我想在切片上添加一些辅助方法。 所以我创建了一个[] * MyType的类型 有没有办法添加到那片MyTypes?追加将无法识别切片。
package main
import "fmt"
type MyType struct{
Name string
Something string
}
type MyTypes []*MyType
func NewMyTypes(myTypes ...*MyType)*MyTypes{
var s MyTypes = myTypes
return &s
}
//example of a method I want to be able to add to a slice
func(m MyTypes) Key() string{
var result string
for _,i := range m{
result += i.Name + ":"
}
return result
}
func main() {
mytype1 ,mytype2 := MyType{Name:"Joe", Something: "Foo"}, MyType{Name:"PeggySue", Something: "Bar"}
myTypes:= NewMyTypes(&mytype1,&mytype2)
//cant use it as a slice sadface
//myTypes = append(myTypes,&MyType{Name:"Random", Something: "asdhf"})
fmt.Println(myTypes.Key())
}
我不想将它包装在另一种类型中,并将该参数命名为即使我正在做它...因为json编组可能会有所不同
添加到MyTypes切片的方法是什么?
我真的希望能够将一个方法添加到切片中,这样它就可以实现一个特定的接口,而不会影响编组。是否有更好的方法?
由于
答案 0 :(得分:2)
更新:这个答案曾经包含两种解决问题的方法:我有点笨重的方式,DaveC更优雅的方式。这是他更优雅的方式:
package main
import (
"fmt"
"strings"
)
type MyType struct {
Name string
Something string
}
type MyTypes []*MyType
func NewMyTypes(myTypes ...*MyType) MyTypes {
return myTypes
}
//example of a method I want to be able to add to a slice
func (m MyTypes) Names() []string {
names := make([]string, 0, len(m))
for _, v := range m {
names = append(names, v.Name)
}
return names
}
func main() {
mytype1, mytype2 := MyType{Name: "Joe", Something: "Foo"}, MyType{Name: "PeggySue", Something: "Bar"}
myTypes := NewMyTypes(&mytype1, &mytype2)
myTypes = append(myTypes, &MyType{Name: "Random", Something: "asdhf"})
fmt.Println(strings.Join(myTypes.Names(), ":"))
}