Question

func GetResult (service interface{}) {
   switch v := service.(type) {
     case services.Account:
       service = service.(services.Account)
     default:
       service = service.(Mock_Account)
    }
   res, err := service.GetAccount()
 }

它说服务是接口的类型没有任何方法。有类型转换不起作用任何想法如何调用GetAccount方法？

Answer 1

注释您的代码示例：

func GetResult (service interface{}) {
    // service is of type interface{}
    switch v := service.(type) {
    case services.Account:
        // This does nothing: it assigns the existing value back into service, which is still of type interface{}
        // v exists in scope and is of type services.Account, but it's unused.
        service = service.(services.Account)
    default:
        // Ditto
        service = service.(Mock_Account)
    }
    // service is still of type interface{}, which defines no methods
    res, err := service.GetAccount()
}

如您所见，此处的变量service 始终类型为interface{}。变量的类型无法更改：Go是静态类型。但是，如果您只想在传入的任何类型上调用方法GetAccount，则可以通过声明接口来完成此类型断言和完全类型安全：

type AccountGetter interface {
    GetAccount() (res ResultType, err error)
}

func GetResult(service AccountGetter) {
    res,err := service.GetAccount()
}

然后，您可以拨打GetResult并传递services.Account或Mock_Account或任何使用相应签名实现GetAccount方法的类型。

接口类型转换不起作用

1 个答案: