代码之家 › 专栏 › 技术社区 › Sedat Kapanoglu johnnywhoop

你能在swift中重载类型的铸造操作员吗?

operator-overloading casting swift

Sedat Kapanoglu johnnywhoop · 技术社区 · 6 年前

我想在swift中实现已知类型之间的自动类型转换。C的方法是重载类型转换运算符。如果我想要我的 X 可以交叉分配的类型,例如, string ,我会写:

public class X 
{
     public static implicit operator string(X value)
     {
          return value.ToString();
     }

     public static implicit operator X(string value)
     {
          return new X(value);
     }
}

在那之后,我可以写一些东西,比如:

string s = new X();
X myObj = s;

它们会被自动转换。这在斯威夫特有可能吗?

1 回复 | 直到 6 年前

akaralar 6 年前

不,该语言没有为自定义类型提供这样的功能。objective-c集合和swift集合之间存在桥接,但这是内置的,不可定制。

// Swift array of `String` elements
let swiftArray: [String] = ["Bob", "John"] 

// Obj-C array of `NSString` elements, but element type information 
// is not known to the compiler, so it behaves like an opaque NSArray
let nsArray: NSArray = ["Kate", "Betty"] 

// Obj-C array of an `NSString` and an `NSNumber`, element type
// information is not known to the compiler
let heterogeneousNSArray: NSArray = ["World", 3]

// Casting with `as` is enough since we're going from Swift array to NSArray 
let castedNSArray: NSArray = swiftArray as NSArray

// Force casting with `as!` is required as element type information 
// of Obj-C array can not be known at compile time
let castedSwiftArray: [String] = nsArray as! [String]

// Obj-C arrays can not contain primitive data types and can only 
// contain objects, so we can cast with `as` without requiring a 
// force-cast with `!` if we want to cast to [AnyObject]
let heterogeneousCastedNSArray: [AnyObject] = heterogeneousNSArray as [AnyObject]

提供类型铸造文件 here 。

我认为你可以实现你想要做的初始化。

extension X {
    init(string: String) {
        self = X(string)
    }        
}

extension String {
    init(x: X) {
        // toString is implemented elsewhere
        self = x.toString
    }        
}

let x = X()
let string = "Bobby"

let xFromString: X = X(string: string)
let stringFromX: String = String(x: x)

与你的问题没有直接关系,但也有一系列的协议 ExpressibleBy... ,使您能够执行以下操作: 假设我们要从整数文本初始化字符串。我们可以通过遵守和实施 ExpressibleByIntegerLiteral

// Strings can not be initialized directly from integer literals
let s1: String = 3 // Error: Can not convert value of type 'Int' to specified type 'String'

// Conform to `ExpressibleByIntegerLiteral` and implement it
extension String: ExpressibleByIntegerLiteral {
  public init(integerLiteral value: Int) {
    // String has an initializer that takes an Int, we can use that to 
    // create a string
    self = String(value)
  }
}

// No error, s2 is the string "4"
let s2: String = 4

一个很好的用例 ExpressibleByStringLiteral 可以找到 here 。