swift 是否允许没有 conditions/loops 的代码块来缩小局部变量范围？

Question

在具有块级作用域的语言中，我有时会创建任意块，这样我就可以封装局部变量，而不是让它们污染它们父级的作用域：

func myFunc() {
  // if statements get block level scope
  if self.someCondition {
    var thisVarShouldntExistElsewhere = true
    self.doSomethingElse(thisVarShouldntExistElsewhere)
  }

  // many languages allow blocks without conditions/loops/etc
  {
    var thisVarShouldntExistElsewhere = false
    self.doSomething(thisVarShouldntExistElsewhere)
  }
}

当我在 Swift 中执行此操作时，它认为我正在创建闭包并且不执行代码。我可以将其创建为闭包并立即执行，但这似乎会带来执行开销（不值得仅仅为了代码清洁）。

func myFunc() {
  // if statements get block level scope
  if self.someCondition {
    var thisVarShouldntExistElsewhere = true
    self.doSomethingElse(thisVarShouldntExistElsewhere)
  }

  // converted to closure
  ({
    var thisVarShouldntExistElsewhere = false
    self.doSomething(thisVarShouldntExistElsewhere)
  })()
}

在Swift中是否支持类似的东西？

Answer 1

您可以使用 do 语句在 Swift 中创建任意范围。例如：

func foo() {
    let x = 5

    do {
        let x = 10
        print(x)
    }
}

foo() // prints "10"

根据The Swift Programming Language：

The do statement is used to introduce a new scope and can optionally contain one or more catch clauses, which contain patterns that match against defined error conditions. Variables and constants declared in the scope of a do statement can be accessed only within that scope.

A do statement in Swift is similar to curly braces ({}) in C used to delimit a code block, and does not incur a performance cost at runtime.

Ref: The Swift Programming Language - Language Guide - Statements - Do Statement

Answer 2

@Jack Lawrence 的答案的替代方法是使用积木；类似于您的第一个代码片段中的块。

func foo () {
    let x = 5

    let block = {
        let x = 10
        print(x)
    }

    block()
}

foo()

swift 是否允许没有 conditions/loops 的代码块来缩小局部变量范围？

Does swift allow code blocks without conditions/loops to reduce local variable scope?

scope

swift

swift2