functional programming - Haskell function that accepts function or value, then calls function or returns value -

how can write type declaration , function in haskell takes either function (that takes no arguments) or value. when given function calls function. when given value returns value.

[edit] give more context, i'm curious how solve problem in haskell without bit twiddling: designing function f(f(n)) == -n

sean

i'm curious how solve problem in haskell without bit twiddling: designing function f(f(n)) == -n

that's quite easy solve:

when :: (a -> bool) -> (a -> a) -> -> when p f x = if p x f x else x  f :: integer -> integer f = (+) <$> when negate <*> signum 

how derive this? consider:

f (f n) = (-n) -- (0) - interview question  f x     = y    -- (1) - assumption  f y     = (-x) -- (2) - (0) , (1), f (f x) = (-x)  f (-x)  = (-y) -- (3) - (0) , (2), f (f y) = (-y)  f (-y)  = x    -- (4) - (0) , (3), f (f (-x)) = x 

now, if see left hand sides of these equations you'll notice there 4 cases:

1. f x.
2. f y.
3. f (-x).
4. f (-y).

notice domain of function f divided positive , negative numbers, x , (-x), , y , (-y). let's assume x , y form set of positive numbers , (-x) , (-y) form set of negative numbers.

the set of positive numbers divided 2 proper disjoint subsets, x , y. how divide set of positive numbers 2 proper disjoint subsets? odd , numbers candidate. hence, let's assume x set of positive odd numbers , y set of positive numbers.

another advantage of using odd , numbers when negated odd numbers remain odd , numbers remain even. hence, (-x) set of negative odd numbers , (-y) set of negative numbers.

now, consider 4 cases again. notice sign changes when number even:

1. f x = y (sign not change).
2. f y = (-x) (sign changes).
3. f (-x) = (-y) (sign not change).
4. f (-y) = x (sign changes).

hence, negate number when (i.e. when negate).

next, need convert odd numbers numbers , vice versa. easiest way add or subtract 1 number. however, care should taken resulting number not 0. consider special case of 0:

f 0    = z    -- (a) - assumption  f z    = (-0) -- (b) - (0) , (a), f (f 0) = (-0)  f (-0) = (-z) -- (c) - (0) , (b), f (f z) = (-z)  (-0)   = 0    -- (d) - reflexivity  f (-0) = f 0  -- (e) - (d)  (-z)   = z    -- (f) - (a) , (c) , (e)  z      = 0    -- (g) - (d) , (f)  f 0    = 0    -- (h) - (a) , (g) 

hence, f n = 0 if , if n = 0. let's consider neighbors of 0, 1 , (-1). both of them odd numbers. hence, not negated. however, need converted number (except 0). hence, 1 converted 2 , (-1) converted (-2).

thus, odd numbers add sign of number number itself.

now, consider numbers. know:

f 1    = 2    -- (a)  f (-1) = (-2) -- (b) 

therefore:

f 2    = (-1) -- (c), (0) , (a), f (f 1) = (-1)  f (-2) = 1    -- (d), (0) , (b), f (f (-1)) = 1 

we know numbers negated. hence, 2 first becomes (-2) , vice versa. let original number n. hence, first negate n , add signum n it:

evenf n    = negate n    + signum n  evenf 2    = negate 2    + signum 2            = (-2)        + 1            = (-1)  evenf (-2) = negate (-2) + signum (-2)            = 2           + (-1)            = 1  evenf 0    = negate 0    + signum 0            = 0           + 0            = 0 

thus, both odd case , case add sign of original number when negate. therefore, f defined as:

f :: integer -> integer f = (+) <$> when negate <*> signum 

hope helps.