[英]How can I implement comparable interface in go?
I've recently started studying Go and faced next issue. 我最近开始研究Go并面临下一期。 I want to implement Comparable interface. 我想实现Comparable接口。 I have next code: 我有下一个代码:
type Comparable interface {
compare(Comparable) int
}
type T struct {
value int
}
func (item T) compare(other T) int {
if item.value < other.value {
return -1
} else if item.value == other.value {
return 0
}
return 1
}
func doComparison(c1, c2 Comparable) {
fmt.Println(c1.compare(c2))
}
func main() {
doComparison(T{1}, T{2})
}
So I'm getting error 所以我收到了错误
cannot use T literal (type T) as type Comparable in argument to doComparison:
T does not implement Comparable (wrong type for compare method)
have compare(T) int
want compare(Comparable) int
And I guess I understand the problem that T
doesn't implement Comparable
because compare method take as a parameter T
but not Comparable
. 我想我理解T
没有实现Comparable
的问题,因为compare方法作为参数T
而不是Comparable
。
Maybe I missed something or didn't understand but is it possible to do such thing? 也许我错过了什么或者不理解但是可以做这样的事情吗?
Your Interface demands a method 您的界面需要一种方法
compare(Comparable) int
but you have implemented 但你已经实施了
func (item T) compare(other T) int {
(other T instead of other Comparable) func (item T) compare(other T) int {
(其他T代替其他可比较)
you should do something like this: 你应该做这样的事情:
func (item T) compare(other Comparable) int {
otherT, ok := other.(T) // getting the instance of T via type assertion.
if !ok{
//handle error (other was not of type T)
}
if item.value < otherT.value {
return -1
} else if item.value == otherT.value {
return 0
}
return 1
}
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