Saya tidak tahu apakah ada satu jawaban untuk ini, seperti yang disebutkan @jozefg. Dan murni dari spekulasi, inilah satu penjelasan yang mungkin:
Saya menduga alasannya adalah karena jenis - (a -> b -> b)
dalam Haskellfoldr
, misalnya - lebih bermakna daripada nama apa pun yang muncul. Karena parameter a
dan b
ketik bisa apa saja , fungsinya bisa melakukan apa saja juga. Jadi kau pergi dengan nama-nama seperti function
, combiner
, morphism
, dan argument
, yang tidak sangat berarti baik. Mungkin juga menggunakan nama pendek, tidak mengganggu.
Contoh lain adalah id :: a -> a
fungsinya: apa yang harus Anda sebut argumennya? Sekali lagi, saya pikir id
tipe ini lebih deskriptif daripada nama argumennya.
Namun, saya setuju dengan Anda - sepertinya harus ada nama umum, mungkin dalam matematika. Saya harap seseorang dapat memperbaiki saya dalam hal ini.
Beberapa contoh namanya dalam kode nyata:
Di perpustakaan Haskell , sebagian besar disebut f
(dan kadang-kadang operator
di komentar):
class Foldable t where
-- | Map each element of the structure to a monoid,
-- and combine the results.
foldMap :: Monoid m => (a -> m) -> t a -> m
foldMap f = foldr (mappend . f) mempty
-- | Right-associative fold of a structure.
--
-- @'foldr' f z = 'Prelude.foldr' f z . 'toList'@
foldr :: (a -> b -> b) -> b -> t a -> b
foldr f z t = appEndo (foldMap (Endo . f) t) z
-- | Right-associative fold of a structure,
-- but with strict application of the operator.
foldr' :: (a -> b -> b) -> b -> t a -> b
foldr' f z0 xs = foldl f' id xs z0
where f' k x z = k $! f x z
-- | Left-associative fold of a structure.
--
-- @'foldl' f z = 'Prelude.foldl' f z . 'toList'@
foldl :: (a -> b -> a) -> a -> t b -> a
foldl f z t = appEndo (getDual (foldMap (Dual . Endo . flip f) t)) z
-- | Left-associative fold of a structure.
-- but with strict application of the operator.
--
-- @'foldl' f z = 'List.foldl'' f z . 'toList'@
foldl' :: (a -> b -> a) -> a -> t b -> a
foldl' f z0 xs = foldr f' id xs z0
where f' x k z = k $! f z x
-- | A variant of 'foldr' that has no base case,
-- and thus may only be applied to non-empty structures.
--
-- @'foldr1' f = 'Prelude.foldr1' f . 'toList'@
foldr1 :: (a -> a -> a) -> t a -> a
foldr1 f xs = fromMaybe (error "foldr1: empty structure")
(foldr mf Nothing xs)
where
mf x Nothing = Just x
mf x (Just y) = Just (f x y)
-- | A variant of 'foldl' that has no base case,
-- and thus may only be applied to non-empty structures.
--
-- @'foldl1' f = 'Prelude.foldl1' f . 'toList'@
foldl1 :: (a -> a -> a) -> t a -> a
foldl1 f xs = fromMaybe (error "foldl1: empty structure")
(foldl mf Nothing xs)
where
mf Nothing y = Just y
mf (Just x) y = Just (f x y)
-- instances for Prelude types
instance Foldable Maybe where
foldr _ z Nothing = z
foldr f z (Just x) = f x z
foldl _ z Nothing = z
foldl f z (Just x) = f z x
instance Ix i => Foldable (Array i) where
foldr f z = Prelude.foldr f z . elems
foldl f z = Prelude.foldl f z . elems
foldr1 f = Prelude.foldr1 f . elems
foldl1 f = Prelude.foldl1 f . elems
-- | Monadic fold over the elements of a structure,
-- associating to the right, i.e. from right to left.
foldrM :: (Foldable t, Monad m) => (a -> b -> m b) -> b -> t a -> m b
foldrM f z0 xs = foldl f' return xs z0
where f' k x z = f x z >>= k
-- | Monadic fold over the elements of a structure,
-- associating to the left, i.e. from left to right.
foldlM :: (Foldable t, Monad m) => (a -> b -> m a) -> a -> t b -> m a
foldlM f z0 xs = foldr f' return xs z0
where f' x k z = f z x >>= k
-- | Map each element of a structure to an action, evaluate
-- these actions from left to right, and ignore the results.
traverse_ :: (Foldable t, Applicative f) => (a -> f b) -> t a -> f ()
traverse_ f = foldr ((*>) . f) (pure ())
-- | Map each element of a structure to a monadic action, evaluate
-- these actions from left to right, and ignore the results.
mapM_ :: (Foldable t, Monad m) => (a -> m b) -> t a -> m ()
mapM_ f = foldr ((>>) . f) (return ())
Itu juga disebut f
dalam Clojure :
(def
^{:arglists '([f coll] [f val coll])
:doc "f should be a function of 2 arguments. If val is not supplied,
returns the result of applying f to the first 2 items in coll, then
applying f to that result and the 3rd item, etc. If coll contains no
items, f must accept no arguments as well, and reduce returns the
result of calling f with no arguments. If coll has only 1 item, it
is returned and f is not called. If val is supplied, returns the
result of applying f to val and the first item in coll, then
applying f to that result and the 2nd item, etc. If coll contains no
items, returns val and f is not called."
:added "1.0"}
reduce
(fn r
([f coll]
(let [s (seq coll)]
(if s
(r f (first s) (next s))
(f))))
([f val coll]
(let [s (seq coll)]
(if s
(if (chunked-seq? s)
(recur f
(.reduce (chunk-first s) f val)
(chunk-next s))
(recur f (f val (first s)) (next s)))
val)))))