Golangmap实现原理浅析

func main(){
  // 第一种
  var a map[string]string
  // 在使用map前，需要先make，make的作用就是给map分配数据空间
  a = make(map[string]string,10)
  a["no1"] = "宋江"
  a["no2"] = "吴用"
  a["no3"] = "武松"
  // 第二种方式
  cities := make(map[string]string)
  cities["no1"] = "北京"
  cities["no2"] = "天津"
  cities["no3"] = "上海"
  fmt.Pringln(cities)
  // 第三种方式
  heroes := map[string]string{
	"hero1":"宋江",
	"hero2":"卢俊义",
	"hero3":"吴用",
  }
  fmt.Println("heroes=",heroes)
}

studentMap := make(map[string]map[string]string)
studentMap["stu01"] = make(map[string]string,3)
studentMap["stu01"]["name"] = "tom"
studentMap["stu01"]["sex"] = "男"
studentMap["stu01"]["address"] = "北京长安街~"
studentMap["stu02"] = make(map[string]string,3)
studentMap["stu02"]["name"] = "mary"
studentMap["stu02"]["sex"] = "女"
studentMap["stu02"]["address"] = "上海~"
fmt.Println(studentMap)
fmt.Println(studentMap["stu02"])
fmt.Println(studentMap["stu02"]["address"])

delete(cities,"no1")
fmt.Println(cities) // 当delete指定的key不存在时，删除不会操作，也不会报错
// 如果希望一次性删除所有的key
// 1.遍历所有的key，如何逐一删除【遍历】
//2.直接make一个新的空间
cities = make(map[string]string)
fmt.Println(cities)

// 演示map的查找
val,ok := cities["no2"]
if ok{
	fmt.Printf("有no1 key 值为%v\n",val)
}else{
	fmt.Printf("没有no1 key \n")
}

// 使用for-range遍历一个结构比较复杂的map
studentMap := make(map[string]map[string]string)
studentMap["stu01"] = make(map[string]string,3)
studentMap["stu01"]["name"] = "tom"
studentMap["stu01"]["sex"] = "男"
studentMap["stu01"]["address"] = "北京长安街~"
studentMap["stu02"] = make(map[string]string,3)
studentMap["stu02"]["name"] = "mary"
studentMap["stu02"]["sex"] = "女"
studentMap["stu02"]["address"] = "上海~"
for k1,v1 := range studentMap{
	fmt.Println("k1=",k1)
	for k2,v2 := range v1{
		fmt.Printf("\t k2=%v v2=%v\n",k2,v2)
	}
	fmt.Println()
}

// 声明一个map切片
var monsters []map[string]string
monsters = make([]map[string]string,2) // 准备放入两个妖怪
// 增加第一个妖怪的信息
if monsters[0] == nil{
   monsters[0] = make(map[string]string,2)
   monsters[0]["name"] = "牛魔王"
   monsters[0]["age"] = "500"
}
if monsters[1] == nil{
   monsters[1] = make(map[string]string,2)
   monsters[1]["name"] = "玉兔精"
   monsters[1]["age"] = "400"
}
// 下面这个写法越界
//if monsters[2] == nil{
//   monsters[2] = make(map[string]string,2)
 //  monsters[2]["name"] = "狐狸精"
 //  monsters[2]["age"] = "300"
//}
//先定义一个monsters信息 可以动态添加monster，append函数
newMonster := map[string]string{
   "name":"新的妖怪~火云邪神",
   "age":"200"
}
monsters = append(monsters,newMonster )
fmt.Println(monsters)

func main(){
	// map 的排序
	map1 := make(map[int]int,10)
	map1[10] = 100
	map1[1] = 13
	map1[4] = 56
	map1[8] = 90
	fmt.Println(map1)
	// 如果按照map的key的顺序进行排序输出
	// 1.先将map的key放入到切片中
	// 2.对切片排序
	// 3.遍历切片，然后按照key来输出map的值
	var keys []int
	for k,_ :=range map1{
		keys = append(keys,k)
	}
	// 排序
	sort.Ints(keys)
	fmt.Println(keys)
	for _,k := range keys{
		fmt.Printf("map1[%v]=%v \n",k,map1[k])
	}
}

func modify(map1 map[int]int){
   map1[10] = 900
}
func main(){
	// map是引用类型，遵守引用类型传递的机制，在一个函数接收map
	// 修改后，会直接修改原来的map
	map1 := make(map[int]int)
	map1[1] = 90
	map1[2] = 88
	map1[10] = 1
	map1[20] = 2
	modify(map1)
	// 看看结果，map1[10] = 900,说明map是引用类型
	fmt.Println(map1)
}

func modifyUser(users map[string]map[string]string,name string){
	// 判断users中是否有name
	if users[name] != nil{
		// 有这个用户
		users[name]["pwd"] = "888888"
	}else{
		// 没有这个用户
		users[name] = make(map[string]string,2)
		users[name]["pwd"] = "888888"
		users[name]["nickname"] = "昵称~"+name // 演示
	}
}
func main(){
	users := make(map[string]map[string]string,10)
	users["smith"] = make(map[string]string,2)
	users["smith"]["pwd"] = "999999"
	users["smith"]["nickname"] = "小花猫"
	modifyUser(users,"tom")
	modifyUser(users,"mary")
	modifyUser(users,"smith")
	fmt.Println(users)
}