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 main
type 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 main
type 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 Graph
g.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 []uint8
Stride int
Rect 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 Graph
g.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 []uint8
Stride int
Rect 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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