go语言数组及结构体继承和初始化示例解析

var a [10]int

 var a [10]int
 fmt.Println(len(a)) //10

var n int = 10
var a [n]int

package main
import "fmt"
func main(){

	// 1.定义结构体
	type Student struct{
		id int
		name string
		age int
		addr string
}
	//2.结构体初始化1，值的顺序与结构体成员的顺序保持一致
	chary := Student{1,"chary",18,"上海"}
	fmt.Println(chary) //{1 chary 18 上海}
	//2.结构体初始化2，没有初始化的成员自动赋值0
	jeff := Student{name:"jeff"}  //{0 jeff 0 }
	fmt.Println(jeff)
	//3.成员使用
	chary.id = 100  //修改
	fmt.Println(chary.id )
	//4.结构体比较,两个结构体可以使用 == 或 != 运算符进行比较，但不支持 > 或 <。
	fmt.Println(jeff==chary) //false
	//5.同类型的两个结构体变量可以相互赋值。
	var user Student
	user = jeff
	fmt.Println(user) //{0 jeff 0 }
}

//人
type Person struct {
    name string
    sex byte
    age int
}
//学生
type Student struct {
    Person    //匿名字段，那么默认Student就包含了Person的所有字段
    id int
    addr string
}

//人
type Person struct {
    name string
    sex byte
    age int
}
//学生
type Student struct {
    Person//匿名字段，那么默认Student就包含了Person的所有字段
    id int
    addr string
}
func main() {
    //顺序初始化
    s1 := Student{Person{"mike",'m',18},1,"sz"}
    //s1 = {Person:{name:mike sex:109 age:18}id:1 addr:sz}
    fmt.Printf("s1=%+v\n",s1)
    //s2 := Student{"mike",'m',18,1,"sz"}//err
    //部分成员初始化1
    s3 := Student{Person:Person{"lily",'f',19},id:2}
    //s3 = {Person:{name:lily sex:102 age:19}id:2 addr:}
    fmt.Printf("s3=%+v\n",s3)
    //部分成员初始化2
    s4 := Student{Person:Person{name:"tom"},id:3}
    //s4 = {Person:{name:tomsex:0age:0}id:3addr:}
    fmt.Printf("s4=%+v\n",s4)
}

var s1 Student//变量声明
//给成员赋值
s1.name = "mike"//等价于s1.Person.name="mike"
s1.sex = 'm'
s1.age = 18
s1.id = 1
s1.addr = "sz"
fmt.Println(s1)    //{{mike 109 18}1 sz}
var s2 Student//变量声明
s2.Person = Person{"lily",'f',19}
s2.id = 2
s2.addr = "bj"
fmt.Println(s2)    //{{lily 102 19}2 bj}

//人
type Person struct{
    name string
    sex byte
    age int
}
//学生
type Student struct{
    Person    //匿名字段，那么默认Student就包含了Person的所有字段
    id int
    addr string
    name string    //和Person中的name同名
}
func main(){
    var s Student//变量声明
    //给Student的name，还是给Person赋值？
    s.name = "mike"
    //{Person:{name:sex:0age:0}id:0addr:name:mike}
    fmt.Printf("%+v\n",s)
    //默认只会给最外层的成员赋值
    //给匿名同名成员赋值，需要显示调用
    s.Person.name = "yoyo"
    //Person:{name:yoyosex:0age:0}id:0addr:name:mike}
    fmt.Printf("%+v\n",s)
}

type mystr string//自定义类型
type Person struct {
    name string
    sex byte
    age int
}
type Student struct {
    Person    //匿名字段，结构体类型
    int    //匿名字段，内置类型
    mystr    //匿名字段，自定义类型
}
func main() {
    //初始化
    s1 := Student{Person{"mike",'m',18},1,"bj"}
    //{Person:{name:mikesex:109age:18}int:1mystr:bj}
    fmt.Printf("%+v\n",s1)
    //成员的操作，打印结果：mike,m,18,1,bj
    fmt.Printf("%s,%c,%d,%d,%s\n",s1.name,s1.sex,s1.age,s1.int,s1.mystr)
}

type Person struct {    //人
    name string
    sex byte
    age int
}
type Student struct {//学生
    *Person    //匿名字段，结构体指针类型
    id int
    addr string
}
func main() {
    //初始化
    s1 := Student{&Person{"mike",'m',18},1,"bj"}
    //{Person:0xc0420023e0id:1addr:bj}
    fmt.Printf("%+v\n",s1)
    //mike,m,18
    fmt.Printf("%s,%c,%d\n",s1.name,s1.sex,s1.age)
    //声明变量
    var s2 Student
    s2.Person = new(Person)//分配空间
    s2.name = "yoyo"
    s2.sex = 'f'
    s2.age = 20
    s2.id = 2
    s2.addr = "sz"
    //yoyo10220220
    fmt.Println(s2.name,s2.sex,s2.age,s2.id,s2.age)
}

package main
import (
	"fmt"
	"sync"
)
type UserAges struct {
	ages map[string] int
	sync.Mutex
}
func (u *UserAges)Add(name string,age int)  {
	u.Lock()
	defer u.Unlock()
	u.ages[name] = age
}
func (u *UserAges)Get(name string)int{
	if age,ok:=u.ages[name];ok{
		return age
	}
	return -1
}
func main() {
	dic:=make(map[string]int)
	dic["age"] = 18
	r:=UserAges{ages: dic}
	r.Add("jeff",20)
	fmt.Println(r)
	age:=r.Get("age")
	fmt.Println(age)
}