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19 votesanswersviews
使用foldr实现zip
我试图绕着implementing foldl in terms of foldr缠身 . (这是他们的代码:) myFoldl :: (a -> b -> a) -> a -> [b] -> a myFoldl f z xs = foldr step id xs z where step x g a = g (f a x) 我以为我会尝试使用相同的技术实... -
98 votesanswersviews
foldl与具有无限列表的foldr行为
this question中myAny函数的代码使用foldr . 当谓词满足时,它会停止处理无限列表 . 我用foldl重写了它: myAny :: (a -> Bool) -> [a] -> Bool myAny p list = foldl step False list where step acc item = p item || acc (请注意,...