Swift 3 不正确的字符串插值与隐式展开的 Optionals

Question

为什么在 Swift 3 中使用字符串插值时隐式解包的可选值不解包？

示例:在 playground 中运行以下代码

var str: String!
str = "Hello"

print("The following should not be printed as an optional: \(str)")

产生这个输出:

The following should not be printed as an optional: Optional("Hello")

当然，我可以使用 + 运算符连接字符串，但我在我的应用程序中几乎无处不在使用字符串插值，由于这个(错误？)，现在它不再起作用了。

这甚至是一个错误，还是他们故意用 Swift 3 改变了这种行为？

Answer 1

根据 SE-0054 , ImplicitlyUnwrappedOptional<T>不再是独特的类型；只有Optional<T>现在。

声明仍然允许被注释为隐式展开的可选类型 T! ，但这样做只是添加了一个隐藏属性，以通知编译器它们的值可能会在需要其解包类型的上下文中被强制解包 T ;它们的实际类型现在是 T? .

所以你可以想到这个声明:

var str: String!

实际看起来像这样:

@_implicitlyUnwrapped // this attribute name is fictitious 
var str: String?

只有编译器才能看到这个 @_implicitlyUnwrapped属性，但它允许的是 str 的隐式展开在需要 String 的上下文中的值(它的展开类型):

// `str` cannot be type-checked as a strong optional, so the compiler will
// implicitly force unwrap it (causing a crash in this case)
let x: String = str

// We're accessing a member on the unwrapped type of `str`, so it'll also be
// implicitly force unwrapped here
print(str.count)

但在所有其他情况下 str可以作为强可选类型进行类型检查，它将是:

// `x` is inferred to be a `String?` (because we really are assigning a `String?`)
let x = str 

let y: Any = str // `str` is implicitly coerced from `String?` to `Any`

print(str) // Same as the previous example, as `print` takes an `Any` parameter.

并且编译器总是更喜欢这样对待它而不是强制解包。

正如提案所说(强调我的):

If the expression can be explicitly type checked with a strong optional type, it will be. However, the type checker will fall back to forcing the optional if necessary. The effect of this behavior is that the result of any expression that refers to a value declared as T! will either have type T or type T?.

当涉及到字符串插值时，编译器在底层使用来自 _ExpressibleByStringInterpolation protocol 的这个初始化程序为了评估字符串插值段:

/// Creates an instance containing the appropriate representation for the
/// given value.
///
/// Do not call this initializer directly. It is used by the compiler for
/// each string interpolation segment when you use string interpolation. For
/// example:
///
///     let s = "\(5) x \(2) = \(5 * 2)"
///     print(s)
///     // Prints "5 x 2 = 10"
///
/// This initializer is called five times when processing the string literal
/// in the example above; once each for the following: the integer `5`, the
/// string `" x "`, the integer `2`, the string `" = "`, and the result of
/// the expression `5 * 2`.
///
/// - Parameter expr: The expression to represent.
init<T>(stringInterpolationSegment expr: T)

因此，当您的代码隐式调用时:

var str: String!
str = "Hello"

print("The following should not be printed as an optional: \(str)")

作为str的实际类型是 String? ，默认情况下，编译器将推断通用占位符 T。成为。因此 str 的值不会被强制展开，您最终会看到可选的描述。

如果您希望 IUO 在用于字符串插值时被强制展开，您可以简单地使用强制展开运算符 ! :

var str: String!
str = "Hello"

print("The following should not be printed as an optional: \(str!)")

或者您可以强制转换为非可选类型(在本例中为 String )，以强制编译器隐式强制为您解包:

print("The following should not be printed as an optional: \(str as String)")

如果str，这两者当然都会崩溃是nil .