当有多个可能的类型构造函数

Question

我有以下数据类型，我想创建一个Sport实例，但我不知道如何从数据类型Sport调用类型Football。

`data Football a  = Football
{ players       :: Players a    
, Stadium       :: Stadium  a
}     

data Tennis a  = Tennis
{ players1      :: Players a    
, Stadium1      :: Stadium  a
}
data Sport a = SF  Football a  | ST Tennis a

当我这样做时：

Instance Show a => Show (Sport a) where
show Football{..} = "<== Football ==>"
show Tennis{..} = "<== Tennis ==>"

我收到一条错误消息：无法匹配预期类型'Sport a'               实际类型'足球t6' 我尝试了其他一些方法并检查了其他一些例子，但我可以弄明白该怎么做...有人有想法吗？

提前致谢:)

Answer 1

你需要括号

data Sport a = SF (Football a) | ST (Tennis a)

然后你需要对运动进行模式匹配，而不是其他类型：

instance Show a => Show (Sport a) where
  show (SF Football{..}) = "<== Football ==>"
  show (ST Tennis{..}) = "<== Tennis ==>"