal types, for example, record.