Adding two 8-Bit numbers in haskell -

i trying define function(s) helped me simulate basic operations when using 8-bit numbers.

i'm having hard time figuring 1 out.. i'm trying keep simple possible without importing anything, started 2 lists of 8 elements in (which 0's , 1's).

if i'm not mistaken should start looking this:

bitsum :: [int] -> [int] -> [int] bitsum [][] = [] 

after last line starts kind of tricky me because can't add 1 one elements lists.

bitsum (x:xs)(y:ys)  

that's have right think correct.

my idea try this:

bitsum :: [int] -> [int] -> [int] bitsum [][] = [] bitsum (x:xs)[] = (x:xs) bitsum [](y:ys) = (y:ys) bitsum (x:xs)(y:ys) | (x:xs) == (y:ys) && < 0 = (x:xs)                     | (x:xs) == (y:ys) && > 0 =  

but think i'm taking wrong turn somewhere.

i appreciate if give me hand problem.

you going need carry bit. can't add column column. let's piece piece:

bitsum :: [int] -> [int] -> [int] bitsum = bitsum' 0       bitsum' _ [] [] = [] 

there's start. start no carry, , deal case of no more bits add. if both bits 0?

    bitsum' 0 (0:xs) (0:ys) = 0 : bitsum' 0 xs ys     bitsum' 1 (0:xs) (0:ys) = 1 : bitsum' 0 xs ys 

okay, if both 0, , carry 0, result bit 0 , no carry. if there carry use it, , continue without one.

    bitsum' 0 (1:xs) (1:ys) = 0 : bitsum' 1 xs ys     bitsum' 1 (1:xs) (1:ys) = 1 : bitsum' 1 xs ys 

if one, similar. except there carry. , if different:

    bitsum' 0 (x:xs) (y:ys) = 1 : bitsum' 0 xs ys     bitsum' 1 (x:xs) (y:ys) = 0 : bitsum' 1 xs ys 

well have 0 , 1 because we've dealt other cases, must add one. can figure out should that. start see patterns above, , can condensed end shorter answer.

bitsum :: [int] -> [int] -> [int] bitsum = bitsum' 0       bitsum' _ [] [] = []     bitsum' carry (x:xs) (y:ys) | x == y = carry : bitsum' x xs ys                                 | otherwise = (1 - carry) : bitsum' carry xs ys 

(1-carry) fancy way of flipping 1 0 or vice versa since in case bit opposite of carry.