在函数参数中不使用*structobject将结构体按引用传递

huangapple
go评论113阅读模式
英文:

Passing a struct by reference without passing as a *structobject in the function parameters

问题

我需要将一个结构体作为引用传递给一个函数，但是我不能在函数参数中使用指针（*）语法：

  1. func abc(client cleintpassedasrefernce) {
  2.     // 在client中进行更改，以便在client的实际内存位置中更新
  3. }
  5. type clienttobepassasref struct {
  6.     data int
  7.     name string
  8. }
  10. // 调用abc函数
  11. object := clienttobepassasref{}
  12. abc(object)

也许我可以创建另一个结构体，像这样传递对象：

  1. type objectPassedStruct struct {
  2.     object *clienttobepassasref
  3. }

然后在函数中稍后访问它？

英文:

I need to pass a struct as a reference to a function; however, I cannot use pointer (*) syntax in my function parameters:

  1. func abc(client cleintpassedasrefernce) {
  2.     // make changes in client so that it will be updated in the client actual memory location
  3. }
  6. type clienttobepassasref struct {
  7.     data int
  8.     name string
  9. } 
  11. // calling abc function 
  12. object = clienttobepassasref{}
  13. abc(object)

Maybe I can create another struct like type object passed struct

  1. {
  2.     object *clienttobepassasref
  3. }

and pass this object and access it later in the function?

答案1

得分: 1

你需要同时更改函数签名和调用函数的方式。此外，你定义了一个名为clienttobepassasref的结构体类型，但是你的函数接受的是一个名为cleintpassedasrefernce的结构体类型，在你的示例中没有定义。我在我的示例中将它们替换为一个名为MyObject的结构体类型。

  1. package main
  3. import (
  4. 	"fmt"
  5. )
  7. type MyObject struct {
  8. 	data int
  9. 	name string
  10. }
  12. // 在这里接受一个指向对象的指针，使用'*'字符
  13. func abc(obj *MyObject) {
  14. 	// 在客户端中进行更改，以便在客户端的实际内存位置中更新
  15. 	obj.name = "foo"
  16. }
  18. func main() {
  19. 	// 调用abc函数
  20. 	object := MyObject{}
  21. 	abc(&object)
  22. 	// 在这里通过引用传递，使用'&'字符
  23. 	fmt.Println(object.name)
  24. }

go playground

英文:

You need to change both the function signature, and the method by which you're calling the function. Also you're defining a struct type called clienttobepassasref but your function is accepting a struct type of cleintpassedasrefernce which is not defined in your example. I replaced these with a struct type in my example called MyObject.

go playground

  1. package main
  3. import (
  4. 	"fmt"
  5. )
  7. type MyObject struct {
  8. 	data int
  9. 	name string
  10. }
  12. // accept a pointer to the object here using the '*' character
  13. func abc(obj *MyObject) {
  14. 	// make changes in client so that it will be updated in the client actual memory location
  15. 	obj.name = "foo"
  16. }
  18. func main() {
  19. 	// calling abc function
  20. 	object := MyObject{}
  21. 	abc(&object)
  22. 	// Pass it by reference here with the '&' character
  23. 	fmt.Println(object.name)
  24. }

答案2

得分: 1

如果您需要更新或更改结构的属性，可以创建一个带有该结构引用的接收器函数（方法），并在该函数（方法）中更新结构的属性。

  1. package main
  3. import "fmt"
  5. type MyObject struct {
  6. 	data int
  7. 	name string
  8. }
  10. func (obj *MyObject) abc() {
  11. 	// 在客户端进行更改，以便在客户端的实际内存位置上进行更新
  12. 	obj.name = "some name"
  13. 	obj.data = 10
  14. }
  16. func main() {
  17. 	// 调用 abc 函数
  18. 	object := &MyObject{}
  19. 	object.abc()
  21. 	fmt.Printf("object = %+v", *object)
  22. 	//object = {data:10 name:some name}
  23. }

在此处运行代码：链接

参考获取更多知识：

英文:

If you need to update, change struct's properties, then create a function (method) with receiver of that struct reference and update struct's properties in that function (method).

  1. package main
  3. import "fmt"
  5. type MyObject struct {
  6. 	data int
  7. 	name string
  8. }
  10. func (obj *MyObject)abc() {
  11. 	// make changes in client so that it will be updated in the client actual memory location
  12. 	obj.name = `some name`
  13. 	obj.data = 10
  14. }
  16. func main() {
  17. 	// calling abc function
  18. 	object := &MyObject{}
  19. 	object.abc()
  21. 	fmt.Printf("object = %+v", *object)
  22. 	//object = {data:10 name:some name}
  23. }

run the code here

reference to get some more knowledge

huangapple
  • 本文由 发表于 2021年7月6日 01:54:07
  • 转载请务必保留本文链接：https://go.coder-hub.com/68260302.html
  • go
  • pointers
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