英文:
evaluation sequence of `switch` in `go`
问题
我正在学习Go语言，通过阅读《Effective Go》。我发现了一个关于类型切换的例子：
var t interface{}
t = functionOfSomeType()
switch t := t.(type) {
default:
fmt.Printf("unexpected type %T\n", t) // %T 打印出 t 的实际类型
case bool:
fmt.Printf("boolean %t\n", t) // t 的类型是 bool
case int:
fmt.Printf("integer %d\n", t) // t 的类型是 int
case *bool:
fmt.Printf("pointer to boolean %t\n", *t) // t 的类型是 *bool
case *int:
fmt.Printf("pointer to integer %d\n", *t) // t 的类型是 *int
}
我的理解是，在switch语句中，case从上到下依次进行匹配，直到找到匹配的条件为止。所以，这个例子不是总是会停在default并打印"unexpected type ..."吗？
英文:
I am learning Go language by reading "Effective Go".
I found a example about type switch:
var t interface{}
t = functionOfSomeType()
switch t := t.(type) {
default:
fmt.Printf("unexpected type %T\n", t) // %T prints whatever type t has
case bool:
fmt.Printf("boolean %t\n", t) // t has type bool
case int:
fmt.Printf("integer %d\n", t) // t has type int
case *bool:
fmt.Printf("pointer to boolean %t\n", *t) // t has type *bool
case *int:
fmt.Printf("pointer to integer %d\n", *t) // t has type *int
}
My understanding is the cases in switch is evaluated from top to bottom and stop at a match condition. So isn't the example about would always stop at default and print "unexpected type ..."?
答案1
得分: 8
从这个Golang教程中：
- 如果没有其他
case块匹配，将执行default的代码块。 default块可以位于switch块的任何位置，不一定是按词法顺序的最后一个。
英文:
From this Golang tutorial:
- The code block of
defaultis executed if none of the othercaseblocks match - the
defaultblock can be anywhere within theswitchblock, and not necessarily last in lexical order
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