Swift中开关盒的穷举条件 [英] Exhaustive condition of switch case in Swift

问题描述

Apple 文档说

每个switch语句必须是详尽的。也就是说，所考虑类型的每个可能的
值必须与
开关案例中的一个匹配。

Every switch statement must be exhaustive. That is, every possible value of the type being considered must be matched by one of the switch cases.

所以在新的Xcode中我放置了这样的代码

So in new Xcode I have placed a code like this

println(UInt16.min); // Output : '0'
println(UInt16.max); // Output : '65535'

var quantity : UInt16 = 10;

switch quantity {
case 0...65535: //OR case UInt16.min...UInt16.max:
    println();
default:
    println();
}

现在，如果我删除默认部分，我会收到编译错误，显示开关必须是详尽的

Now if i remove the default section I get a compile error showing that switch must be exhaustive

所以我的问题是我提到的一个案例案例0 ... 65535：我没有提到 UInt16 的所有案例值？但我仍然得到一个错误？为什么我收到这个错误，我错过了什么？

So my question is for a case that I have mentioned as case 0...65535: have I not mentioned all the case values for an UInt16 ?? But still I am getting an error ?? Why am I getting this error, Did i miss something ??

推荐答案

Swift只能真正验证使用枚举类型时，切换块是详尽无遗的。即使开启 Bool ，除了 true 默认块c>和 false ：

Swift only truly verifies that a switch block is exhaustive when working with enum types. Even a switching on Bool requires a default block in addition to true and false:

var b = true
switch b {
case true:  println("true")
case false: println("false")
}
// error: switch must be exhaustive, consider adding a default clause

但是，使用枚举，编译器很乐意只看两种情况：

With an enum, however, the compiler is happy to only look at the two cases:

enum MyBool {
    case True
    case False
}

var b = MyBool.True
switch b {
case .True:  println("true")
case .False: println("false")
}

如果你需要包含默认阻止编译器的缘故，但没有任何操作， break 关键字派上用场：

If you need to include a default block for the compiler's sake but don't have anything for it to do, the break keyword comes in handy:

var b = true
switch b {
case true:  println("true")
case false: println("false")
default: break
}

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