## Exercise 5.8

```encrypt :: [Char] -> [Char] encrypt cs = map encChar cs where encChar c = toEnum((fromEnum c + 1) `mod` 256)   decrypt :: [Char] -> [Char] decrypt cs = map decChar cs where decChar c = toEnum((fromEnum c + 256 - 1) `mod` 256)```

The addition of 256 ensures a proper wraparound on decryption. This exercise seemed a bit silly to me: why not have one function to do both, and allow an arbitrary key?

```caesar :: [Char] -> Int -> [Char] caesar cs k = map caesarChar cs where caesarChar c = toEnum((fromEnum c + 256 + k) `mod` 256)```

### 4 Responses to “Exercise 5.8”

1. Wieslaw Poszewiecki says:

Examples are allowed to be silly.
Paul Hudak said that frogs have small brains 🙂

IMHO
is equal to
beacuse
(a +k ) `mod` k
is equal to:
a `mod` k

My solution:
shift s x = toEnum (mod (fromEnum x+s) 256)::Char
encrypt,decrypt:: String->String
encrypt = map (shift 1)
decrypt = map (shift (-1))

2. Mike says:

c+k mod 256
in general is not equal to
c + 256 + k mod 256.

Consider what happens when c + k is negative. In Python, the mod operator always returns a positive value, but not in Haskell.

3. Conor Hackett says:

This code seems complicated to me..

how come there are no paramaters to the encrypt and decrypt functions?

Where does s come from in the shift function?

4. admin says:

s is the first parameter to shift – the amount to shift by. Inverting the order of the parameters from my original solution allows a more functional style.

It looks like there are no parameters to encrypt/decrypt because they are written in point free style.

See section 9.1 of the book.