元组“上转”在Swift [英] Tuple "upcasting" in Swift

问题描述

如果我有一个签名（String，Bool）的元组我不能把它转换为（String，Any）。编译器说：

If I have a tuple with signature (String, Bool) I cannot cast it to (String, Any). The compiler says:

错误：无法表示元组转换'（String，Bool）'到'（String，
Any） '

error: cannot express tuple conversion '(String, Bool)' to '(String, Any)'

但这应该可以工作，因为 Bool 可以安全地转换为任何与为。如果你这样做，几乎相同的错误抛出：

But this should work since Bool can be casted safely to Any with as. Almost the same error gets thrown if you do something like that:

let any: Any = ("String", true)
any as! (String, Any) // error
any as! (String, Bool) // obviously succeeds

错误：

无法将类型（Swift.String，Swift.Bool）的值转换为
'（协议<>，协议<>）'

Could not cast value of type '(Swift.String, Swift.Bool)' to '(protocol<>, protocol<>)'

那么有什么解决方法，特别是对于第二种情况？因为你甚至不能将任何转换到任何可以分别投射元素的任何元组（Any，Any） p>

So is there any workaround especially for the second scenario? Because you cannot even cast Any to any tuple (Any, Any) where you could cast the elements separately.

推荐答案

元组不能被转换，即使它们包含的类型也可以。例如：

Tuples cannot be cast, even if the types they contain can. For example:

let nums = (1, 5, 9)
let doubleNums = nums as (Double, Double, Double) //fails

但是：

let nums : (Double, Double, Double) = (1, 5, 9) //succeeds

您的情况下的解决方法是转换单个元素，而不是元组本身：

The workaround in your case is to cast the individual element, not the Tuple itself:

let tuple = ("String", true)
let anyTuple = (tuple.0, tuple.1 as Any)
// anyTuple is (String, Any)

这是原因之一 Swift文档注释：

元组对于相关值的临时组很有用。它们不适合于创建复杂的数据结构。如果您的数据结构可能超出临时范围，则将其建模为类或结构，而不是元组。

Tuples are useful for temporary groups of related values. They are not suited to the creation of complex data structures. If your data structure is likely to persist beyond a temporary scope, model it as a class or structure, rather than as a tuple.

我认为这是一个实现限制，因为元组是复合类型像函数。同样，您不能创建元组的扩展（例如 extension（String，Bool）{...} ）。

I think this is an implementation limitation because Tuples are compound types like functions. Similarly, you cannot create extensions of Tuples (e.g. extension (String, Bool) { … }).

如果你实际上使用返回（String，Any）的API，请尝试将其改为使用类或结构。但如果您无权改进API，您可以在第二个元素的类型上切换：

If you're actually working with an API that returns (String, Any), try to change it to use a class or struct. But if you're powerless to improve the API, you can switch on the second element's type:

let tuple : (String, Any) = ("string", true)

switch tuple.1 {

case let x as Bool:
    print("It's a Bool")
    let boolTuple = (tuple.0, tuple.1 as! Bool)

case let x as Double:
    print("It's a Double")
    let doubleTuple = (tuple.0, tuple.1 as! Double)

case let x as NSDateFormatter:
    print("It's an NSDateFormatter")
    let dateFormatterTuple = (tuple.0, tuple.1 as! NSDateFormatter)

default:
    print("Unsupported type")
}

如果API返回 Any 保证（String，Any），你运气不好。

If the API returns Any and the tuple isn't guaranteed to be (String, Any), you're out of luck.

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