英文:
Iterate Struct (map key) value in golang
问题
我已经查阅了在Golang中将结构体作为map键。
我了解在Golang中遍历map时没有保证的顺序。我按照Golang博客中的示例尝试使用struct作为map的键。
以下是我的代码：
package main
import (
"fmt"
"sort"
)
func main() {
req := make(map[mapKey]string)
req[mapKey{1, "r"}] = "rob pike"
req[mapKey{2, "gri"}] = "robert griesemer"
req[mapKey{3, "adg"}] = "andrew gerrand"
req[mapKey{4, "rsc"}] = "russ cox"
var keys []mapKey
for k := range req {
keys = append(keys, k)
}
for _, k := range keys {
fmt.Printf("short name: %s, long name: %s\n", k.Option, req[k])
}
sort.Slice(keys, func(i, j int) bool {
return keys[i].Key < keys[j].Key
})
}
我希望结果如下所示：
short name: r, long name: rob pike
short name: gri, long name: robert griesemer
short name: adg, long name: andrew gerrand
short name: rsc, long name: russ cox
我不知道如何将结构体的值和键分别迭代到不同的数据结构中。
英文:
I've looked up Structs as keys in Golang maps
I understand iteration over the maps in golang has no guaranteed order. I've followed the example in golang blog, and tried using a struct as a map key.
Here's my code
package main
func main() {
req := make(map[mapKey]string)
req[mapKey{1, "r"}] = "robpike"
req[mapKey{2, "gri"}] = "robert griesemer"
req[mapKey{3, "adg"}] = "andrew gerrand"
req[mapKey{4, "rsc"}] = "russ cox"
var keys []int
for k := range req {
keys = append(keys, k.Key)
}
for _, k := range keys {
fmt.Printf("short name : %s , long name : %s\n",req[k], req[k]) // How do I iterate here
}
sort.Ints(keys)
}
type mapKey struct {
Key int
Option string
}
What I want the results to be is
short name : r , long name : rob pike
short name : gri , long name : robert griesemer
short name : adg , long name : andrew gerrand
short name : rsc , long name : russ cox
And I don't know how I can get the struct value and key iterated by separated data structure.
答案1
得分: 3
简短版本？你不能用那种方式做。
长版本中，你可以使用自定义排序器：
func main() {
req := make(map[mapKey]string)
req[mapKey{1, "r"}] = "robpike"
req[mapKey{2, "gri"}] = "robert griesemer"
req[mapKey{3, "adg"}] = "andrew gerrand"
req[mapKey{4, "rsc"}] = "russ cox"
var keys mapKeys
for k := range req {
keys = append(keys, k)
}
sort.Sort(keys)
for _, k := range keys {
fmt.Printf("short name : %s , long name : %s\n", k.Option, req[k])
}
}
type mapKey struct {
Key int
Option string
}
type mapKeys []mapKey
func (mk mapKeys) Len() int { return len(mk) }
func (mk mapKeys) Swap(i, j int) { mk[i], mk[j] = mk[j], mk[i] }
func (mk mapKeys) Less(i, j int) bool { return mk[i].Key < mk[j].Key }
请注意，如果你的mapKey结构体有一个不支持相等性比较的字段（比如结构体或切片），使用req[k]将无法工作。
在这种情况下，你可以切换到type mapKeys []*mapKey和map[*mapKey]string。
英文:
Short version? You can't do it that way.
Long version you, you can use a custom sorter:
func main() {
req := make(map[mapKey]string)
req[mapKey{1, "r"}] = "robpike"
req[mapKey{2, "gri"}] = "robert griesemer"
req[mapKey{3, "adg"}] = "andrew gerrand"
req[mapKey{4, "rsc"}] = "russ cox"
var keys mapKeys
for k := range req {
keys = append(keys, k)
}
sort.Sort(keys)
for _, k := range keys {
fmt.Printf("short name : %s , long name : %s\n", k.Option, req[k])
}
}
type mapKey struct {
Key int
Option string
}
type mapKeys []mapKey
func (mk mapKeys) Len() int { return len(mk) }
func (mk mapKeys) Swap(i, j int) { mk[i], mk[j] = mk[j], mk[i] }
func (mk mapKeys) Less(i, j int) bool { return mk[i].Key < mk[j].Key }
<kbd>play</kbd>
Keep in mind that if your mapKey struct have a field that doesn't support equality (aka a struct or a slice) using req[k] won't work.
In that case you can switch to type mapKeys []*mapKey and map[*mapKey]string.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论