Haskell,String vs ... String？

Question

我目前正在关注他们网站上的Happstack lite教程.http://happstack.com/page/view-page-slug/9/happstack-lite-tutorial

现在,我正在实现该echo功能,编译器给我一个我不太懂的错误信息.这是我的代码:

echo :: ServerPart Response
echo =
    path $ \(msg :: String) ->
        ok $ template "echo" $ do
            h1 "Echo service"
            p "Giant, Haskell style Papagallo"
            p msg

这是错误信息:

src/motiondude.hs:35:15:
    Couldn't match type `[Char]' with `Text.Blaze.Internal.MarkupM ()'
    Expected type: Html
      Actual type: String
    In the first argument of `p', namely `msg'
    In a stmt of a 'do' block: p msg
    In the second argument of `($)', namely
      `do { h1 "Echo service";
            p "Giant, Haskell style Papagallo";
            p msg }'

我认为这是一个引用封闭的"东西",就像

"Giant, Haskell style Papagallo"

是一个字符串.然而,根据我对编译器错误的理解,p不会接受字符串作为参数.谁可以给我解释一下这个？

Answer 1

问题是你似乎启用了OverloadedString扩展,这意味着

"foo" :: IsString a => a

强迫一个普通的老人String做任何IsString只使用的东西fromString.然而,在这种情况下,Happstack具有html看起来更合适的功能.