Xcode warning: Immutable property will not be decoded because it is declared with an initial value which cannot be overwritten

问题

Running Xcode 12, my Swift 5 Xcode project now has warnings whenever a Decodable or Codable type declares a let constant with an initial value.

struct ExampleItem: Decodable {
    let number: Int = 42 // warning
}

Immutable property will not be decoded because it is declared with an initial value which cannot be overwritten

Xcode suggests changing the let to a var:

Fix: Make the property mutable instead

    var number: Int = 42

It also suggests the fix:

Fix: Set the initial value via the initializer or explicitly define a CodingKeys enum including a 'title' case to silence this warning

What is the purpose of this new warning? Should it be heeded, or ignored? Can this type of warning be silenced?

Should Xcode's fix be implemented? Or is there a better solution?

回答1:

Noah's explanation is correct. It’s a common source of bugs and it's not immediately obvious what’s happening due to the “magical” behaviour of Codable synthesis, which is why I added this warning to the compiler, since it brings your attention to the fact that the property won't be decoded and makes you explicitly call it out if that's the expected behaviour.

As the fix-it explains, you have a couple of options if you want to silence this warning - which one you choose depends on the exact behaviour you want:

Pass the initial value via an init:

struct ExampleItem: Decodable {
  let number: Int
    
  init(number: Int = 42) {
    self.number = number
  }
}

This will allow number to be decoded, but you can also pass around instances of ExampleItem where the default value is used.

You can also use it directly inside init instead, during decoding:

struct ExampleItem: Decodable {
  let number: Int
    
  private enum CodingKeys: String, CodingKey {
    case number
  }
    
  init(from decoder: Decoder) throws {
    let container = try decoder.container(keyedBy: CodingKeys.self)
    number = try container.decodeIfPresent(Int.self, forKey: .number) ?? 42
  }
}

This will allow number to be decoded, but use 42 as the default value if the decoding fails.

Make the property a var, although you can also make it a private(set) var:

struct ExampleItem: Decodable {
  var number: Int = 42
}

Making it a var will allow number to be decoded, but it will also allow callers to modify it. By marking it as private(set) var instead, you can disallow this if you want.

Define an explicit CodingKeys enum:

struct ExampleItem: Decodable {
  let number: Int = 42
  
  private enum CodingKeys: CodingKey {}
}

This will prevent number from being decoded. Since the enum has no cases, this makes it clear to the compiler that there are no properties that you want to decode.

回答2:

This warning appears because immutable properties with initial values don't participate in decoding - after all, they're immutable and they have an initial value, which means that initial value will never be changed.

For example, consider this code:

struct Model: Decodable {
    let value: String = "1"
}

let json = """
{"value": "2"}
"""
let decoder = JSONDecoder()
let model = try! decoder.decode(Model.self, from: json.data(using: .utf8)!)
print(model)

This will actually print Model(value: "1"), even though the json we gave it had value as "2".

In fact, you don't even need to provide the value in the data you're decoding, since it has an initial value anyway!

let json = """
{}
"""
let decoder = JSONDecoder()
let model = try! decoder.decode(Model.self, from: json.data(using: .utf8)!)
print(model) // prints "Model(value: "1")"

Changing the value to a var means it will decode correctly:

struct VarModel: Decodable {
    var value: String = "1"
}
let json = """
{"value": "2"}
"""
let varModel = try! decoder.decode(VarModel.self, from: json.data(using: .utf8)!)
print(varModel) // "VarModel(value: "2")"

If you're seeing this error, it means your code has never correctly parsed the property in question when decoding. If you change it to a var, the property will be parsed correctly, which might be what you want - however, you should make sure that the data you're decoding always has that key set. For example, this will throw an exception (and crash since we're using try!):

let json = """
{}
"""
let decoder = JSONDecoder()

struct VarModel: Decodable {
    var value: String = "1"
}

let varModel = try! decoder.decode(VarModel.self, from: json.data(using: .utf8)!)

In conclusion, Xcode's suggestion is probably viable in many cases, but you should evaluate on a case by case basis whether changing the property to a var will break your app's functionality.

If you want the property to always return the hard-coded initial value (which is what's happening right now), consider making it a computed property or a lazy var.

回答3:

Solution: define an explicit CodingKeys enum to prevent id from decoded. For example,

struct Course: Identifiable, Decodable {
  let id = UUID()
  let name: String

  private enum CodingKeys: String, CodingKey {
    case name
  }
  
  init(name: String) { self.name = name }
  init(from decoder: Decoder) throws {
    let container = try decoder.container(keyedBy: CodingKeys.self)
    let name = try container.decodeIfPresent(String.self, forKey: .name)
    self.name = name ?? "default-name"
  }
}

来源：https://stackoverflow.com/questions/62545594/xcode-warning-immutable-property-will-not-be-decoded-because-it-is-declared-wit

标签

swift

xcode

warnings

xcode12