Question

嵌入式结构Struct1和结构Struct2定义为字段。两个fmt.Printf()都是相同的结果，而初始化有所不同。我对此感到困惑。对不起。

Struct1和Struct2之间有什么不同？
在什么情况下应该使用哪种？

脚本

type sample1 struct {
    Data string
}

type sample2 struct {
    Data string
}

type Struct1 struct {
    *sample1
    *sample2
}

type Struct2 struct {
    Sample1 sample1
    Sample2 sample2
}

func main() {
    s1 := &Struct1{
        &sample1{},
        &sample2{},
    }
    s1.sample1.Data = "s1 sample1 data"
    s1.sample2.Data = "s1 sample2 data"

    s2 := &Struct2{}
    s2.Sample1.Data = "s2 sample1 data"
    s2.Sample2.Data = "s2 sample2 data"

    fmt.Printf("%s, %s\n", s1.sample1.Data, s1.sample2.Data)
    fmt.Printf("%s, %s\n", s2.Sample1.Data, s2.Sample2.Data)
}

https://play.golang.org/p/gUy6gwVJDP

非常感谢你的时间和建议。我很抱歉我的问题不成熟。

Answer 1

关于第二个问题，我个人主要使用嵌入来推广方法

//having
type Doer interface{
    Do()
}
func DoWith(d Doer){}
func (s sample1)Do(){} //implemented
type Struct1 struct {
    sample1
}
type Struct2 struct {
    Sample1 sample1
}
var s1 Struct1
var s2 Struct2
//you can call
DoWith(s1) //method promoted so interface satisfied
//but cannot
DoWith(s2)

并解码JSON，

//having
type Sample1 struct {
    Data string `json:"data"`
}
type Sample2 struct {
    Number int `json:"number"`
}
//you can easy and handy compose
type Struct1 struct {
    Sample1
    Sample2
}
var s1 Struct1
json.Unmarshal([]byte(`{"data": "foo", "number": 5}`), &s1)
fmt.Println(s1.Data, s1.Number) //print foo 5

嵌入式和现场的差异

脚本

1 个答案: