Swift 5.9: Type of expression is ambiguous without more context

Using Swift
1

Hi all,

I have the following code which does not compile:

private func handleOnText(ws: WebSocket) -> Void {
   ws.onText { (_: WebSocket, text: String) in
      guard let payload = text.data(using: .utf8) else {
         ws.send(generateSerializationErrorMessage())
         return
      }
      
      let message: String = switch Result { try decoder.decode(ConsumerMessage.self, from: payload) } {
         case .success(let msg):
            "Success"
         case .failure():
            "Failed"
      }
      
      ws.send(message)
   }
}

The compiler complains:

Type of expression is ambiguous without more context

As you can see, I am using the new switch expression which is provided by version 5.9 and is causing the compiler error.

I will be very appreciate for the help.

Best regards

2

I thought you couldn’t use trailing closure syntax in an if, switch, or while like that.

2 Likes
3

Thanks. True.

4

The error message doesn't point that out, though.

5

The use of the trailing closure isn't the main issue here, the compiler can actually parse it (though it'll emit a warning asking you to enclose in parens), it looks like the issue is with:

case .failure():

You need to either write a pattern to match against the error associated value, or write case .failure: instead. With the latest 5.9 snapshot you'll get a marginally better error message (in that it will actually point to the problematic bit of code):

error: type '()' cannot conform to 'Error' [type_cannot_conform]
    case .failure():
          ^

But we still ought to be able to improve this, I've filed Improve enum element pattern matching diagnostics when the arity of arguments mismatch · Issue #66750 · apple/swift · GitHub to track that.