纯脚本行多态性。正确的语法是什么？

Question

使用基于“Purescript by Example”第 5 章中示例的函数，我对如何声明多态行类型有点困惑。

以下编译正常

type Student = {
              first :: String,
              last :: String,
              class :: String
            }

type GymMember = {
              first :: String,
              last :: String,
              benchPressPB :: Int
            }

daveG :: GymMember
daveG = {
        first: "Dave",
        last: "Bro",
        benchPressPB: 300
    }

philS :: Student
philS = {
        first : "Dave",
        last : "Swat",
        class : "1A"
      }

 schoolRollName :: forall t15.
  { last :: String
  , first :: String
  | t15
  } -> String
 schoolRollName rec = rec.last <> ", " <> rec.first

firstAndSurname :: forall t82.
{ first :: String
, last :: String
| t82
}
-> String
firstAndSurname rec =  rec.first <> " " <> rec.last

daveFandS :: String
daveFandS = firstAndSurname daveG

daveSR :: String
daveSR = schoolRollName daveG

philFandS :: String
philFandS = firstAndSurname philS

philSR :: String
philSR = schoolRollName philS

但是如何删除 schoolRollName 和 firstAndSurname 类型签名中的重复项。

我认为下面的方法可行，但类型不匹配:

type NamedThing = forall t15.
                { last :: String
                , first :: String
                | t15
                }

schoolRollName :: NamedThing -> String
schoolRollName rec = rec.last <> ", " <> rec.first

firstAndSurname :: NamedThing -> String
firstAndSurname rec =  rec.first <> " " <> rec.last

-- !! Could not match type
daveFandS :: String
daveFandS = firstAndSurname daveG

Answer 1

NamedThing 未正确声明。要匹配 NamedThing 类型，您必须提供一个值，该值对 所有 至少有一个 first 的可能记录有效和最后 字段。由于 daveG 不是这样的值，编译器会提示 - 更进一步，这种类型没有值。

将 t 移动到类型别名:

type NamedThing t = {first :: String, last :: String | t}

现在 firstAndSurname 确实必须提供一个工作函数 forall NamedThings 以及任何额外的字段。简单的解决方案:

firstAndSurname :: forall t. NamedThing t -> String
firstAndSurname rec =  rec.first <> " " <> rec.last

最后编译器对此感到满意:

daveFandS :: String
daveFandS = firstAndSurname daveG