blocks|key|2934736|text|之所以会出现这个错误，部分原因是编译器在不运行代码的情况下无法验证开关是否详尽。表达式0...65535创建一个ClosedInterval结构，当执行switch语句时，它必须询问该结构的值quantity是否在间隔内。这是可以在运行时更改的，因此编译器无法在编译时对其进行检查。(请参阅Halting+Problem。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2934737|更广泛地说，编译器无法检测到整数值的详尽开关--即使您为每个整数值(case+0:+...+case+1:+...+...+case+65535:)添加特定的情况，编译器也不知道您的开关是详尽的。(但从理论上讲是可以的:如果您希望看到这一点，请考虑使用filing+a+feature+request。)|2934738|目前，Swift可以在两种情况下检测完整性并允许您省略default子句:枚举和元组中的值绑定。@NateCook的答案涵盖了枚举-如果你打开一个枚举值，并且在你的switch中为枚举中的每个case都有一个case，那么你就不需要default了。如果您打开一个元组并绑定所有可能的值组合，也不需要default标签，如in+the+Swift+book所示|2934739|switch+anotherPoint+{
case+(let+x,+0):
++++println("on+the+x-axis+with+an+x+value+of+\(x)")
case+(0,+let+y):
++++println("on+the+y-axis+with+a+y+value+of+\(y)")
case+let+(x,+y):
++++println("somewhere+else+at+(\(x),+\(y))")
}|code-block|syntax|javascript|2934740|你可以将这条规则概括为“如果类型系统知道你的类型的可能值，它就可以检测switch的完整性”，但是在某种程度上，类型系统不知道可能的(例如)+UInt32值的范围是有问题的……|2934741|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Halting_problem|1|http://bugreport.apple.com/|2|https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/ControlFlow.html#//apple_ref/doc/uid/TP40014097-CH9-XID_205^0|17|9|1K|E|2O|8|41|F|0|0|Y|13|3I|O|1|0|R|7|2A|6|2O|4|2W|4|38|7|46|7|4H|H|2|0|0|Z|6|1Z|6|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|G|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|H|3|I|5|6|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]|$9|1T|A|1U|B|C]]|D|@$9|1V|A|1W|1|1X]]|E|$]]|$1|J|3|K|5|L|7|1Y|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1Z|8|@$9|20|A|21|B|C]|$9|22|A|23|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|24|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]|10|$5|T|U|V|E|$W|11]]]]

Part of why you see that error because the compiler can't verify that switch is exhaustive without running code. The expression <code>0...65535</code> creates a <code>ClosedInterval</code> struct, and when the switch statement executes it has to ask that struct if the value <code>quantity</code> is in the interval. There's room for that to change at run time, so the compiler can't check it at compile time. (See the <a href="https://en.wikipedia.org/wiki/Halting_problem" rel="noreferrer">Halting Problem</a>.)

More generally, the compiler can't detect an exhaustive switch for integer values — even if you add specific cases for every integer value (<code>case 0: ... case 1: ... ... case 65535:</code>), it doesn't know your switch is exhaustive. (Theoretically it could, though: consider <a href="http://bugreport.apple.com" rel="noreferrer">filing a feature request</a> about this if it's something you'd like to see.)

As it stands, there are two scenarios where Swift can detect completeness and allow you to omit the <code>default</code> clause: enums and value binding in tuples. @NateCook's answer covers enums — if you switch on an enum value and have a <code>case</code> in your <code>switch</code> for every <code>case</code> in the enum, you don't need a <code>default</code>. You also don't need a <code>default</code> label if you switch on a tuple and bind every possible combination of values, as seen <a href="https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/ControlFlow.html#//apple_ref/doc/uid/TP40014097-CH9-XID_205" rel="noreferrer">in the Swift book</a>:

<pre><code>switch anotherPoint {
case (let x, 0):
 println("on the x-axis with an x value of \(x)")
case (0, let y):
 println("on the y-axis with a y value of \(y)")
case let (x, y):
 println("somewhere else at (\(x), \(y))")
}
</code></pre>

You might generalize this rule as "if the type system knows about the possible values of your type, it can detect <code>switch</code> completeness", but the fact that there's a level on which the type system doesn't know the range of possible (e.g.) <code>UInt32</code> values is sort of splitting hairs...

