Xcode 箱子和开关在swift中_Xcode_Swift

Xcode 箱子和开关在swift中

xcode swift

Xcode 箱子和开关在swift中,xcode,swift,Xcode,Swift,在下面所示的代码中，蔬菜多次声明为常量。但是Xcode没有得到一个错误。为什么Xcode成功地编译了它，但没有得到错误信息？设为常数 let vegetable = "red pepper" switch vegetable { case "celery": let vegetableComment = "Add some raisins and make ants on a log." case "cucumber", "watercres

在下面所示的代码中，蔬菜多次声明为常量。但是Xcode没有得到一个错误。为什么Xcode成功地编译了它，但没有得到错误信息？设为常数

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."
}

摘自：苹果公司的《Swift编程语言》

我想你实际上是在说

vegetableComment

被定义为常数，这就是你所困惑的

在Swift中，

开关

语句中的每个

大小写：

块都有自己的词法范围。这意味着您可以在每个变量中为所有变量命名相同的名称，并且它们不会冲突。几乎就好像它们有不同的功能

同样，您也不能访问其他数据库中的变量。以下是一些例子：

let vegetable = "red pepper"
var comment = ""
switch vegetable {
case "celery":
    comment = "Add some raisins and make ants on a log."
    // This is only defined here
    var favoriteVegetable = "celery"

case "cucumber", "watercress":
    comment = "That would make a good tea sandwich."
    // This will be an error, because `favoriteVegetable` is only valid inside the celery case block
    // favoriteVegetable = "either cucumber or watercress"

case let x where x.hasSuffix("pepper"):
    comment = "Is it a spicy \(x)?"

    // We can redefine favoriteVegetable here, because it has nothing to do with the one in the celery block
    let favoriteVegetable = "a pepper"

default:
    comment = "Everything tastes good in soup."
}

// Similarly, we can't access `favoriteVegetable` here
// println(favoriteVegetable)

// This was defined before the switch statement, so we can get the value that was calculated
println(comment)

我假设您实际上是在谈论将

vegetableComment

定义为常量，这就是您所困惑的

在Swift中，

开关

语句中的每个

大小写：

块都有自己的词法范围。这意味着您可以在每个变量中为所有变量命名相同的名称，并且它们不会冲突。几乎就好像它们有不同的功能

同样，您也不能访问其他数据库中的变量。以下是一些例子：

let vegetable = "red pepper"
var comment = ""
switch vegetable {
case "celery":
    comment = "Add some raisins and make ants on a log."
    // This is only defined here
    var favoriteVegetable = "celery"

case "cucumber", "watercress":
    comment = "That would make a good tea sandwich."
    // This will be an error, because `favoriteVegetable` is only valid inside the celery case block
    // favoriteVegetable = "either cucumber or watercress"

case let x where x.hasSuffix("pepper"):
    comment = "Is it a spicy \(x)?"

    // We can redefine favoriteVegetable here, because it has nothing to do with the one in the celery block
    let favoriteVegetable = "a pepper"

default:
    comment = "Everything tastes good in soup."
}

// Similarly, we can't access `favoriteVegetable` here
// println(favoriteVegetable)

// This was defined before the switch statement, so we can get the value that was calculated
println(comment)

我假设您实际上是在谈论将

vegetableComment

定义为常量，这就是您所困惑的

在Swift中，

开关

语句中的每个

大小写：

块都有自己的词法范围。这意味着您可以在每个变量中为所有变量命名相同的名称，并且它们不会冲突。几乎就好像它们有不同的功能

同样，您也不能访问其他数据库中的变量。以下是一些例子：

let vegetable = "red pepper"
var comment = ""
switch vegetable {
case "celery":
    comment = "Add some raisins and make ants on a log."
    // This is only defined here
    var favoriteVegetable = "celery"

case "cucumber", "watercress":
    comment = "That would make a good tea sandwich."
    // This will be an error, because `favoriteVegetable` is only valid inside the celery case block
    // favoriteVegetable = "either cucumber or watercress"

case let x where x.hasSuffix("pepper"):
    comment = "Is it a spicy \(x)?"

    // We can redefine favoriteVegetable here, because it has nothing to do with the one in the celery block
    let favoriteVegetable = "a pepper"

default:
    comment = "Everything tastes good in soup."
}

// Similarly, we can't access `favoriteVegetable` here
// println(favoriteVegetable)

// This was defined before the switch statement, so we can get the value that was calculated
println(comment)

我假设您实际上是在谈论将

vegetableComment

定义为常量，这就是您所困惑的

在Swift中，

开关

语句中的每个

大小写：

块都有自己的词法范围。这意味着您可以在每个变量中为所有变量命名相同的名称，并且它们不会冲突。几乎就好像它们有不同的功能

同样，您也不能访问其他数据库中的变量。以下是一些例子：

let vegetable = "red pepper"
var comment = ""
switch vegetable {
case "celery":
    comment = "Add some raisins and make ants on a log."
    // This is only defined here
    var favoriteVegetable = "celery"

case "cucumber", "watercress":
    comment = "That would make a good tea sandwich."
    // This will be an error, because `favoriteVegetable` is only valid inside the celery case block
    // favoriteVegetable = "either cucumber or watercress"

case let x where x.hasSuffix("pepper"):
    comment = "Is it a spicy \(x)?"

    // We can redefine favoriteVegetable here, because it has nothing to do with the one in the celery block
    let favoriteVegetable = "a pepper"

default:
    comment = "Everything tastes good in soup."
}

// Similarly, we can't access `favoriteVegetable` here
// println(favoriteVegetable)

// This was defined before the switch statement, so we can get the value that was calculated
println(comment)

你认为什么会失败<代码>蔬菜从未重新定义或更改

vegetableComment

被重新声明，但这没关系，因为每个

case

语句都有自己的作用域。对于每个块

蔬菜评论。因此，没有错误。让veg{let veg}没有错误。你认为什么会失败<代码>蔬菜

从未重新定义或更改

vegetableComment

被重新声明，但这没关系，因为每个

case

语句都有自己的作用域。对于每个块

蔬菜评论。因此，没有错误。让veg{let veg}没有错误。你认为什么会失败<代码>蔬菜

从未重新定义或更改

vegetableComment

被重新声明，但这没关系，因为每个

case

语句都有自己的作用域。对于每个块

蔬菜评论。因此，没有错误。让veg{let veg}没有错误。你认为什么会失败<代码>蔬菜

从未重新定义或更改

vegetableComment

被重新声明，但这没关系，因为每个

case

语句都有自己的作用域。对于每个块

蔬菜评论。因此，没有错误。让veg{let veg}没有错误。