Golang基础——数据类型:结构体

结构体:

type student struct {
	name string
	age  int
}
type teacher struct {
	name string
}
func main(){
    s1 := student{"aaa", 1} // 必须全部赋值
    s2 := student{name: "bbb", age: 1}
    var s3 student // 地址为第一个成员的地址
    s3.name = "ccc"
    // s3.age = 1  // 可以只赋值一个,其他是各个类型的零值
    fmt.Println(s1.name)
    fmt.Println(s2)
    fmt.Println(s3)
    fmt.Println(s1 == s2) // false,各个属性必须相同
    // s4 := teacher{"aaad"}
    // fmt.Println(s1 == s4)  // invalid operation: s1 == s4 (mismatched types student and teacher)
    var s5 student
    s5 = s1 // 同类型可以赋值
    fmt.Println(s5)
    var s_list [3]student
    // append(s_list, s1)  // first argument to append must be slice; have [3]student
    s_list[0] = s1
    fmt.Println(s_list)
    var a_l = [...]student{s1, s2}
    fmt.Println(a_l)
    // var a [...]int // 不行
}

结构体是值传递:map是引用传递,作为map的值,改变是因为map是引用传递

type student struct {
	name string
	age  int
}
type teacher struct {
	name string
}

func good(m map[int]student) {
	// m[1].name = "eeeee"  /*cannot assign to struct field m[1].name in map",*/
	stu1 := m[1]
	stu1.name = "eeeee"
	m[1] = stu1
	fmt.Println(m)
}
func good2(s student) {
	s.name = "eeeee"
	fmt.Println(s)
}
func main() {
	// var map1 map[int]student     // 这种初始化方式会报错误
	// map1[1] = student{"kkk", 1}  // panic: assignment to entry in nil map
	map1 := make(map[int]student)
	map1[1] = student{name: "aaa", age: 1}
	good(map1)
	fmt.Println(map1)
	/*结果:作为map的值是引用传递,map是引用传递
	map[1:{eeeee 1}]
	map[1:{eeeee 1}]
	*/
	s1 := student{"aaa", 1}
	good2(s1)
	fmt.Println(s1)
	/*结果:结构体是值传递
	{eeeee 1}
	{aaa 1}
	*/
}

posted @ 2021-03-30 21:23  pythoner_wl  阅读(56)  评论(0编辑  收藏  举报