go - 为什么我在 golang 的类型断言上得到奇怪的结果？

Question

此代码工作正常，但是，当我在一种情况下绑定(bind)它们时，例如:

var i interface{}
s := make([]map[string]interface{}, 5, 5)
i = s
switch x := i.(type) {
    case []interface{}:
        fmt.Println("type is an array")
        fmt.Println("length is: ")
        fmt.Println(len(x))
    case []map[string]interface{}:
        fmt.Println("type is an array")
        fmt.Println("length is: ")
        fmt.Println(len(x))
  }

但是当我在一个case语句中绑定(bind)它们时不起作用

switch x := i.(type) {
case []interface{}, []map[string]interface{}:
    fmt.Println("type is an array")
    fmt.Println("length is: ")
    fmt.Println(len(x))
}

Answer 1

请参阅语言规范:

The TypeSwitchGuard may include a short variable declaration. When that form is used, the variable is declared at the end of the TypeSwitchCase in the implicit block of each clause. In clauses with a case listing exactly one type, the variable has that type; otherwise, the variable has the type of the expression in the TypeSwitchGuard.

另外，请记住 Go 是静态类型的。您无法决定在运行时声明的变量的类型。因此，如果您列出多种类型，它只是使用与开关变量相同的类型来声明变量。