Tôi đã tự do lắp ráp các câu trả lời vào một chương trình haskell rất đơn giản, mà chỉ thông qua việc khớp mẫu cố gắng dịch mã haskell sang tiếng Anh. Tôi gọi nó letterator
bởi vì nó dịch các ký hiệu thành các chữ cái
-- letterator
main = translateLn <$> getLine >>= putStrLn
translateLn :: String -> String
translateLn = unwords . map t . words
t :: String -> String -- t(ranslate)
-- historical accurate naming
t "=" = "is equal too" -- The Whetstone of Witte - Robert Recorde (1557)
-- proposed namings
-- src http://stackoverflow.com/a/7747115/1091457
t ">>=" = "bind"
t "*>" = "then"
t "->" = "to" -- a -> b: a to b
t "<$" = "map-replace by" -- 0 <$ f: "f map-replace by 0"
t "<*>" = "ap(ply)" -- (as it is the same as Control.Monad.ap)
t "!!" = "index"
t "!" = "index/strict" -- a ! b: "a index b", foo !x: foo strict x
t "<|>" = "or/alternative" -- expr <|> term: "expr or term"
t "[]" = "empty list"
t ":" = "cons"
t "\\" = "lambda"
t "@" = "as" -- go ll@(l:ls): go ll as l cons ls
t "~" = "lazy" -- go ~(a,b): go lazy pair a, b
-- t ">>" = "then"
-- t "<-" = "bind" -- (as it desugars to >>=)
-- t "<$>" = "(f)map"
-- t "$" = "" -- (none, just as " " [whitespace])
-- t "." = "pipe to" -- a . b: "b pipe-to a"
-- t "++" = "concat/plus/append"
-- t "::" = "ofType/as" -- f x :: Int: f x of type Int
-- additional names
-- src http://stackoverflow.com/a/16801782/1091457
t "|" = "such that"
t "<-" = "is drawn from"
t "::" = "is of type"
t "_" = "whatever"
t "++" = "append"
t "=>" = "implies"
t "." = "compose"
t "<=<" = "left fish"
-- t "=" = "is defined as"
-- t "<$>" = "(f)map"
-- src http://stackoverflow.com/a/7747149/1091457
t "$" = "of"
-- src http://stackoverflow.com/questions/28471898/colloquial-terms-for-haskell-operators-e-g?noredirect=1&lq=1#comment45268311_28471898
t ">>" = "sequence"
-- t "<$>" = "infix fmap"
-- t ">>=" = "bind"
--------------
-- Examples --
--------------
-- "(:) <$> Just 3 <*> Just [4]"
-- meaning "Cons applied to just three applied to just list with one element four"
t "(:)" = "Cons"
t "Just" = "just"
t "<$>" = "applied to"
t "3" = "three" -- this is might go a bit too far
t "[4]" = "list with one element four" -- this one too, let's just see where this gets us
-- additional expressions to translate from
-- src http://stackoverflow.com/a/21322952/1091457
-- delete (0, 0) $ (,) <$> [-1..1] <*> [-1..1]
-- (,) <$> [-1..1] <*> [-1..1] & delete (0, 0)
-- liftA2 (,) [-1..1] [-1..1] & delete (0, 0)
t "(,)" = "tuple constructor"
t "&" = "then" -- flipped `$`
-- everything not matched until this point stays at it is
t x = x