英文:
golang another func change map of a struct
问题
我是你的中文翻译助手,以下是翻译好的内容:
我是一个新手,现在有一个关于函数传递变量的问题。以下是代码:
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"
}
上述代码的输出结果为:
before --------
0xc04203a4c0
0xc04203a4d0
{Leto map[]}
in func --------
0xc04203a500
0xc04203a510
after --------
{Leto map[t:t]}
在Modify
函数中,我知道User
是一个副本,所以更改Name
不起作用是可以理解的,但为什么更改Map
会影响到外部的User
结构呢?
英文:
I'm a new one on go.Now I have a question about functoin pass variable.Here is the code:
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"
}
code above output:
before --------
0xc04203a4c0
0xc04203a4d0
{Leto map[]}
in func --------
0xc04203a500
0xc04203a510
after --------
{Leto map[t:t]}
in Modify func i know user is a copy ,so change name not work is ok ,but why change Map effect out user struct ?
答案1
得分: 0
我们需要了解每次调用时内存分配的工作原理:
u := User{Name: "Leto"}
// u 是 User 类型的对象
// 在这行代码之后,内存已经分配给了 u.Name(字符串类型)和 u.Map(引用类型)的属性
// 假设 u.Name 的分配内存地址以 x 开头
// u.Map 的分配内存地址以 y 开头,并且请注意,u.Map 是一个引用,即它所包含的值将是一个不同的内存地址,该地址将是实际 map 的起始内存地址
// 此时,y 处的值为 nil,因为它没有指向任何内存地址
u.Map = make(map[string]string)
// 现在,y 的值已更新为 z(这是使用 make 调用初始化的 map 的起始内存地址)
fmt.Println("before --------")
fmt.Println(unsafe.Pointer(&u))
fmt.Println(unsafe.Pointer(&(u.Map)))
fmt.Println(u)
// 在这里,您将通过值传递对象
// 因此,基本上会创建一个新的 User 类型的对象副本
// u.Name 的副本将有一个新的地址
// 让我们称之为 x1,并且值 "Leto" 也将被复制到以 x1 开头的内存地址
// u.Map 的副本也将有一个新的地址,让我们称之为 y1,并且它的值 z 也将被复制到以 y1 开头的内存地址
// 我想您现在已经得到了答案。
// 基本上,新复制的对象的属性 u.Map 和旧对象的 u.Map 都指向相同的内存地址 "z",因此无论哪个对象更新了 map,另一个对象都会看到它
Modify(u)
fmt.Println("after --------")
fmt.Println(u)
英文:
We need to understand how memory allocation is working here in each call:
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
你应该在赋值和修改操作中分别使用 &User 和 *User。
请查看此链接:https://play.golang.org/p/tDh1JBpK-t
使用指向结构体的指针总是更好的选择。
英文:
You should use &User and *User respectively in your assignment and modify operations.
Check this https://play.golang.org/p/tDh1JBpK-t
Its always better to work with pointers to structs.
答案3
得分: -3
切片(Slices)、映射(Maps)和通道(Channels)都是引用类型,因此它们总是通过引用传递。
英文:
Slices, maps and channels are reference types. so they are always passed by reference.
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