Вот моя реализация:
Сначала некоторые полезные функции:
(* Pad the string s with x on the right/left so it has length n. *)
strpadr[s_, n_, x_] := StringJoin@PadRight[Characters[s], n, x]
strpadl[s_, n_, x_] := StringJoin@PadLeft[Characters[s], n, x]
(* hex representation; optionally pad to length n with zeros *)
hex[x_, n_:Null]:= If[n===Null, Identity, strpadl[#,n,"0"]&]@IntegerString[x,16]
(* parse hex representation (return an integer) *)
unhex[x_] := FromDigits[x, 16]
(* Concatenate all the arguments as strings (if they're not already). *)
cat = StringJoin @@ (ToString /@ {##}) &;
(* Takes a string like "xy" and returns the hex representation (also a string,
twice as long) of the bytes (ascii codes). *)
tobytes[s_] := cat @@ (hex[#, 2] & /@ ToCharacterCode[s])
(* Takes a length n string like "0a10" and returns a length n/2 string where in
this example the first character is whatever has ascii code 10 ("a" in hex)
and the second is whatever has ascii code 16 ("10" in hex). *)
frombytes[hs_] := FromCharacterCode[unhex /@ cat@@@Partition[Characters[hs], 2]]
(* Bitwise-xor of two integers given in hex. *)
hexbitxor[a_String, b_String] := hex@BitXor[unhex@a, unhex@b]
(* Repeat the string s, n times. *)
strrpt[s_, n_] := cat @@ ConstantArray[s, n]
(* Byte length of a hex string is half the string length. *)
bytelen[s_] := Ceiling[StringLength[s]/2]
Реализация основных хеш-функций в Mathematica не совсем тривиальна. Смотрите этот вопрос:
Криптографический хеш (sha1 или md5) данных, представленных в Mathematica в виде строки .
(* FileHash is the only way to hash data given as a string in Mma. *)
hash[s_String, h_:"SHA"] := Module[{stream = StringToStream[s], result},
result = FileHash[stream, h];
Close[stream];
hex@result];
sha1[s_] := hash[s]
md5[s_] := hash[s, "MD5"]
Наконец, вот функция hmac:
(* Return the hmac digest using hash function h for string s with key k. *)
hmac[s_, k_, h_:sha1] := Module[{b, key, ipad, opad},
b = 64; (* block size for both md5 and sha1 *)
key = tobytes[k];
key = If[bytelen[key] > b, h[frombytes@key], key];
key = strpadr[key, 2 b, "0"];
ipad = hexbitxor[strrpt["36", b], key];
opad = hexbitxor[strrpt["5c", b], key];
h[frombytes[opad <> h[frombytes[ipad <> tobytes@s]]]]]
Я подтвердил, что это соответствует всем примерам, приведенным в RFC 2104 . Например:
hmac["what do ya want for nothing?", "Jefe", md5]
возвращает
"750c783e6ab0b503eaa86e310a5db738"