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：
package maintype Vertex struct {label string}type Graph struct {connections map[Vertex][]Vertex}func main() {v1 := Vertex{"v1"}v2 := Vertex{"v2"}g := new(Graph)g.connections[v1] = append(g.coonections[v1], v2)g.connections[v2] = append(g.connections[v2], v1)}
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：
func (g *Graph) add_connection(v1, v2 Vertex) {if g.connections == nil {g.connections = make(map[Vertex][]Vertex)}g.connections[v1] = append(g.connections[v1], v2)g.connections[v2] = append(g.connections[v2], v1)}
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:
package maintype Vertex struct {label string}type Graph struct {connections map[Vertex][]Vertex}func main() {v1 := Vertex{"v1"}v2 := Vertex{"v2"}g := new(Graph)g.connections[v1] = append(g.coonections[v1], v2)g.connections[v2] = append(g.connections[v2], v1)}
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:
func (g *Graph) add_connection(v1, v2 Vertex) {if g.connections == nil {g.connections = make(map[Vertex][]Vertex)}g.connections[v1] = append(g.connections[v1], v2)g.connections[v2] = append(g.connections[v2], v1)}
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：
func NewGraph() *Graph {var g Graphg.connections = make(map[Vertex][]Vertex)return &g}
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）：
type Alpha struct {Pix []uint8Stride intRect Rectangle}func NewAlpha(r Rectangle) *Alpha {w, h := r.Dx(), r.Dy()pix := make([]uint8, 1*w*h)return &Alpha{pix, 1 * w, r}}
英文:
I would probably use a constructor to do this:
func NewGraph() *Graph {var g Graphg.connections = make(map[Vertex][]Vertex)return &g}
I've found this example in the standard image/jpeg package (not with a map though, but with a slice):
type Alpha struct {Pix []uint8Stride intRect Rectangle}func NewAlpha(r Rectangle) *Alpha {w, h := r.Dx(), r.Dy()pix := make([]uint8, 1*w*h)return &Alpha{pix, 1 * w, r}}
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。
g := &Graph{connections: make(map[Vertex][]Vertex),}
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
g := &Graph{connections: make(map[Vertex][]Vertex),}
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）：
func NewGraph(v1 Vertex, v2 Vertex) *Graph {return &Graph{map[Vertex][]Vertex{v1: []Vertex{v2}, v2: []Vertex{v1}}}}func main() {v1 := Vertex{"v1"}v2 := Vertex{"v2"}g := NewGraph(v1, v2)fmt.Println(g)}
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):
func NewGraph(v1 Vertex, v2 Vertex) *Graph {return &Graph{ map[Vertex][]Vertex{ v1: []Vertex{v2}, v2: []Vertex{v1} }}}func main() {v1 := Vertex{"v1"}v2 := Vertex{"v2"}g := NewGraph(v1, v2)fmt.Println(g)}
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