英文:
slice of struct != slice of interface it implements?
问题
我有一个接口Model
,由结构体Person
实现。
为了获得一个模型实例,我有以下辅助函数:
func newModel(c string) Model {
switch c {
case "person":
return newPerson()
}
return nil
}
func newPerson() *Person {
return &Person{}
}
上述方法允许我返回一个正确类型的Person实例(可以使用相同的方法轻松添加新模型)。
当我尝试类似的方式返回一个模型切片时,我遇到了错误。代码如下:
func newModels(c string) []Model {
switch c {
case "person":
return newPersons()
}
return nil
}
func newPersons() *[]Person {
var models []Person
return &models
}
Go报错:cannot use newPersons() (type []Person) as type []Model in return argument
我的目标是返回一个请求的任何模型类型的切片(无论是[]Person
,[]FutureModel
,[]Terminator2000
等)。我漏掉了什么,如何正确实现这样的解决方案?
英文:
I have an interface Model
, which is implemented by struct Person
.
To get a model instance, I have the following helper functions:
func newModel(c string) Model {
switch c {
case "person":
return newPerson()
}
return nil
}
func newPerson() *Person {
return &Person{}
}
The above approach allows me to return a properly typed Person instance (can easily add new models later with same approach).
When I attempted to do something similar for returning a slice of models, I get an error. Code:
func newModels(c string) []Model {
switch c {
case "person":
return newPersons()
}
return nil
}
func newPersons() *[]Person {
var models []Person
return &models
}
Go complains with: cannot use newPersons() (type []Person) as type []Model in return argument
My goal is to return a slice of whatever model type is requested (whether []Person
, []FutureModel
, []Terminator2000
, w/e). What am I missing, and how can I properly implement such a solution?
答案1
得分: 129
这与我刚刚回答的一个问题非常相似:https://stackoverflow.com/a/12990540/727643
简短的答案是你是正确的。一个结构体切片与一个接口切片是不相等的,即使结构体实现了该接口。
[]Person
和[]Model
有不同的内存布局。这是因为它们所包含的类型具有不同的内存布局。Model
是一个接口值,这意味着在内存中它占用两个字长的空间。一个字长用于类型信息,另一个字长用于数据。Person
是一个结构体,其大小取决于它包含的字段。为了将[]Person
转换为[]Model
,你需要遍历数组并对每个元素进行类型转换。
由于这个转换是一个O(n)的操作,并且会创建一个新的切片,Go不会隐式地执行它。你可以使用以下代码显式地执行它。
models := make([]Model, len(persons))
for i, v := range persons {
models[i] = Model(v)
}
return models
正如dskinner指出的,你很可能想要一个指针切片而不是一个指向切片的指针。通常不需要指向切片的指针。
*[]Person // 指向切片的指针
[]*Person // 指针切片
英文:
This is very similar to a question I just answered: https://stackoverflow.com/a/12990540/727643
The short answer is that you are correct. A slice of structs is not equal to a slice of an interface the struct implements.
A []Person
and a []Model
have different memory layouts. This is because the types they are slices of have different memory layouts. A Model
is an interface value which means that in memory it is two words in size. One word for the type information, the other for the data. A Person
is a struct whose size depends on the fields it contains. In order to convert from a []Person
to a []Model
, you will need to loop over the array and do a type conversion for each element.
Since this conversion is an O(n) operation and would result in a new slice being created, Go refuses to do it implicitly. You can do it explicitly with the following code.
models := make([]Model, len(persons))
for i, v := range persons {
models[i] = Model(v)
}
return models
And as dskinner pointed out, you most likely want a slice of pointers and not a pointer to a slice. A pointer to a slice is not normally needed.
*[]Person // pointer to slice
[]*Person // slice of pointers
答案2
得分: 8
Maybe this is an issue with your return type *[]Person
, where it should actually be []*Person
so to reference that each index of the slice is a reference to a Person
, and where a slice []
is in itself a reference to an array.
Check out the following example:
package main
import (
"fmt"
)
type Model interface {
Name() string
}
type Person struct {}
func (p *Person) Name() string {
return "Me"
}
func NewPersons() (models []*Person) {
return models
}
func main() {
var p Model
p = new(Person)
fmt.Println(p.Name())
arr := NewPersons()
arr = append(arr, new(Person))
fmt.Println(arr[0].Name())
}
英文:
Maybe this is an issue with your return type *[]Person
, where it should actually be []*Person
so to reference that each index of the slice is a reference to a Person
, and where a slice []
is in itself a reference to an array.
