在Golang中解引用地图索引。

huangapple
go评论193阅读模式
英文:

Dereferencing a map index in Golang

问题

我正在学习Go语言，目前我制作了这个简单而粗糙的库存程序，只是为了玩弄结构体和方法，以理解它们的工作原理。在驱动文件中，我尝试从Cashier类型的items映射中调用item类型的方法。我的方法使用指针接收器直接使用结构体，而不是创建副本。当我运行程序时，我得到了这个错误.\driver.go:11: cannot call pointer method on f[0] .\driver.go:11: cannot take the address of f[0]。

Inventory.go:

  1. package inventory
  3. type item struct{
  4.     itemName string
  5.     amount int
  6. }
  8. type Cashier struct{
  9.     items map[int]item
  10.     cash int
  11. }
  13. func (c *Cashier) Buy(itemNum int){
  14.     item, pass := c.items[itemNum]
  16.     if pass{
  17.         if item.amount == 1{
  18.             delete(c.items, itemNum)
  19.         } else{
  20.             item.amount--
  21.             c.items[itemNum] = item 
  22.         }
  23.         c.cash++
  24.     }
  25. }
  27. func (c *Cashier) AddItem(name string, amount int){
  28.     if c.items == nil{
  29.         c.items = make(map[int]item)
  30.     }
  31.     temp := item{name, amount}
  32.     index := len(c.items)
  33.     c.items[index] = temp
  34. }
  36. func (c *Cashier) GetItems() map[int]item{
  37.     return c.items;
  38. }
  40. func (i *item) GetName() string{
  41.     return i.itemName
  42. }
  44. func (i *item) GetAmount() int{
  45.     return i.amount
  46. }

Driver.go:

  1. package main
  3. import "fmt"
  4. import "inventory"
  6. func main() {
  7.     x := inventory.Cashier{}
  8.     x.AddItem("item1", 13)
  9.     f := x.GetItems()
  11.     fmt.Println(f[0].GetAmount())
  12. }

与我的问题相关的代码部分是inventory.go中的GetAmount函数和driver.go中的打印语句。

英文:

I'm learning Go currently and I made this simple and crude inventory program just to tinker with structs and methods to understand how they work. In the driver file I try to call a method from and item type from the items map of the Cashier type. My method have pointer reciever to use the structs directly instead of making copies. When I run the program I get this error .\driver.go:11: cannot call pointer method on f[0]
.\driver.go:11: cannot take the address of f[0]

Inventory.go:

  1. package inventory
  4. type item struct{
  5.     itemName string
  6.     amount int
  7. }
  9. type Cashier struct{
  10.     items map[int]item
  11.     cash int
  12. }
  14. func (c *Cashier) Buy(itemNum int){
  15.     item, pass := c.items[itemNum]
  17.     if pass{
  18. 	    if item.amount == 1{
  19. 		    delete(c.items, itemNum)
  20. 	    } else{
  21. 		    item.amount--
  22. 		    c.items[itemNum] = item 
  23. 	    }
  24. 	    c.cash++
  25.     }
  26. }
  29. func (c *Cashier) AddItem(name string, amount int){
  30.     if c.items == nil{
  31. 	    c.items = make(map[int]item)
  32.     }
  33.     temp := item{name, amount}
  34.     index := len(c.items)
  35.     c.items[index] = temp
  36. }
  38. func (c *Cashier) GetItems() map[int]item{
  39.     return c.items;
  40. }
  42. func (i *item) GetName() string{
  43.     return i.itemName
  44. }
  46. func (i *item) GetAmount() int{
  47.     return i.amount
  48. }

Driver.go:

  1. package main
  3. import "fmt"
  4. import "inventory"
  6. func main() {
  7.     x := inventory.Cashier{}
  8.     x.AddItem("item1", 13)
  9.     f := x.GetItems()
  11.     fmt.Println(f[0].GetAmount())
  12. }

The part of the code that really pertains to my problem is the GetAmount function in inventory.go and print statement in the driver.go

答案1

得分: 23

地图条目无法被寻址（因为在地图的增长/缩小过程中，它们的地址可能会发生变化），因此无法在它们上面调用指针接收器方法。

详细信息请参阅：https://groups.google.com/forum/?fromgroups=#!topic/golang-nuts/4_pabWnsMp0

英文:

A map entry cannot be addressed (as its address might change during map growth/shrink), so you cannot call pointer receiver methods on them.

