Swift 3中，为什么在使用字符串插值时隐式解包可选项不会被解包？
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)")
生成此输出：
The following should not be printed as an optional: Optional("Hello")
+操作符来连接字符串，但我在我的应用程序中几乎到处都在使用字符串插值，现在由于这个（错误？）导致它不再起作用。ImplicitlyUnwrappedOptional<T>不再是一个独立的类型；现在只有Optional<T>。T!，但这样做只是添加了一个隐藏属性来告诉编译器它们的值可能会在需要其解包类型T的上下文中被强制解包；它们的实际类型现在是T?。var str: String!
@_implicitlyUnwrapped // this attribute name is fictitious
var str: String?
只有编译器能看到这个@_implicitlyUnwrapped属性，但它允许在需要String（它的解包类型）的上下文中隐式地解包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)
但在所有其他情况下，如果str可以被类型检查为强制可选项，那么它将是一个强制可选项：
// `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.
编译器总是更倾向于将其视为强制解包，而不是强制展开。
正如提案所说（重点在于我）：
如果表达式可以使用强可选类型进行显式类型检查，那么就会这样做。但是，如果必要的话，类型检查器将退回到强制使用可选项。这种行为的效果是，任何引用声明为
T!的值的表达式的结果都将具有类型T或类型T?。
当涉及到字符串插值时，在幕后编译器使用_ExpressibleByStringInterpolation 协议中的此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，当由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)")
str 的实际类型是 String?，默认情况下编译器将推断泛型占位符 T 为此类型。因此，str 的值不会被强制解包，你最终会看到一个可选项的描述。! 即可：var str: String!
str = "Hello"
print("The following should not be printed as an optional: \(str!)")
String），以便强制编译器隐式地强制解包它：print("The following should not be printed as an optional: \(str as String)")
当然，如果str为nil，这两者都会崩溃。
str为nil时会发生什么：print("The following should not be printed as an optional: \(str ?? "")")。 - hasheminil检查。而且 - 如其名所示 - 编译器应该为您完成所有的解包操作，但它没有这么做。 - Keiwan