结构体类型嵌入字段访问

huangapple
go评论114阅读模式
英文:

struct type embedded fields access

问题

我正在学习golang，目前我正在尝试理解指针。我定义了三个结构体类型：

  1. type Engine struct {
  2.     power int
  3. }
  5. type Tires struct {
  6.     number int
  7. }
  9. type Cars struct {
  10.     *Engine
  11.     Tires
  12. }

如你所见，在Cars结构体中，我定义了一个嵌入类型指针*Engine。看一下main函数：

  1. func main() {
  2.     car := new(Cars)
  3.     car.number = 4
  4.     car.power = 342
  5.     fmt.Println(car)
  6. }

当我尝试编译时，我得到了以下错误：

  1. panic: runtime error: invalid memory address or nil pointer dereference
  2. [signal 0xb code=0x1 addr=0x0 pc=0x23bb]

如何访问power字段呢？

英文:

I am trying to learn golang and at moment I am trying to understand pointers. I defined three struct type

  1. type Engine struct {      
  2. 	power int             
  3. }                         
  4.                           
  5. type Tires struct {       
  6. 	number int            
  7. }                         
  8.                           
  9.                           
  10. type Cars struct {           
  11. 	*Engine               
  12. 	Tires                 
  13. }

As you can see, in the Cars struct I defined an embedded type pointer *Engine. Look in the main.

  1. func main() {                    
  2.                                      
  3. 	car := new(Cars)             
  4. 	car.number = 4               
  5.     car.power = 342     
  6. 	fmt.Println(car)            
  7. }

When I try to compile, I've got the following errors

  1. panic: runtime error: invalid memory address or nil pointer dereference
  2. [signal 0xb code=0x1 addr=0x0 pc=0x23bb]

How can I access the power field?

答案1

得分: 10

例如，

  1. package main
  3. import "fmt"
  5. type Engine struct {
  6.     power int
  7. }
  9. type Tires struct {
  10.     number int
  11. }
  13. type Cars struct {
  14.     *Engine
  15.     Tires
  16. }
  18. func main() {
  19.     car := new(Cars)
  20.     car.Engine = new(Engine)
  21.     car.power = 342
  22.     car.number = 4
  23.     fmt.Println(car)
  24.     fmt.Println(car.Engine, car.power)
  25.     fmt.Println(car.Tires, car.number)
  26. }

输出：

  1. &{0x10328100 {4}}
  2. &{342} 342
  3. {4} 4

未经限定的类型名称 Engine 和 Tires 充当了各自匿名字段的字段名。

Go编程语言规范

结构体类型

声明了类型但没有显式字段名的字段是匿名字段，也称为嵌入字段或将类型嵌入结构体中。嵌入类型必须指定为类型名 T 或非接口类型名 *T 的指针，并且 T 本身不能是指针类型。未经限定的类型名称充当字段名。

英文:

For example,

  1. package main
  3. import "fmt"
  5. type Engine struct {
  6. 	power int
  7. }
  9. type Tires struct {
  10. 	number int
  11. }
  13. type Cars struct {
  14. 	*Engine
  15. 	Tires
  16. }
  18. func main() {
  19. 	car := new(Cars)
  20. 	car.Engine = new(Engine)
  21. 	car.power = 342
  22. 	car.number = 4
  23. 	fmt.Println(car)
  24. 	fmt.Println(car.Engine, car.power)
  25. 	fmt.Println(car.Tires, car.number)
  26. }

Output:

  1. &{0x10328100 {4}}
  2. &{342} 342
  3. {4} 4

The unqualified type names Engine and Tires act as the field names of the respective anonymous fields.

> The Go Programming Language Specification
>
> Struct types
>
> A field declared with a type but no explicit field name is an
> anonymous field, also called an embedded field or an embedding of the
> type in the struct. An embedded type must be specified as a type name
> T or as a pointer to a non-interface type name *T, and T itself may
> not be a pointer type. The unqualified type name acts as the field
> name.

答案2

得分: 8

尝试这样做：

  1. type Engine struct {
  2.     power int
  3. }
  5. type Tires struct {
  6.     number int
  7. }
  9. type Cars struct {
  10.     *Engine
  11.     Tires
  12. }
  14. func main() {
  15.     car := Cars{&Engine{5}, Tires{10}}
  16.     fmt.Println(car.number)
  17.     fmt.Println(car.power)
  18. }

如果你想要一个指向Engine的指针，你必须将你的Car结构初始化为：

  1. car := Cars{&Engine{5}, Tires{10}}
  2. fmt.Println(car.number)
  3. fmt.Println(car.power)

你可以在这里查看代码运行的示例：http://play.golang.org/p/_4UFFB7OVI

英文:

Try this:

  1. type Engine struct {      
  2.     power int             
  3. }                         
  5. type Tires struct {       
  6.     number int            
  7. }                         
  10. type Cars struct {           
  11.     Engine               
  12.     Tires                 
  13. }

and than:

  1. car := Cars{Engine{5}, Tires{10}}
  2. fmt.Println(car.number)
  3. fmt.Println(car.power)

http://play.golang.org/p/_4UFFB7OVI

If you want a pointer to the Engine, you must initialize your Car structure as:

  1. car := Cars{&Engine{5}, Tires{10}}
  2. fmt.Println(car.number)
  3. fmt.Println(car.power)

答案3

得分: 0

另一个例子，当字段名不唯一时

  1. package embeded
  3. import "fmt"
  5. type Engine struct {
  6.     id    int
  7.     power int
  8. }
  10. type Tires struct {
  11.     id     int
  12.     number int
  13. }
  15. type Cars struct {
  16.     id int
  17.     Engine
  18.     Tires
  19. }
  21. func Embed() Cars {
  22.     car := Cars{
  23.         id:     3,
  24.         Engine: Engine{id: 1, power: 5},
  25.         Tires:  Tires{id: 2, number: 10},
  26.     }
  27.     fmt.Println(car.number)
  28.     fmt.Println(car.power)
  29.     fmt.Println(car.id)
  30.     fmt.Println(car.Engine.id)
  31.     fmt.Println(car.Tires.id)
  32.     return car
  33. }
英文:

One more example, in case of not unique field names

  1. package embeded
  3. import "fmt"
  5. type Engine struct {
  6. 	id    int
  7. 	power int
  8. }
  10. type Tires struct {
  11. 	id     int
  12. 	number int
  13. }
  15. type Cars struct {
  16. 	id int
  17. 	Engine
  18. 	Tires
  19. }
  21. func Embed() Cars {
  22. 	car := Cars{
  23. 		id:     3,
  24. 		Engine: Engine{id: 1, power: 5},
  25. 		Tires:  Tires{id: 2, number: 10},
  26. 	}
  27. 	fmt.Println(car.number)
  28. 	fmt.Println(car.power)
  29. 	fmt.Println(car.id)
  30. 	fmt.Println(car.Engine.id)
  31. 	fmt.Println(car.Tires.id)
  32. 	return car
  33. }

huangapple
  • 本文由 发表于 2014年7月21日 02:30:20
  • 转载请务必保留本文链接：https://go.coder-hub.com/24853604.html
  • go
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