英文:
Using interfaces correctly when creating structs
问题
我正在尝试编写一个小程序,其中包含几个包,每个包都有一个实现接口的结构体。我的想法是根据用户输入,我可以选择使用哪个包来构建特定的结构体,然后调用它们都应该具有的函数。由于我事先不知道类型,我以为可以使用interface{}
并将其用作前向声明(请参见最后的代码片段)。我有一个类似这样的东西:
在foo
包中,我有以下内容:
package foo
type FooInput struct {
Bar string
Baz int
}
type Foo interface {
Ding()
Dong()
}
在另一个包bob
中,我有以下内容:
package bob
import "your-package/foo"
type Bob struct {
foo.FooInput
}
func (mybob *Bob) Ding() {}
func (mybob *Bob) Dong() {}
func MakeBob(foo_input foo.FooInput) (*Bob, error) {
my_bob := Bob{foo_input}
return &my_bob, nil
}
在我的主包中,我有以下内容:
package main
import (
"your-package/foo"
"your-package/bob"
)
func main() {
foo_input := foo.FooInput{Bar: "awyiss", Baz: 1}
var something foo.Foo
var err error
some_string := "foo"
switch some_string {
case "foo":
something, err = bob.MakeBob(foo_input)
case "bar":
// imagine bar is like foo
// something, err = bar.MakeBar(foo_input)
// imagine other cases
}
something.Dong()
}
当运行/构建等操作时,我收到以下错误:
something.Dong undefined (type interface {} is interface with no methods)
我有点困惑我做错了什么...关于如何使用interface{}
(如果需要的话)的任何解释都将非常有帮助!
英文:
I'm trying to write a small program in which I have a few packages, each with a struct that implements an interface. The idea is that based on user input, I can choose what package to use to build a particular struct and then call a function on it that they're all supposed to have. Since I don't know the type ahead of time, I was under the impression that I could use a interface{}
and use that as a forward declaration (see the last code snippet). I have something that looks like this:
package foo
type FooInput struct {
Bar string
Baz int
}
type Foo interface {
Ding()
Dong()
}
In another package, bob
, I have something like:
type Bob struct {
foo.FooInput
}
func (mybob *Bob) Ding() {}
func (mybob *Bob) Dong() {}
func MakeBob(foo_input foo.FooInput) (*Bob, error) {
my_bob := Bob{foo_input}
return &my_bob, nil
}
In my main package, I have something that looks like so:
data = foo.FooInput("awyiss", 1}
var something interface{}
var err error
switch some_string {
case "foo":
something, err = bob.MakeBob(foo_input)
case "bar":
// imagine bar is like foo
something, err = bar.MakeBar(foo_input)
// imagine other cases
}
something.Dong()
When running / building / etc, I get the following error:
something.Dong undefined (type interface {} is interface with no methods)
I'm a bit confused as to what I'm doing wrong... any clarifiers on how I should use interface{}
(if at all) would be extremely helpful!
答案1
得分: 1
变量 something
被声明为空接口。空接口上没有任何方法。要调用 Dong
方法,请将 something
声明为 foo.Foo
。
将代码中的
var something interface{}
改为
var something foo.Foo
这假设 bar
类似于 foo
,意味着 bar
满足 foo.Foo
接口的要求。
英文:
The variable something
is declared as the empty interface. There are no methods on the empty interface. To call the Dong
method, declare something
as a foo.Foo
.
Change
var something interface{}
to
var something foo.Foo
This assumes that bar
being like foo
means that bar
satisfies the foo.Foo
interface.
答案2
得分: 1
如果看起来你有点困惑,请选择两种可能性之一。如果你想将所有要放入变量"something"的类型都具有Ding和Dong方法,那么请为其定义接口。
Foo不是最好的名称,最好使用DingDonger(https://golang.org/doc/effective_go.html#interface-names)。
之后:
var something DingDonger
对于每个分配给something的值,都会检查类型是否真的具有所需的方法,因此something.Ding()不会引发错误。
如果任何变量的类型是interface{},则无法确定它是否具有该方法,你必须使用类型断言(https://golang.org/doc/effective_go.html#interface_conversions)来验证方法是否被实现。
英文:
If looks, that you are a bit confused. Please, choose - one of two possibilities. If all the types, that you want to put into variable "something" has methods Ding, and Dong - than define interface for it.
Foo is not the best name, better is DingDonger (https://golang.org/doc/effective_go.html#interface-names).
After:
var something DingDonger
each assign to something will be checked, whether type really has required methods, and therefore something.Ding() can not cause an error.
If any variable is type interface{} no one knows, whether it has or not such method and you have to use type assertion (https://golang.org/doc/effective_go.html#interface_conversions) to verify whether method is implemented.
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