Swift 中的元组 "upcasting"
Tuple "upcasting" in Swift
如果我有一个带有签名 (String, Bool)
的元组,我不能将它转换为 (String, Any)
。编译器说:
error: cannot express tuple conversion '(String, Bool)' to '(String,
Any)'
但这应该可以工作,因为可以使用 as
将 Bool
安全地转换为 Any
。如果你这样做,几乎会抛出同样的错误:
let any: Any = ("String", true)
any as! (String, Any) // error
any as! (String, Bool) // obviously succeeds
错误:
Could not cast value of type '(Swift.String, Swift.Bool)' to
'(protocol<>, protocol<>)'
那么对于第二种情况有什么解决方法吗?因为您甚至不能将 Any
转换为任何元组 (Any, Any)
,您可以在其中单独转换元素。
Swift 编译器将强制要求您明确指定类型。因此,如果您将其声明为 (String, Bool)
,它将不允许转换。
以下在 Playground 中按预期工作:
var specificTuple : (String, Bool) = ("Hi", false)
var generalTuple : (Any, Any) = ("Hi", false)
var gany = generalTuple
gany.1 = "there"
gany // (.0 "Hi", .1 "there")
var spany = specificTuple
spany.1 = "there" // error
您可以临时创建 (Any, Any)
元组,但您需要将其分解
var any : (Any, Any) = (specificTuple.0, specificTuple.1)
any.1 = "there"
any // (.0 "Hi", .1 "there")
不能强制转换元组,即使它们包含的类型可以。例如:
let nums = (1, 5, 9)
let doubleNums = nums as (Double, Double, Double) //fails
但是:
let nums : (Double, Double, Double) = (1, 5, 9) //succeeds
您的解决方法是转换单个元素,而不是元组本身:
let tuple = ("String", true)
let anyTuple = (tuple.0, tuple.1 as Any)
// anyTuple is (String, Any)
这是原因之一the Swift documentation注:
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,类似于函数。同样,您不能创建元组的扩展(例如 extension (String, Bool) { … }
)。
如果您实际上正在使用 returns (String, Any)
的 API,请尝试将其更改为使用 class 或结构。但是如果你无力改进 API,你可以 switch
第二个元素的类型:
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 returns Any
并且元组不能保证是 (String, Any)
,那你就不走运了。
如果我有一个带有签名 (String, Bool)
的元组,我不能将它转换为 (String, Any)
。编译器说:
error: cannot express tuple conversion '(String, Bool)' to '(String, Any)'
但这应该可以工作,因为可以使用 as
将 Bool
安全地转换为 Any
。如果你这样做,几乎会抛出同样的错误:
let any: Any = ("String", true)
any as! (String, Any) // error
any as! (String, Bool) // obviously succeeds
错误:
Could not cast value of type '(Swift.String, Swift.Bool)' to '(protocol<>, protocol<>)'
那么对于第二种情况有什么解决方法吗?因为您甚至不能将 Any
转换为任何元组 (Any, Any)
,您可以在其中单独转换元素。
Swift 编译器将强制要求您明确指定类型。因此,如果您将其声明为 (String, Bool)
,它将不允许转换。
以下在 Playground 中按预期工作:
var specificTuple : (String, Bool) = ("Hi", false)
var generalTuple : (Any, Any) = ("Hi", false)
var gany = generalTuple
gany.1 = "there"
gany // (.0 "Hi", .1 "there")
var spany = specificTuple
spany.1 = "there" // error
您可以临时创建 (Any, Any)
元组,但您需要将其分解
var any : (Any, Any) = (specificTuple.0, specificTuple.1)
any.1 = "there"
any // (.0 "Hi", .1 "there")
不能强制转换元组,即使它们包含的类型可以。例如:
let nums = (1, 5, 9)
let doubleNums = nums as (Double, Double, Double) //fails
但是:
let nums : (Double, Double, Double) = (1, 5, 9) //succeeds
您的解决方法是转换单个元素,而不是元组本身:
let tuple = ("String", true)
let anyTuple = (tuple.0, tuple.1 as Any)
// anyTuple is (String, Any)
这是原因之一the Swift documentation注:
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,类似于函数。同样,您不能创建元组的扩展(例如 extension (String, Bool) { … }
)。
如果您实际上正在使用 returns (String, Any)
的 API,请尝试将其更改为使用 class 或结构。但是如果你无力改进 API,你可以 switch
第二个元素的类型:
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 returns Any
并且元组不能保证是 (String, Any)
,那你就不走运了。