Go：嵌入类型在派生类型中的字段初始化

Question

这是有效的代码：

package main

import (
    "fmt"
)

type Base struct {
    Field  int
}

type Derived struct {
    Base
}

func main() {
    d := &Derived{}
    d.Field = 10
    fmt.Println(d.Field)
}

下面是无法通过 ./main.go:17: unknown Derived field 'Field' in struct literal

编译的代码

package main

import (
    "fmt"
)

type Base struct {
    Field  int
}

type Derived struct {
    Base
}

func main() {
    d := &Derived{
        Field: 10,
    }
    fmt.Println(d.Field)
}

这里到底发生了什么？对不起，如果它很明显，但我就是不明白。

Answer 1

要初始化组合对象，您必须像任何其他对象一样初始化嵌入字段：

package main

import (
    "fmt"
)

type Base struct {
    Field int
}

type Derived struct {
    Base
}

func main() {
    d := &Derived{
        Base{10},
    }
    fmt.Println(d.Field)
}

Answer 2

来自语言specification:

Promoted fields act like ordinary fields of a struct except that they cannot be used as field names in composite literals of the struct.

所以这就是它不的原因。

这里有两种可能的方法来解决该限制，每种方法都在以下函数中进行了说明：

func main() {
    d := &Derived{
        Base{Field: 10},
    }

    e := new(Derived)
    e.Field = 20

    fmt.Println(d.Field)
    fmt.Println(e.Field)
}

Go：嵌入类型在派生类型中的字段初始化

Go: embedded type's field initialization in derived type

embedding

go