Swift 3 隐式解包选项导致错误的字符串插值

Question

为什么在 Swift 3 中使用字符串插值时，隐式解开的 optional 没有解开？

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)")

产生以下输出:

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，但我在应用程序中几乎到处都使用字符串插值，由于这个(错误？)，现在它不再工作了。

这到底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>不再是一个独特的类型；只有Optional<T>现在。

仍然允许将声明注释为隐式解包选项 T! ，但这样做只是添加一个隐藏属性来通知编译器，它们的值可能会在需要其解包类型的上下文中强制解包 T ;他们的实际类型现在是 T? .

所以你可以想到这个声明:

var str: String!

实际上看起来像这样:

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

只有编译器才能看到此@_implicitlyUnwrapped属性，但它允许隐式展开 str在需要 String 的环境中的值(value)(其未包装类型):

// `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可以作为强选项进行类型检查，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.

与强制展开相比，编译器总是更喜欢将其视为此类。

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 _ExpressibleByStringInterpolation protocol 中的这个初始化程序。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)")

如str的实际类型是 String? ，默认情况下，编译器将推断出通用占位符 T成为。因此 str 的值不会被强制打开，并且您最终会看到可选的描述。

如果您希望在字符串插值中使用 IUO 时强制展开，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!)")

或者您可以强制为其非可选类型(在本例中为 String )，以强制编译器隐式强制为您解开它:

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

如果str，当然，这两个都会崩溃。是 nil .