这是一个通过foldl连接的无用案例,纯粹是教育(对我来说):
foldl
foldl (\xs x -> xs ++ [x]) [1,2] [11,12,13] [1,2,11,12,13]
有没有办法把它包装得更紧凑,使用组合而不是lambda?
发布于 2017-12-21 13:16:25
这只是从HTNW和Will Ness的评论中提取的可读性更好的摘要
-- Reduction to poinfree a = \xs x -> xs ++ [x] b = \xs x -> xs ++ return x c = \xs x -> ((xs ++) . return) x d = \xs x -> ((. return) (xs ++)) x e = \xs x -> ((. return) . (++)) xs x
https://stackoverflow.com/questions/47908295
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