Learn You a Haskell presents the Caesar Cipher:

The Caesar cipher is a primitive method of encoding messages by shifting each character in them by a fixed number of positions in the alphabet


Since I'm not limiting this cipher to letters, it's not actually the Caesarian Cipher. Thanks to Anonymous. This implementation is only a Substitution Cipher.

Here's my implementation:

import Data.Char (ord, chr)

encode :: Int -> String -> String
encode n = map $ shift' n
                 where shift' x =  chr . (+ x) . ord

decode :: Int -> String -> String
decode n = map $ shift' n
                 where shift' x =  chr . abs . ((-) x) . ord


*Main Data.Char> encode 1 "AAA"
*Main Data.Char>
*Main Data.Char> decode 1 "BBB"

I don't like that I'm using abs in my decode. How can I write a function (through currying) that will subtract X from it?

Also, please critique my implementation.

  • \$\begingroup\$ Hey 200_Success, thanks for editing to include the proper tag. But, since Anonymous pointed out that it's only a Substitution Cipher, I edited the title and removed the tag. \$\endgroup\$ May 18, 2014 at 2:03
  • 1
    \$\begingroup\$ I suggest fixing the cipher to actually be Caesarean rather than just documenting the bug. It's a good exercise. Hint: isAsciiUpper, isAsciiLower and mod. \$\endgroup\$
    – Anonymous
    May 18, 2014 at 2:12
  • 1
    \$\begingroup\$ I've put the tag back, since it's still Caesar-inspired in nature. As the tag description says, caesar-cipher also applies to similar toy-grade ciphers. \$\endgroup\$ May 18, 2014 at 2:31

1 Answer 1


This is a legitimate substitution cipher, but it's not a Caesarean cipher. A Caesarian cipher affects only letters, and it's modulo the alphabet size: caesar 13 "Hello, world!""Urzzb, jbeyq!"

encode and decode can be the same function (caesar?) called with opposite shifts. (Or with the same shift, when it's 13.)

There's no need for a shift' helper.

How can I write a function (through currying) that will subtract X from it?

(x -)

  • \$\begingroup\$ thanks for mentioning that I'm not using the actual Caesarian Cipher. When you say that shift' is not needed, you recommend just putting the composed function directly in the map? \$\endgroup\$ May 18, 2014 at 2:00
  • \$\begingroup\$ Sorry, I didn't notice the map. shift is OK — yes, you could just write map (chr . (+ n) . ord) (as LYaH suggests). It's not necessary to pass n to shift, because it's in scope. \$\endgroup\$
    – Anonymous
    May 18, 2014 at 2:04

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