Check out the following example:
package main
import (
"fmt"
)
type Model interface {
Name() string
}
type Person struct {}
func (p *Person) Name() string {
return "Me"
}
func NewPersons() (models []*Person) {
return models
}
func main() {
var p Model
p = new(Person)
fmt.Println(p.Name())
arr := NewPersons()
arr = append(arr, new(Person))
fmt.Println(arr[0].Name())
}
答案3
得分: 7
由于Stephen已经回答了这个问题,而你是一个初学者,我强调给出建议。
在处理go的接口时,更好的方式不是像其他语言(如Java)那样有一个返回接口的构造函数,而是为每个对象单独提供一个构造函数,因为它们隐式地实现了接口。
不要使用
newModel(type string) Model { ... }
而应该使用
newPerson() *Person { ... }
newPolitician() *Politician { ... }
其中Person
和Politician
都实现了Model
的方法。
你仍然可以在任何接受Model
的地方使用Person
或Politician
,但你也可以实现其他接口。
使用你的方法,你将被限制在Model
上,直到你手动转换为另一个接口类型。
假设我有一个实现了Walk()
方法的Person
和一个实现了ShowOff()
方法的Model
,以下代码将无法直接工作:
newModel("person").ShowOff()
newModel("person").Walk() // 无法编译,Model没有Walk方法
然而,以下代码可以工作:
newPerson().ShowOff()
newPerson().Walk()
英文:
As Stephen already answered the question and you're a beginner I emphasize on giving advises.
A better way of working with go's interfaces is not to have a constructor returning
the interface as you might be used to from other languages, like java, but to have
a constructor for each object independently, as they implement the interface implicitly.
Instead of
newModel(type string) Model { ... }
you should do
newPerson() *Person { ... }
newPolitician() *Politician { ... }
with Person
and Politician
both implementing the methods of Model
.
You can still use Person
or Politician
everywhere where a Model
is accepted, but you can also implement other interfaces.
With your method you would be limited to Model
until you do a manual conversion to
another interface type.
Suppose I have a Person
which implements the method Walk()
and a Model
implements ShowOff()
, the following would not work straight forward:
newModel("person").ShowOff()
newModel("person").Walk() // Does not compile, Model has no method Walk
However this would:
newPerson().ShowOff()
newPerson().Walk()
答案4
得分: 3
正如其他人已经回答的那样,[]T是一种独特的类型。我只想补充一点,可以使用一个简单的工具来通用地将它们转换。
import "reflect"
// 将特定类型的切片或数组转换为interface{}类型的数组
func ToIntf(s interface{}) []interface{} {
v := reflect.ValueOf(s)
// 没有必要进行检查,如果不是切片或数组,我们希望引发panic
intf := make([]interface{}, v.Len())
for i := 0; i < v.Len(); i++ {
intf[i] = v.Index(i).Interface()
}
return intf
}
现在,你可以像这样使用它:
ToIntf([]int{1,2,3})
英文:
As others have already answered, []T is a distinct type. I'd just like to add that a simple utility can be used to convert them generically.
import "reflect"
// Convert a slice or array of a specific type to array of interface{}
func ToIntf(s interface{}) []interface{} {
v := reflect.ValueOf(s)
// There is no need to check, we want to panic if it's not slice or array
intf := make([]interface{}, v.Len())
for i := 0; i < v.Len(); i++ {
intf[i] = v.Index(i).Interface()
}
return intf
}
Now, you can use it like this:
ToIntf([]int{1,2,3})
答案5
得分: 2
类型T和[]T是不同的类型,它们的方法也是不同的,即使它们满足相同的接口。换句话说,每个满足Model接口的类型必须自己实现Model的所有方法 - 方法的接收者只能是一个特定的类型。
英文:
Types T and []T are distinct types and distinct are their methods as well, even when satisfying the same interface. IOW, every type satisfying Model must implement all of the Model's methods by itself - the method receiver can be only one specific type.
答案6
得分: 2
即使Go的实现允许这样做,不幸的是这是不安全的:你不能将[]Person
赋值给类型为[]Model
的变量,因为[]Model
具有不同的能力。例如,假设我们还有实现了Model
的Animal
:
var people []Person = ...
var models []Model = people // 在真正的Go中不允许
models[0] = Animal{..} // ???
var person Person = people[0] // !!!
如果我们允许第2行,那么第3行也应该可以工作,因为models
可以完全存储一个Animal
。而第4行仍然应该工作,因为people
存储的是Person
。但是,我们最终得到的是一个类型为Person
的变量持有一个Animal
!
Java实际上允许类似于第2行的操作,但这被广泛认为是一个错误。(错误在运行时被捕获;第3行将抛出ArrayStoreException
。)
英文:
Even if Go's implementation allowed this, it's unfortunately unsound: You can't assign a []Person
to a variable of type []Model
because a []Model
has different capabilities. For example, suppose we also have Animal
which implements Model
:
var people []Person = ...
var models []Model = people // not allowed in real Go
models[0] = Animal{..} // ???
var person Person = people[0] // !!!
If we allow line 2, then line 3 should also work because models
can perfectly well store an Animal
. And line 4 should still work because people
stores Person
s. But then we end up with a variable of type Person
holding an Animal
!
Java actually allows the equivalent of line 2, and it's widely considered a mistake. (The error is caught at run time; line 3 would throw an ArrayStoreException
.)
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库,让每个人都能够通过互相帮助和分享经验来进步。
评论