Swift中开关盒的穷举条件 | Swift中开关盒的穷举条件

本文介绍了Swift中开关盒的穷举条件的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

Apple 说

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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