我有一個字串串列 ["Hi", "hELLO", "nO"]
我想將所有內容都轉換為小寫。我為此寫了一個函式：
import Data.Char(toLower)
makeSmall xs = map toLower xs
它編譯，但是當我這樣做時，makeSmall ["Hi", "hELLO", "nO"]它給了我這個錯誤
<interactive>:2:12: error:
? Couldn't match expected type ‘Char’ with actual type ‘[Char]’
? In the expression: "Hi"
In the first argument of ‘makeSmall’, namely
‘["Hi", "hELLO", "nO"]’
In the expression: makeSmall ["Hi", "hELLO", "nO"]
<interactive>:2:17: error:
? Couldn't match expected type ‘Char’ with actual type ‘[Char]’
? In the expression: "hELLO"
In the first argument of ‘makeSmall’, namely
‘["Hi", "hELLO", "nO"]’
In the expression: makeSmall ["Hi", "hELLO", "nO"]
<interactive>:2:25: error:
? Couldn't match expected type ‘Char’ with actual type ‘[Char]’
? In the expression: "nO"
In the first argument of ‘makeSmall’, namely
‘["Hi", "hELLO", "nO"]’
In the expression: makeSmall ["Hi", "hELLO", "nO"]
我試圖了解如何使我的函式適用于字串串列，而不僅僅是字串
uj5u.com熱心網友回復：
toLower :: Char -> Char將 a 映射Char到 a Char，這意味著map toLower將采用單個 String，它將回傳 a String。
如果你想處理一個串列的StringS，你應該使用第二個map：
makeSmall :: [String] -> [String]
makeSmall xs = map (map toLower) xs或更短：
makeSmall :: [String] -> [String]
makeSmall = map (map toLower)
因此，我們制作了一個map toLower用作地圖功能的映射器。因此，此功能將在單個 String. 因為我們這樣做了這樣的映射，我們對一個作業串列的String秒。
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標籤：哈斯克尔
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