Missing compiler warning

[bastie]

typo - see error description 2 days later 🥺

Description

In result of overload functions / methods with same name but more parameters and these parameters get default values (line 23) the "simple" function (line 20) appears uncallable.

Reproduction

//
//  main.swift
//  MissingInitInfo
//
//  Created by Sebastian Ritter on 26.12.23.
//

import Foundation

// ✅ expected error
//let _ = Printable ()

// 🆘 name is not set, but called is init with param *as name* and ignoring init without second parameter
let _ = Printable (from: "Sebastian")

// ✅ expected result
let _ = Printable (from: "Sebastian", as: "Mr. Ritter")

public class Printable {
  public init (form sender : String) { // 🆘 uncallable (?) init method but no compiler message
    print ("Hello from \(sender)!")
  }
  public init (from sender : String, as name : String = "Bastie") {
    print ("Hello from \(name)?")
  }
}

Expected behavior

Compiler warning at line 20 as uncallable method / function

Environment

swift-driver version: 1.87.3 Apple Swift version 5.9.2 (swiftlang-5.9.2.2.56 clang-1500.1.0.2.5)
Target: arm64-apple-macosx14.0

Additional information

No response

[msrutek-paylocity]

Is the typo

init (form sender : String)

vs

init (from sender : String, as name : String = "Bastie")

form vs from intentional?

[bastie]

@msrutek-paylocity Thanks!!! I like my German "Autokorrektur"...
and so I see the missing compiler message some lines below, or please tell how default value can be used.

//
//  main.swift
//  MissingInitInfo
//
//  Created by Sebastian Ritter on 26.12.23.
//

import Foundation

// ✅ expected error
//let _ = Printable ()

// ✅ expected result
let _ = Printable (from: "Sebastian")

// ✅ expected result
let _ = Printable (from: "Sebastian", as: "Mr. Ritter")

public class Printable {
  public init (from sender : String) {
    print ("Hello from \(sender)!")
  }
  public init (from sender : String, as name : String = "Bastie") { // 🆘  in result of init with one parameter unusable default value "Bastie" from init method
    print ("Hello from \(name)?")
  }
}

[Kyle-Ye]

This is expected behavior currently. See https://github.com/apple/swift/blob/main/docs/ReferenceGuides/UnderscoredAttributes.md#_disfavoredoverload and there is some great forum posts discussing the issue.

Suggestion 1:

• Call init(from sender: String) in your init(from sender: String, as name: String) method.

Suggestion 2:

• Swift prefer a less constraint overload method. If you like your second init to be called. You can add @_disfavoredoverload to your second init method.

@_disfavoredOverload
public init(from sender: String) {
    print("Hello from \(sender)!")
}
_ = Printable(from: "Sebastian") // Hello from Bastie?

[Kyle-Ye]

Feel free to ask further. If no reply is received, I think we can close it later.

[AnthonyLatsis]

Per Swift overload ranking rules the function that best matches the Printable(from: "Sebastian") call is Printable.init(form:). If you want to call Printable.init(from:as:) with the default argument without rethinking anything, you must disambiguate:

let _ = Printable.init(from:as:)("Sebastian")

@_disfavoredOverload, as Kyle suggests, is one other option that requires minimal change, but please note that this is not a public feature! There are no promises made about its stability or existence.

Important

Whether any of this is reasonable depends on the actual use case.

[Kyle-Ye]

Thanks for sharing the Printable.init(from:as:)("Sebastian") tip. Don't know it before.

[AnthonyLatsis]

Compiler warning at line 20 as uncallable method / function

I do not think a warning is justified here. The API design expectation behind this sort of overloading is that Printable(from:) behaves exactly like Printable(from:as:) in most, if not all, sensible implementations.

[Kyle-Ye]

let fn = Printable.init(from:as:)
let _ = fn("Sebastian")

// or

let _ = Printable.init(from:as:)("Sebastian")

The second way can be compiled, and the former will report a compiler error - "Missing argument for parameter #2 in call"

[AnthonyLatsis]

the former will report a compiler error

Right, thanks. Function values do not retain default arguments.