Swift 3 个不正确的字符串插值与隐式解包选项

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：运行 playground 中的以下代码

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> 不再是不同的类型；现在只有Optional<T>。

声明仍然允许被注释为隐式解包的可选值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解包类型的上下文中被强制解包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!

实际看起来像这样：

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

只有编译器可以看到此 @_implicitlyUnwrapped 属性，但它允许在需要 String（其展开类型）的上下文中隐式展开 str 的值:

// `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)

但是在 str 可以作为强可选类型进行类型检查的所有其他情况下，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.

而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?.

当谈到字符串插值时，编译器在后台使用 _ExpressibleByStringInterpolation protocol 中的这个初始化程序来评估字符串插值段：

/// 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)")

因为 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 的类型。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)")

当然，如果 str 是 nil，两者都会崩溃。

Swift 3 个不正确的字符串插值与隐式解包选项

Swift 3 incorrect string interpolation with implicitly unwrapped Optionals

string

string-interpolation

swift

swift3