Haskell 带参数的拟群
我想用Haskell写一篇引文。name参数需要传递到gen函数中才能生成声明Haskell 带参数的拟群,haskell,template-haskell,Haskell,Template Haskell,我想用Haskell写一篇引文。name参数需要传递到gen函数中才能生成声明 quote :: String -> QuasiQuoter quote name = QuasiQuoter { quoteExp = undefined, quotePat = undefined, quoteType = undefined, quoteDec = \jsonStr -> gen name (getValue str)
quote :: String -> QuasiQuoter
quote name = QuasiQuoter {
quoteExp = undefined,
quotePat = undefined,
quoteType = undefined,
quoteDec = \jsonStr -> gen name (getValue str)
}
但是,我似乎不能这样引用
[quote "Hello"| from x to y |]
由于Haskell不允许quote声明和Quotes位于同一个文件中,这很烦人,我该如何将参数从外部传递到Quotes中?您有两个选项:
$(…)
quote :: String -> String -> Q [Dec]
quote name jsonStr = gen name (getValue jsonStr)
调用它看起来像:$(从x到y引用“Hello”)
为了演示选项2,这里有一个简单的引号,它用一个字符围绕一个文字字符串:
import Language.Haskell.TH (litE, stringL)
import Language.Haskell.TH.Quote
surround :: QuasiQuoter
surround = QuasiQuoter
{ quoteExp = litE . stringL . (\(c:s) -> [c] ++ s ++ [c])
, quotePat = undefined
, quoteType = undefined
, quoteDec = undefined
}
-- in another file:
main = print [surround|_some text|] -- prints "_some text_"
输入字符串的第一个字符被解释为要使用的括号字符。实际上,我们已经将Char
参数传递给了Char->quasikoter
类型的函数
对于更复杂的参数或多个参数,您必须创建自己的语法和解析器来解码它们
更新:下面是一个稍微复杂的示例,其中调用[foo | var xyz |]
将var
视为变量名称,将xyz
视为文本字符串:
-- [foo| var xyz|] is translated to: var ++ "xyz"
foo :: QuasiQuoter
foo = QuasiQuoter
{ quoteExp = go
, quotePat = undefined
, quoteType = undefined
, quoteDec = undefined
}
where go str = infixE (Just $ varE (mkName var))
(varE $ mkName "++")
(Just $ litE (stringL arg1))
where (var:arg1:_) = words str