Detail here: https://groups.google.com/forum/?fromgroups=#!topic/golang-nuts/4_pabWnsMp0

答案2

得分: 20

如Volker在他的回答中所说，你不能获取映射中项目的地址。你应该存储指向映射中项目的指针，而不是存储项目的值：

  1. package main
  3. import "fmt"
  5. type item struct {
  6. 	itemName string
  7. 	amount   int
  8. }
  10. type Cashier struct {
  11. 	items map[int]*item
  12. 	cash  int
  13. }
  15. func (c *Cashier) Buy(itemNum int) {
  16. 	item, pass := c.items[itemNum]
  18. 	if pass {
  19. 		if item.amount == 1 {
  20. 			delete(c.items, itemNum)
  21. 		} else {
  22. 			item.amount--
  23. 		}
  24. 		c.cash++
  25. 	}
  26. }
  28. func (c *Cashier) AddItem(name string, amount int) {
  29. 	if c.items == nil {
  30. 		c.items = make(map[int]*item)
  31. 	}
  32. 	temp := &item{name, amount}
  33. 	index := len(c.items)
  34. 	c.items[index] = temp
  35. }
  37. func (c *Cashier) GetItems() map[int]*item {
  38. 	return c.items
  39. }
  41. func (i *item) GetName() string {
  42. 	return i.itemName
  43. }
  45. func (i *item) GetAmount() int {
  46. 	return i.amount
  47. }
  49. func main() {
  50. 	x := Cashier{}
  51. 	x.AddItem("item1", 13)
  52. 	f := x.GetItems()
  53. 	fmt.Println(f[0].GetAmount()) // 13
  54. 	x.Buy(0)
  55. 	f = x.GetItems()
  56. 	fmt.Println(f[0].GetAmount()) // 12
  57. }

你可以在这里运行代码：http://play.golang.org/p/HkIg668fjN

英文:

As Volker said in his answer - you can't get address of an item in the map. What you should do - is to store pointers to items in your map, instead of storing item values:

  1. package main
  2. import "fmt"
  3. type item struct {
  4. itemName string
  5. amount   int
  6. }
  7. type Cashier struct {
  8. items map[int]*item
  9. cash  int
  10. }
  11. func (c *Cashier) Buy(itemNum int) {
  12. item, pass := c.items[itemNum]
  13. if pass {
  14. if item.amount == 1 {
  15. delete(c.items, itemNum)
  16. } else {
  17. item.amount--
  18. }
  19. c.cash++
  20. }
  21. }
  22. func (c *Cashier) AddItem(name string, amount int) {
  23. if c.items == nil {
  24. c.items = make(map[int]*item)
  25. }
  26. temp := &item{name, amount}
  27. index := len(c.items)
  28. c.items[index] = temp
  29. }
  30. func (c *Cashier) GetItems() map[int]*item {
  31. return c.items
  32. }
  33. func (i *item) GetName() string {
  34. return i.itemName
  35. }
  36. func (i *item) GetAmount() int {
  37. return i.amount
  38. }
  39. func main() {
  40. x := Cashier{}
  41. x.AddItem("item1", 13)
  42. f := x.GetItems()
  43. fmt.Println(f[0].GetAmount()) // 13
  44. x.Buy(0)
  45. f = x.GetItems()
  46. fmt.Println(f[0].GetAmount()) // 12
  47. }

http://play.golang.org/p/HkIg668fjN

答案3

得分: 3

虽然其他答案很有用，但我认为在这种情况下最好的做法是使非变异函数不接受指针：

  1. func (i item) GetName() string{
  2.     return i.itemName
  3. }
  5. func (i item) GetAmount() int{
  6.     return i.amount
  7. }
英文:

While the other answers are useful, I think in this case it is best just to make non-mutating functions not take a pointer:

  1. func (i item) GetName() string{
  2. return i.itemName
  3. }
  4. func (i item) GetAmount() int{
  5. return i.amount
  6. }

huangapple
  • 本文由 发表于 2013年11月27日 01:55:31
  • 转载请务必保留本文链接：https://go.coder-hub.com/20224478.html
  • go
  • pointers
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