英文:
Why isn't fallthrough allowed in a type switch?
问题
我想知道为什么在golang的类型switch语句中不允许使用fallthrough。
根据规范：“在类型switch中不允许使用“fallthrough”语句。”，这并没有解释为什么不允许。
附上的代码是为了模拟可能需要在类型switch语句中使用fallthrough的情况。
**注意！**这段代码无法运行，会产生错误：“无法在类型switch中使用fallthrough”。我只是想知道为什么不允许在类型switch中使用fallthrough语句的可能原因。
//一个类型switch的问题package mainimport "fmt"//为什么不允许在类型switch中使用fallthrough？func main() {//空接口var x interface{}x = //一个int、float64、bool或string值switch i := x.(type) {case int:fmt.Println(i + 1)case float64:fmt.Println(i + 2.0)case bool:fallthroughcase string:fmt.Printf("%v", i)default:fmt.Println("未知类型。抱歉！")}}
英文:
I'm wondering why fallthrough isn't allowed in a type switch statement in golang.
According to the specification: "The "fallthrough" statement is not permitted in a type switch.", which doesn't explain much about WHY it isn't allowed.
The code attached is to simulate a possible scenario were a fallthrough in a type switch statement might have been useful.
Notice! This code doesn't work, it will produce the error: "cannot fallthrough in type switch". I'm just wondering what possible reasons might have been for not allowing the fallthrough statement in a type switch.
//A type switch questionpackage mainimport "fmt"//Why isn't fallthrough in type switch allowed?func main() {//Empty interfacevar x interface{}x = //A int, float64, bool or string valueswitch i := x.(type) {case int:fmt.Println(i + 1)case float64:fmt.Println(i + 2.0)case bool:fallthroughcase string:fmt.Printf("%v", i)default:fmt.Println("Unknown type. Sorry!")}}
答案1
得分: 48
你希望fallthrough如何工作？在这个类型开关中，i变量的类型取决于所调用的特定情况。所以在case bool中，i变量的类型是bool。但在case string中，它的类型是string。所以要么你希望i魔法般地改变它的类型，这是不可能的，要么你希望它被一个新的变量i string遮蔽，但它没有值，因为它的值来自于x，而x实际上不是一个string。
这里有一个示例来试图说明这个问题：
switch i := x.(type) {case int:// i是一个intfmt.Printf("%T\n", i); // 输出 "int"case bool:// i是一个boolfmt.Printf("%T\n", i); // 输出 "bool"fallthroughcase string:fmt.Printf("%T\n", i);// 这个类型是什么？它应该是 "string"，但如果类型是bool并且我们遇到了fallthrough，那么它会怎么做呢？}
唯一可能的解决方案是让fallthrough导致后续的case表达式将i保持为interface{}，但这将是一个令人困惑和糟糕的定义。
如果你真的需要这种行为，你可以通过现有的功能来实现：
switch i := x.(type) {case bool, string:if b, ok := i.(bool); ok {// b是一个bool}// i是一个包含bool或string的interface{}}
英文:
How would you expect fallthrough to work? In this type switch, the i variable has a type that depends on the particular case that's invoked. So in the case bool the i variable is typed as bool. But in case string it's typed as string. So either you're asking for i to magically morph its type, which isn't possible, or you're asking for it to be shadowed by a new variable i string, which will have no value because its value comes from x which is not, in fact, a string.
Here's an example to try and illustrate the problem:
switch i := x.(type) {case int:// i is an intfmt.Printf("%T\n", i); // prints "int"case bool:// i is a boolfmt.Printf("%T\n", i); // prints "bool"fallthroughcase string:fmt.Printf("%T\n", i);// What does that type? It should type "string", but if// the type was bool and we hit the fallthrough, what would it do then?}
The only possible solution would be to make the fallthrough cause the subsequent case expression to leave i as an interface{}, but that would be a confusing and bad definition.
If you really need this behavior you can already accomplish this with the existing functionality:
switch i := x.(type) {case bool, string:if b, ok := i.(bool); ok {// b is a bool}// i is an interface{} that contains either a bool or a string}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论