golang – how to initialize a map field within a struct?

huangapple
go评论117阅读模式
英文:

golang - how to initialize a map field within a struct?

问题

我对于初始化包含映射的结构体的最佳方法感到困惑。运行这段代码会产生panic: runtime error: assignment to entry in nil map的错误：

  1. package main
  3. type Vertex struct {
  4.    label string
  5. } 
  7. type Graph struct {
  8.   connections map[Vertex][]Vertex
  9. } 
  11. func main() {
  12.   v1 := Vertex{"v1"}
  13.   v2 := Vertex{"v2"}
  15.   g := new(Graph)
  16.   g.connections[v1] = append(g.coonections[v1], v2)
  17.   g.connections[v2] = append(g.connections[v2], v1)
  18. }

一个想法是创建一个构造函数，就像这个答案中所示。

另一个想法是使用一个add_connection方法，如果映射为空，则可以初始化它：

  1. func (g *Graph) add_connection(v1, v2 Vertex) {
  2.   if g.connections == nil {
  3.     g.connections = make(map[Vertex][]Vertex)
  4.   }
  5.   g.connections[v1] = append(g.connections[v1], v2)
  6.   g.connections[v2] = append(g.connections[v2], v1)
  7. }

还有其他选项吗？我只是想知道是否有一种常见的方法来做到这一点。

英文:

I'm confused about the best way to initialize a struct that contains a map. Running this code produces panic: runtime error: assignment to entry in nil map:

  1. package main
  3. type Vertex struct {
  4.    label string
  5. } 
  7. type Graph struct {
  8.   connections map[Vertex][]Vertex
  9. } 
  11. func main() {
  12.   v1 := Vertex{"v1"}
  13.   v2 := Vertex{"v2"}
  15.   g := new(Graph)
  16.   g.connections[v1] = append(g.coonections[v1], v2)
  17.   g.connections[v2] = append(g.connections[v2], v1)
  18. }

One idea is to create a constructor, as in this answer.

Another idea is to use an add_connection method that can initialize the map if it's empty:

  1. func (g *Graph) add_connection(v1, v2 Vertex) {
  2.   if g.connections == nil {
  3.     g.connections = make(map[Vertex][]Vertex)
  4.   }
  5.   g.connections[v1] = append(g.connections[v1], v2)
  6.   g.connections[v2] = append(g.connections[v2], v1)
  7. }

Are there other options? Just wanted to see if there is a commonly-accepted way to do this.

答案1

得分: 50

我可能会使用构造函数来完成这个任务：

  1. func NewGraph() *Graph {
  2.     var g Graph
  3.     g.connections = make(map[Vertex][]Vertex)
  4.     return &g
  5. }

我在标准的image/jpeg包中找到了这个例子（不过没有使用map，而是使用了slice）：

  1. type Alpha struct {
  2.     Pix    []uint8
  3.     Stride int
  4.     Rect   Rectangle
  5. }
  7. func NewAlpha(r Rectangle) *Alpha {
  8.     w, h := r.Dx(), r.Dy()
  9.     pix := make([]uint8, 1*w*h)
  10.     return &Alpha{pix, 1 * w, r}
  11. }
英文:

I would probably use a constructor to do this:

  1. func NewGraph() *Graph {
  2.     var g Graph
  3.     g.connections = make(map[Vertex][]Vertex)
  4.     return &g
  5. }

I've found this example in the standard image/jpeg package (not with a map though, but with a slice):

  1. type Alpha struct {
  2.     Pix []uint8
  3.     Stride int
  4.     Rect Rectangle
  5. }
  7. func NewAlpha(r Rectangle) *Alpha {
  8. 	w, h := r.Dx(), r.Dy()
  9. 	pix := make([]uint8, 1*w*h)
  10. 	return &Alpha{pix, 1 * w, r}
  11. }

答案2

得分: 19

很常见的情况是代码（尤其是完全由你控制的代码）假设你正确初始化了数据结构。在这种情况下，通常会使用结构体字面量。

  1. g := &Graph{
  2.     connections: make(map[Vertex][]Vertex),
  3. }
英文:

It's very common for code (especially code fully under your control) to assume you initialize the data structure correctly. A struct literal is usually used in this case

  1. g := &Graph{
  2.     connections: make(map[Vertex][]Vertex),
  3. }

答案3

得分: 4

复合字面量在构造函数内部完全正常工作。通过使用初始问题来构造一个示例（并且天真地将 Vertices 的副本存储在映射中）：

  1. func NewGraph(v1 Vertex, v2 Vertex) *Graph {
  2.     return &Graph{map[Vertex][]Vertex{v1: []Vertex{v2}, v2: []Vertex{v1}}}
  3. }
  5. func main() {
  6.     v1 := Vertex{"v1"}
  7.     v2 := Vertex{"v2"}
  9.     g := NewGraph(v1, v2)
  10.     fmt.Println(g)
  11. }

点击此处查看示例代码。

英文:

Composite literals work just fine inside a constructor. Contriving an example using the initial question (and naively storing copies of Vertices in the map):

  1. func NewGraph(v1 Vertex, v2 Vertex) *Graph {
  2.     return &Graph{ map[Vertex][]Vertex{ v1: []Vertex{v2}, v2: []Vertex{v1} }}
  3. }
  5. func main() {
  6.   v1 := Vertex{"v1"}
  7.   v2 := Vertex{"v2"}
  9.   g := NewGraph(v1, v2)
  10.   fmt.Println(g)
  11. }

https://play.golang.org/p/Lf4Gomp4tJ

huangapple
  • 本文由 发表于 2014年12月19日 02:35:16
  • 转载请务必保留本文链接：https://go.coder-hub.com/27553399.html
  • dictionary
  • go
  • struct
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定