blocks|key|145075|text|Swift仅在处理enum类型时才真正验证switch块是否详尽。即使打开Bool，除了true和false之外，还需要一个default块|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|145076|var+b+=+true
switch+b+{
case+true:++println("true")
case+false:+println("false")
}
//+error:+switch+must+be+exhaustive,+consider+adding+a+default+clause|code-block|syntax|javascript|145077|然而，对于enum，编译器很乐意只考虑这两种情况：|145078|enum+MyBool+{
++++case+True
++++case+False
}

var+b+=+MyBool.True
switch+b+{
case+.True:++println("true")
case+.False:+println("false")
}|145079|如果您需要为了编译器的缘故而包含一个default块，但又没有任何事情可做，那么break关键字就派上用场了：|145080|var+b+=+true
switch+b+{
case+true:++println("true")
case+false:+println("false")
default:+break
}|145081|entityMap^0|9|4|L|6|11|4|18|4|1D|5|1Q|7|0|0|5|4|0|0|I|7|14|5|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|T|$]]

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

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

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

<pre><code>enum MyBool {
 case True
 case False
}

var b = MyBool.True
switch b {
case .True: println("true")
case .False: println("false")
}
</code></pre>

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

<pre><code>var b = true
switch b {
case true: println("true")
case false: println("false")
default: break
}
</code></pre>

blocks|key|146070|text|Swift+4.1。要么需要指定所有情况，要么只在switch语句中包含缺省块。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|146071|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Swift 4.1. Either you need to specify all cases or Just include default block inside switch statement.

blocks|key|2722561|text|(从Swift+4.2开始，可能更早)：我有一个助手函数，可以将Bool?转换为具有2个段的UISegmentedControl的selectedSegmentIndex。如果该值为nil，则不应选择这两个线段。我的函数使用了一个开关，它为true或false值返回适当的段索引，并使用它显式地测试nil，并满足编译器对它的穷举的需求：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2722562|case+nil:++//+only+remaining+possible+value
++++fallthrough
default:
++++return+UISegmentedControl.noSegment|code-block|syntax|javascript|2722563|从技术上讲，不需要case+nil:+fallthrough，因为default:就足够了，但是如果您想显式测试值以使代码更具自文档性，或者在另一种情况下，这种语法可能会很有用。|2722564|entityMap^0|W|5|1T|K|2J|3|45|3|0|0|9|9|J|B|X|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|N|$]]

(As of Swift 4.2, and probably earlier): I have a helper function that converts a <code>Bool?</code> into the <code>selectedSegmentIndex</code> for a UISegmentedControl with 2 segments. If the value is <code>nil</code> then neither segment should be selected. My function uses a switch, which returns the appropriate segment index for true or false values, and uses this to explicitly test for the <code>nil</code> and satisfy the compiler's need for it to be exhaustive:

<pre><code>case nil: // only remaining possible value
 fallthrough
default:
 return UISegmentedControl.noSegment
</code></pre>

Technically, the <code>case nil:</code> <code>fallthrough</code> isn't required because the <code>default:</code> will suffice, but this syntax may be useful if you want to explicitly test a value to make the code more self-documenting, or perhaps in another situation.

blocks|key|2722616|text|检查您的枚举是否初始化为可选的，可以是case或nil|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2722617|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Check if your enum was initialised as an optional which could be either case or nil

The Apple <a href="https://developer.apple.com/library/ios/documentation/swift/conceptual/Swift_Programming_Language/ControlFlow.html" rel="noreferrer">documentation</a> says

<blockquote>
 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.
</blockquote>

So in new Xcode I have placed a code like this

<pre><code>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();
}
</code></pre>

Now if i remove the default section I get a compiler error:

<blockquote>
 Switch must be exhaustive 
 Do you want to add missing cases? Fix
</blockquote>

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

Exhaustive condition of switch case in Swift

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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苹果称每条switch语句都必须是详尽的。也就是说，所考虑的类型的每个可能值都必须与其中一个切换情况相匹配。因此，在新的Xcode中，我放置了如下代码println(UInt16.min); // Output : '0'println(UInt16.max); // Output : '65535'var quant...

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