alt*_*ler 12 haskell pattern-matching
是否有可能创建一个泛型函数,将Foo或Bar作为参数,并返回一个在模式匹配中使用该参数的函数?
例如,如果我有
isFoo :: SomeData -> Bool
isFoo (Foo _) = True
isFoo _ = False
isBar :: SomeData -> Bool
isBar (Bar _) = True
isBar _ = False
Run Code Online (Sandbox Code Playgroud)
有没有办法创建一个通用函数,如
checkType :: SomeClass -> SomeData -> Bool
checkType (SomeClass _) = True
checkType _ = False
Run Code Online (Sandbox Code Playgroud)
我意识到情况看起来有点奇怪,实际的用例有点复杂,但问题是相同的.
我正在尝试重构的实际代码如下
isString :: [LispVal] -> ThrowsError LispVal
isString [(String _)] = return $ Bool True
isString ((String _):xs) = isString xs >>= unpackBool >>= return . Bool
isString _ = return $ Bool False
isSymbol :: [LispVal] -> ThrowsError LispVal
isSymbol [(Atom _)] = return $ Bool True
isSymbol ((Atom _):xs) = isSymbol xs >>= unpackBool >>= return . Bool
isSymbol _ = return $ Bool False
isNumber :: [LispVal] -> ThrowsError LispVal
isNumber [(Number _)] = return $ Bool True
isNumber ((Number _):xs) = isNumber xs >>= unpackBool >>= return . Bool
isNumber _ = return $ Bool False
Run Code Online (Sandbox Code Playgroud)
所以我想用一些方法让它更干燥
dan*_*iaz 27
Prism来自lens图书馆的s可以充当"一流的模式".要为您的数据类型定义棱镜:
{-# LANGUAGE TemplateHaskell #-}
import Control.Lens
data SomeData = Foo Int
| Bar Char
-- Will create prisms named _Foo and _Bar
$(makePrisms ''SomeData)
Run Code Online (Sandbox Code Playgroud)
由于Prisms是有效的Folds,我们可以将它们传递给has函数Control.Lens.Fold:
*Main> has _Foo (Foo 5)
True
*Main> has _Bar (Foo 5)
False
Run Code Online (Sandbox Code Playgroud)
棱镜作为第一类模式的另一个有趣的应用是"覆盖"函数的行为,用于参数与棱镜匹配的情况.您可以使用outsidefrom Control.Lens.Prism来做到这一点.outside是一个函数,它接受Prism并返回一个Lensfor函数,允许你"设置"特殊情况.例如:
functionToOverride :: SomeData -> Int
functionToOverride = const 5
-- If the arg is a Foo, return the contained int + 1
newFunction :: SomeData -> Int
newFunction = functionToOverride & outside _Foo .~ succ
Run Code Online (Sandbox Code Playgroud)
测试这两个功能:
*Main> functionToOverride (Foo 77)
5
*Main> newFunction (Bar 'a')
5
*Main> newFunction (Foo 77)
78
Run Code Online (Sandbox Code Playgroud)