CODEWITHSUNDEEP
haskell
Arseny Nerinovsky
Arseny Nerinovsky

Reputation: 91

Using Bool in haskell with other numerical types

Is there any 'elegant' way to use Bool type in numerical expressions ? My current solution is using fromIntegral.fromEnum but is seems long. Example:

lst::[(Double, Double)]
lst = [(1, -1 * (odd i)) | i<- [1..]]

Upvotes: 0

Views: 160

Answers (3)

erisco
erisco

Reputation: 14329

I do not understand the general purpose of interpreting Bools as Integers from the question, so I am only answering to the example given. Let me know how I can expand my answer.

lst::[(Double, Double)]
lst = [(1, -1 * (odd i)) | i<- [1..]]

The first element of the tuple is always 1, so this example can be simplified.

lst :: [Double]
lst = [-1 * (odd i) | i<- [1..]]

For the reader, odd :: Integral a => a -> Bool, and fromEnum True = 1; fromEnum False = 0.

Under the fromEnum interpretation, lst is an infinite list of -1 interspersed with zeroes. How else can we implement this list?

intersperse 0 (repeat (-1))

Then how do we get back the desired tuples? Apply fmap (1,) to the list.

Upvotes: 1

dfeuer
dfeuer

Reputation: 48591

I don't believe there's a standard way. You can write your own, of course:

enumToNum :: (Enum e, Num n) => e -> n
enumToNum = fromIntegral . fromEnum

The existence of this and similar functions really points to the silly nature of the Enum class.

Upvotes: 1

user2512323
user2512323

Reputation:

Package Foreign.Marshal.Utils contains fromBool function with the following definition:

-- |Convert a Haskell 'Bool' to its numeric representation
--
fromBool       :: Num a => Bool -> a
fromBool False  = 0
fromBool True   = 1

While I don't recommend to import this package for fromBool alone, you can copy it into your code.

By the way, good way to discover functions by type in the standard library is Hoogle.

Upvotes: 1

Related Questions