我是一个新的人。现在我对functoin传递变量有疑问。这是代码:
type User struct {
Name string
Map map[string]string
}
func main() {
u := User{Name: "Leto"}
u.Map = make(map[string]string)
fmt.Println("before --------")
fmt.Println(unsafe.Pointer(&u))
fmt.Println(unsafe.Pointer(&(u.Map)))
fmt.Println(u)
Modify(u)
fmt.Println("after --------")
fmt.Println(u)
}
func Modify(u User) {
fmt.Println("in func --------")
fmt.Println(unsafe.Pointer(&u))
fmt.Println(unsafe.Pointer(&(u.Map)))
u.Name = "Paul"
u.Map["t"] = "t"
}
输出上面的代码:
before --------
0xc04203a4c0
0xc04203a4d0
{Leto map[]}
in func --------
0xc04203a500
0xc04203a510
after --------
{Leto map[t:t]}
在修改功能我知道用户是副本,所以更改名称不起作用,但为什么要更改地图效果用户结构?
答案 0 :(得分:0)
答案 1 :(得分:0)
我们需要了解每次调用中内存分配的工作方式:
u := User{Name: "Leto"}
// u is an object of type User
// after this line memory has been allocated to both the
// properties u.Name(string) and u.Map(reference)
// lets say allocated memory address for u.Name starts with x
// for u.Map it starts with y, and note that u.Map is a reference i.e.
// the value contained in it will be a different memory address which
// will be the starting memory address of the actual map
// right now the value written at y is nil since it
// does not point to any memory address
u.Map = make(map[string]string)
// now value of y has been updated to z (which is the
// memory address of the beginning of the map initialized
// with make call)
fmt.Println("before --------")
fmt.Println(unsafe.Pointer(&u))
fmt.Println(unsafe.Pointer(&(u.Map)))
fmt.Println(u)
// here you are about to pass object by value
// so basically a new object will be created of type User
// lets talk about how copy of this object will be created
// copy of u.Name will have a new address
// lets call it x1 and the value "Leto" too will be
// copied to memory address starting with x1
// copy of u.Map will have a new address too lets call it
// y1 and its value z will be copied too to the memory address y1
// I think you must have got your answer by now.
// Basically the newly copied object's property u.Map and
// the old one's u.Map both points to the same memory address "z"
// and hence whosoever updates the map the other one will see it
Modify(u)
fmt.Println("after --------")
fmt.Println(u)
答案 2 :(得分:-1)
切片,贴图和通道是参考类型。所以他们总是通过参考传递。