Swift：switch语句中未解析的标识符

Question

我正在 Xcode6-Beta4 的操场上尝试以下操作，遵循苹果的 swift tour：

let vegetable = "red pepper"
switch vegetable{
case "celery":
    let vegetableComment = "Add some raisins and make ants on a log."
case "cucumber", "watercress":
    let vegetableComment = "That would make a good tea sandwich."
case let x where x.hasSuffix("pepper"):
    let vegetableComment = "Is it a spicy \(x)"
default:
    let vegetableComment = "Everything tastes good in soup."
}

然后我尝试调用switch语句中定义的变量vegetableComment，我得到了Use of unresolved identifier 'vegetableComment'的错误

这是否与swift中switch语句的范围/关闭有关？

Answer 1

这是否与swift中switch语句的范围/关闭有关？

是的，它与变量的范围有关。您有四个名为vegetableComment 的常量。每一个都限定在switch 语句中的case。

为了访问您在switch 中分配的变量，您需要在进入开关之前将其声明为var：

var vegetableComment = String()
let vegetable = "red pepper"
switch vegetable{
case "celery":
    vegetableComment = "Add some raisins and make ants on a log."
case "cucumber", "watercress":
    vegetableComment = "That would make a good tea sandwich."
case let x where x.hasSuffix("pepper"):
    vegetableComment = "Is it a spicy \(x)"
default:
    vegetableComment = "Everything tastes good in soup."
}