英文:
Calling method of named type
问题
我有一个命名类型，我需要创建它来进行一些JSON解组：
type StartTime time.Time
func (st *StartTime) UnmarshalJSON(b []byte) error {...}
由于StartTime是一个time.Time类型，我以为我可以调用属于time.Time的方法，比如Date()：
myStartTime.Date() // myStartTime.Date未定义（类型my_package.StartTime没有字段或方法Date）
我如何在现有类型上添加方法，同时保留其原始方法？
英文:
I have a named type, which I needed to make to do some JSON unmarshmaling:
type StartTime time.Time
func (st *StartTime) UnmarshalJSON(b []byte) error {...}
Since StartTime is a time.Time, I thought that I would be able to call methods that belong to time.Time, such as Date():
myStartTime.Date() // myStartTime.Date undefined (type my_package.StartTime has no field or method Date)
How can I add methods to an existing type, while preserving its original methods?
答案1
得分: 5
使用type关键字创建一个新类型，因此它不会具有底层类型的方法。
使用嵌入：
type StartTime struct {
time.Time
}
引用自规范：结构类型：
在结构体
x中，匿名字段的字段或方法f如果x.f是合法的选择器，表示该字段或方法f，则称为_提升_。
因此，嵌入字段的所有方法和字段都会被提升，并且可以引用。
以下是一个使用示例：
type StartTime struct {
time.Time
}
func main() {
s := StartTime{time.Now()}
fmt.Println(s.Date())
}
输出结果（在Go Playground上尝试）：
2009 November 10
英文:
Using the type keyword you are creating a new type and as such it will not have the methods of the underlying type.
Use embedding:
type StartTime struct {
time.Time
}
Quoting from the Spec: Struct types:
> A field or method f of an anonymous field in a struct x is called promoted if x.f is a legal selector that denotes that field or method f.
So all the methods and fields of the embedded (anonymous) fields are promoted and can be referred to.
Example using it:
type StartTime struct {
time.Time
}
func main() {
s := StartTime{time.Now()}
fmt.Println(s.Date())
}
Output (try it on the Go Playground):
2009 November 10
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