Swift 协议只能设置吗？

为什么我可以毫无错误地执行此操作：

var testDto = ModelDto(modelId: 1)
testDto.objectId = 2

当我定义这个时：

protocol DataTransferObject {
    var objectType: DtoType { get }
    var parentObjectId: Int { get set }
    var objectId: Int { get }
    var objectName: String { get set }
}

struct ModelDto: DataTransferObject {
    var objectType: DtoType
    var parentObjectId: Int
    var objectId: Int
    var objectName: String

    init(modelId: Int) {
        self.objectType = DtoType.Model
        self.objectId = modelId
        self.parentObjectId = -1
        self.objectName = String()
    }
}

如果我的协议中的定义大部分被忽略（getter、setter 定义），那么我为什么要使用它们呢？

苹果公司在《Swift 编程语言文档》 https://docs.swift.org/swift-book/documentation/the-swift-programming-language/protocols/:

如果协议仅要求属性可获取，则任何类型的属性都可以满足该要求，并且如果这对您自己的代码有用，则该属性也是可设置的也是有效的。

因此，以下五个 Playground 代码片段都是有效的：

示例#1：常量属性

protocol FullyNamed {
    var fullName: String { get }
}

struct Duck: FullyNamed {
    let fullName: String
}

let scrooge = Duck(fullName: "Scrooge McDuck")
print(scrooge.fullName) // returns "Scrooge McDuck"

示例#2：变量属性

protocol FullyNamed {
    var fullName: String { get }
}

struct Duck: FullyNamed {
    var fullName: String        
}    

var scrooge = Duck(fullName: "Scrooge McDuck")
print(scrooge.fullName) // returns "Scrooge McDuck"

scrooge.fullName = "Scrooge H. McDuck"
print(scrooge.fullName) // returns "Scrooge H. McDuck"

示例#3：计算属性（仅获取）

protocol FullyNamed {
    var fullName: String { get }
}

struct Duck: FullyNamed {
    private var name: String
    var fullName: String {
        return name
    }
}

let scrooge = Duck(name: "Scrooge McDuck")
print(scrooge.fullName) // returns "Scrooge McDuck"

示例#4：计算属性（获取和设置）

protocol FullyNamed {
    var fullName: String { get }
}

struct Duck: FullyNamed {
    private var name: String
    var fullName: String {
        get {
            return name
        }
        set {
            name = newValue
        }
    }
}

var scrooge = Duck(name: "Scrooge McDuck")
print(scrooge.fullName) // returns "Scrooge McDuck"

scrooge.fullName = "Scrooge H. McDuck"
print(scrooge.fullName) // returns "Scrooge H. McDuck"

示例#5：private(set)可变属性

/* Duck.swift located in Sources folder */

protocol FullyNamed {
    var fullName: String { get }
}

public struct Duck: FullyNamed {
    public private(set) var fullName: String
    
    public init(fullName: String) {
        self.fullName = fullName
    }

    public mutating func renameWith(fullName: String) {
        self.fullName = fullName
    }
}

/* Playground file */

var scrooge = Duck(fullName: "Scrooge McDuck")
print(scrooge.fullName) // returns "Scrooge McDuck"

scrooge.renameWith("Scrooge H. McDuck")
print(scrooge.fullName) // returns "Scrooge H. McDuck"

苹果还表示：

如果协议要求属性可获取和可设置，则常量存储属性或只读计算属性无法满足该属性要求。

因此，以下两个 Playground 代码片段ARE NOT valid:

示例#1：常量属性

protocol FullyNamed {
    var fullName: String { get set }
}

struct Duck: FullyNamed {
    let fullName: String
}

let scrooge = Duck(fullName: "Scrooge McDuck")
// Error message: Type 'Duck' does not conform to protocol 'FullyNamed'

示例#2：计算属性（仅获取）

protocol FullyNamed {
    var fullName: String { get set }
}

struct Duck: FullyNamed {
    private var name: String
    var fullName: String {
        return name
    }
}

var scrooge = Duck(name: "Scrooge McDuck")
// Error message: Type 'Duck' does not conform to protocol 'FullyNamed'

示例#3：计算属性（仅获取）

protocol FullyNamed {
    var fullName: String { get }
}

struct Duck: FullyNamed {
    var fullName: String {return "Scrooge McDuck"}
    
    init(fullName: String) {
        self.fullName = fullName 
  // Error Message Cannot assign to Property: "FullName" is get only
    }
}