GoLang：内部结构的泛型和动态类型

huangapple
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英文:

GoLang: Generic In Inner Struct and Dynamic Type

问题

我想创建一个子结构体，使用泛型类型作为变量，根据情况可以是两种类型之一。我有以下解决方案：

  1. type Resource interface {
  2.     Phone | Computer
  3. }
  5. type MetricData[R Resource] struct {
  6.     ResourceName string      `json:"resourceName"`
  7.     Metrics      []Metric[R] `json:"metrics"`
  8. }
  10. type Metric[R Resource] struct {
  11.     Resource    R   `json:"resource"`
  12.     Occurrences int `json:"occurrences"`
  13.     Change      int `json:"change"`
  14. }

然而，我在运行时不知道Resource的类型。当我创建一个新的结构体实例时，我需要提供R的类型：

  1. model.MetricData[model.Phone]{ResourceName: resource, Metrics: tm}

有没有办法在运行时给MetricData指定类型，并通过if/else语句根据某些条件确定类型？

英文:

I want to make a child struct use a generic type as variable can be one of two types depending on the situation. I have the following solution

  1. type Resource interface {
  2. 	Phone | Computer
  3. }
  5. type MetricData[R Resource] struct {
  6. 	ResourceName string      `json:"resourceName"`
  7. 	Metrics      []Metric[R] `json:"metrics"`
  8. }
  10. type Metric[R Resource] struct {
  11. 	Resource    R   `json:"resource"`
  12. 	Occurrences int `json:"occurrences"`
  13. 	Change      int `json:"change"`
  14. }

However, I won't know the type of Resource at runtime. When I create a new struct instance I need to provide the type of R

  1. model.MetricData[model.Phone]{ResourceName: resource, Metrics: tm}

Is there a way to give MetricData the type at runtime and determine it via if/else statement depending on some condition?

答案1

得分: 2

也许这会有所帮助（playground）：

  1. func resourceType[T Resource](r T) string {
  2.     switch any(r).(type) {
  3.     case Phone:
  4.         return "phone"
  5.     case Computer:
  6.         return "computer"
  7.     default:
  8.         return ""
  9.     }
  10. }
英文:

May be this will help (playground):

  1. func resourceType[T Resource](r T) string {
  2. 	switch any(r).(type) {
  3. 	case Phone:
  4. 		return "phone"
  5. 	case Computer:
  6. 		return "computer"
  7. 	default:
  8. 		return ""
  9. 	}
  10. }

答案2

得分: 0

你可以像这样修改你的结构：

  1. type Resource interface {}
  3. type MetricData struct {
  4.     ResourceName string      `json:"resourceName"`
  5.     Metrics      []Metric    `json:"metrics"`
  6. }
  8. type Metric struct {
  9.     Resource    Resource   `json:"resource"`
  10.     Occurrences int        `json:"occurrences"`
  11.     Change      int        `json:"change"`
  12. }

现在，要知道Resource的确切类型，你可以这样做：

  1. dummyMetric := {Resource: models.Phone{}, Occurrence: 1, Change: 1}
  2. resourceType := ""
  4. switch dummyMetric.Resource.(type) {
  5. case models.Phone:
  6.     resourceType = "phone"
  7. case models.Computer:
  8.     resourceType = "computer"
  9. }

希望对你有所帮助！

英文:

You can modify your structures like this:

  1. type Resource interface {}
  3. type MetricData struct {
  4.     ResourceName string      `json:"resourceName"`
  5.     Metrics      []Metric    `json:"metrics"`
  6. }
  8. type Metric struct {
  9.     Resource    Resource   `json:"resource"`
  10.     Occurrences int        `json:"occurrences"`
  11.     Change      int        `json:"change"`
  12. }

Now, to know the exact type of Resource, you can do something like this:

  1. dummyMetric := {Resource: models.Phone{}, Occurrence: 1, Change: 1}
  2. resourceType := ""
  4. switch dummyMetric.Resource.(type) {
  5. case models.Phone:
  6.     resourceType = "phone"
  7. case models.Computer:
  8.     resourceType = "computer"
  9. }

Hope it helps!

huangapple
  • 本文由 发表于 2023年7月25日 07:54:38
  • 转载请务必保留本文链接：https://go.coder-hub.com/76758667.html
  • generics
  • go
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