给定以下 Go 代码：package maintype CatToy interface { Rattle() string}type Cat struct {}func (cat *Cat) Play(catToy CatToy) { println("The cat is playing!", catToy.Rattle())}type DogToy interface { Roll() string}type Dog struct {}func (dog *Dog) Play(dogToy DogToy) { println("The dog is playing!", dogToy.Roll())}type SuperToy struct {}func (toy *SuperToy) Rattle() string { return "Rattle!!!"}func (toy *SuperToy) Roll() string { return "Rolling..."}type Pet interface { Play(toy interface{})}func main() { cat := &Cat{} dog := &Dog{} superToy := &SuperToy{} // Working cat.Play(superToy) dog.Play(superToy) // Not Working pets := []Pet{cat, dog} for _, pet := range pets { pet.Play(superToy) }}我收到这些错误：# command-line-arguments./main.go:65:16: cannot use cat (type *Cat) as type Pet in array or slice literal: *Cat does not implement Pet (wrong type for Play method) have Play(CatToy) want Play(interface {})./main.go:65:21: cannot use dog (type *Dog) as type Pet in array or slice literal: *Dog does not implement Pet (wrong type for Play method) have Play(DogToy) want Play(interface {})实现SuperToy了CatToy和DogToy。但是，当我创建一个接口Pet作为参数的接口时，我得到一个错误。我可以知道我怎样才能得到一个里面有猫和狗的数组/切片吗？我想遍历这个切片并为每个切片调用一个函数。我还想保留CatToy和DogToy接口。我也可以删除Pet界面。更多信息：未来我更有可能添加更多pets。我认为我不会添加更多操作，例如Play.
3 回答
万千封印
TA贡献1891条经验 获得超3个赞
我明白你想做什么，但这是不可能的：你的Cat和Dog类型没有实现Pet接口，因为它们的Play方法采用不同的类型，所以你不能只用Play你的SuperToy.
要解决此问题，您需要创建一个Toy接口，该接口同时具有Roll和Rattle方法以及 make Pet.Play，Cat.Play并将Dog.Play此接口作为参数。
package main
type Cat struct {
}
func (cat *Cat) Play(catToy Toy) {
println("The cat is playing!", catToy.Rattle())
}
type Dog struct {
}
func (dog *Dog) Play(dogToy Toy) {
println("The dog is playing!", dogToy.Roll())
}
type Toy interface {
Roll() string
Rattle() string
}
type SuperToy struct {
}
func (toy *SuperToy) Rattle() string {
return "Rattle!!!"
}
func (toy *SuperToy) Roll() string {
return "Rolling..."
}
type Pet interface {
Play(toy Toy)
}
func main() {
cat := &Cat{}
dog := &Dog{}
superToy := &SuperToy{}
// Working
cat.Play(superToy)
dog.Play(superToy)
// Not Working
pets := []Pet{cat, dog}
for _, pet := range pets {
pet.Play(superToy)
}
}
给出输出
The cat is playing! Rattle!!!
The dog is playing! Rolling...
The cat is playing! Rattle!!!
The dog is playing! Rolling...
皈依舞
TA贡献1851条经验 获得超3个赞
您可以让 Play 方法接受一个interface{}，然后在方法内进行类型断言：
func (dog *Dog) Play(toy interface{}) {
dogToy, isDogToy := toy.(DogToy)
if !isDogToy {
println("The dog does not know what to do with this toy!")
return
}
println("The dog is playing!", dogToy.Roll())
}
Go Playground 上的完整可执行示例：
https://play.golang.org/p/LZZ-HqpzR-Z
12345678_0001
TA贡献1802条经验 获得超5个赞
这是另一个可行的解决方案：
package main
type CatToy interface {
Rattle() string
}
type Cat struct {
Toy CatToy
}
func (cat *Cat) Play() {
println("The cat is playing!", cat.Toy.Rattle())
}
type DogToy interface {
Roll() string
}
type Dog struct {
Toy DogToy
}
func (dog *Dog) Play() {
println("The dog is playing!", dog.Toy.Roll())
}
type SuperToy struct {
}
func (toy *SuperToy) Rattle() string {
return "Rattle!!!"
}
func (toy *SuperToy) Roll() string {
return "Rolling..."
}
type Pet interface {
Play()
}
func main() {
superToy := &SuperToy{}
cat := &Cat{superToy}
dog := &Dog{superToy}
// Working
cat.Play()
dog.Play()
// Working also
pets := []Pet{cat, dog}
for _, pet := range pets {
pet.Play()
}
}
该解决方案使Cat + CatToy和Dog + DogToy独立于main + SuperToy. 这允许提取以分离包。
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