Golang中的通用方法参数

我需要帮助才能为任何类型的工作.

我有一个函数我需要接受其他具有ID属性的类型.

我尝试过使用接口,但这对我的ID属性案例不起作用.这是代码:

package main


import (
  "fmt"
  "strconv"
  )

type Mammal struct{
  ID int
  Name string 
}

type Human struct {  
  ID int
  Name string 
  HairColor string
}

func Count(ms []Mammal) *[]string { // How can i get this function to accept any type not just []Mammal
   IDs := make([]string, len(ms))
   for i, m := range ms {
     IDs[i] = strconv.Itoa(int(m.ID))
   }
   return &IDs
}

func main(){
  mammals := []Mammal{
    Mammal{1, "Carnivorious"},
    Mammal{2, "Ominivorious"},
  }

  humans := []Human{
    Human{ID:1, Name: "Peter", HairColor: "Black"},
    Human{ID:2, Name: "Paul", HairColor: "Red"},
  } 
  numberOfMammalIDs := Count(mammals)
  numberOfHumanIDs := Count(humans)
  fmt.Println(numberOfMammalIDs)
  fmt.Println(numberOfHumanIDs)
}

我明白了

错误prog.go:39:不能使用人类(类型[]人类)作为类型[]哺乳动物参数计数

有关详细信息,请参阅Go Playground http://play.golang.org/p/xzWgjkzcmH

你无法在Go中完全按照你的要求去做.在Go做最好的事情就是这样的

type Ids interface{
  Id() int
}

func (this Mammal) Id() int{
  return this.ID
} 

func (this Human) Id() int{
  return this.ID
} 


func Count(ms []Ids) *[]string {
...
    IDs[i] = strconv.Itoa(int(m.Id()))
...
}

func main(){
  mammals := []Ids{
    Mammal{1, "Carnivorious"},
    Mammal{2, "Ominivorious"},
  }

  humans := []Ids{
    Human{ID:1, Name: "Peter", HairColor: "Black"},
    Human{ID:2, Name: "Paul", HairColor: "Red"},
  }
  ...
}  

这是工作的例子