如何在函数外部和内部退出 GUARD - Swift

Question

下面的代码我在练习GUARD的使用（书籍：OReilly Learning Swift）

guard 2+2 == 4 else {
print("The universe makes no sense")
return // this is mandatory!
}
print("We can continue with our daily lives")

为什么会出现以下代码错误？

error: return invalid outside of a func

或者 GUARD 只在函数内使用？

Answer 1

如果不满足guard 语句中的条件，则else 分支必须退出当前作用域。 return 只能在函数内部使用，如错误消息所示，但 return 不是退出范围的唯一方法。

你也可以使用throw外部函数，如果guard语句在循环中，你也可以使用break或continue.

return 在函数中有效：

func testGuard(){
    guard 2+2 == 4 else {
        print("The universe makes no sense")
        return // this is mandatory!
    }
    print("We can continue with our daily lives")
}

throw 在函数外也有效：

guard 2+2 == 4 else { throw NSError() }

break 在循环中有效：

for i in 1..<5 {
    guard i < 5 else {
        break
    }
}

Answer 2

Or is GUARD only used within functions?

不一定。

来自documentation:

If that condition is not met, the code inside the else branch is executed. That branch must transfer control to exit the code block in which the guard statement appears. It can do this with a control transfer statement such as return, break, continue, or throw, or it can call a function or method that doesn’t return, such as fatalError(_:file:line:).

例如，这可以用于顶层的 Playgound

for i in 0...10 {
    guard i % 2 == 0 else { continue }
    print(i)
}

错误信息实际上与guard无关。它只是说明 return 不能在函数 之外使用

如何在函数外部和内部退出 GUARD - Swift

How to exit GUARD outside and inside a function - Swift

guard

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