```
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”

Examples are allowed to be silly.

Paul Hudak said that frogs have small brains 🙂

IMHO

fromEnum c+256-1 `mod` 256

is equal to

fromEnum c-1 `mod` 256

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))

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.

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?

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.