Go:在结构中声明一个 slice ？

Question

我有以下代码:

type room struct {
    width float32
    length float32
}
type house struct{
    s := make([]string, 3)
    name string
    roomSzSlice := make([]room, 3)
} 

func main() {


}

当我尝试构建和运行它时，出现以下错误:

c:\go\src\test\main.go:10: syntax error: unexpected :=
c:\go\src\test\main.go:11: non-declaration statement outside function body
c:\go\src\test\main.go:12: non-declaration statement outside function body
c:\go\src\test\main.go:13: syntax error: unexpected }

我做错了什么？

谢谢!

Answer 1

您可以在结构声明中声明一个 slice ，但不能对其进行初始化。你必须通过不同的方式来做到这一点。

// Keep in mind that lowercase identifiers are
// not exported and hence are inaccessible
type House struct {
    s []string
    name string
    rooms []room
}

// So you need accessors or getters as below, for example
func(h *House) Rooms()[]room{
    return h.rooms
}

// Since your fields are inaccessible, 
// you need to create a "constructor"
func NewHouse(name string) *House{
    return &House{
        name: name,
        s: make([]string, 3),
        rooms: make([]room, 3),
    }
}

请将以上内容视为 runnable example on Go Playground

---

编辑

要按照评论中的要求部分初始化结构，可以简单地更改

func NewHouse(name string) *House{
    return &House{
        name: name,
    }
}

请将以上内容视为 runnable example on Go Playground ，再次