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下面哪个定义实现了函数map :: (a -> b) -> [a] -> [b]


A、map f ys = foldr (\x xs -> xs ++ [f x]) [] ys;

B、map f ys = foldr (\x xs -> f x ++ xs) [] ys;

C、 map f ys = foldl (\xs x -> f x : xs) [] ys;

D、map f ys = foldl (\xs x -> xs ++ [f x]) [] ys

发布时间:2025-05-20 21:26:15
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答案:map f ys = foldl (\xs x -> xs ++ [f x]) [] ys
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