Below is the hash function implementation of the roguewave library. It returns a 32 bit hash. The core operation is A = B ^ ((C << 5) | (C >> 27)). Is this hash safe to use as a password check or is it possible to retrieve all used B's by reversing it?

const unsigned RW_HASH_SHIFT = 5;

inline static void mash(unsigned& hash, unsigned chars)
  hash = (chars ^
       ((hash << RW_HASH_SHIFT) |
        (hash >> (RWBITSPERBYTE*sizeof(unsigned) - RW_HASH_SHIFT))));

RWCStringRef::hash() const
  unsigned hv       = (unsigned)length(); // Mix in the string length.
  unsigned i        = length()*sizeof(char)/sizeof(unsigned);
  const unsigned* p = (const unsigned*)data();
    while (i--)
      mash(hv, *p++);           // XOR in the characters.
  // XOR in any remaining characters:
  if ((i = length()*sizeof(char)%sizeof(unsigned)) != 0) {
    unsigned h = 0;
    const char* c = (const char*)p;
    while (i--) 
      h = ((h << RWBITSPERBYTE*sizeof(char)) | *c++);
    mash(hv, h);
  return hv;
  • Possibly this would be better asked on Cryptography Aug 22, 2014 at 12:37
  • 8
    This is not safe for a password hash. 32-bits of check is not nearly enough for hashing passwords. It doesn't matter that I don't know what the password is, I can just continuously generate a bunch of strings until a collision occurs. And collisions will occur quite a bit in 32-bits of space.
    – RoraΖ
    Aug 22, 2014 at 12:40
  • 2
    sizeof(char) is 1. Always. By design. By definition. Don't compute sizeof(char).
    – curiousguy
    Aug 22, 2014 at 14:48
  • 1
    @curiousguy Any sane compiler would macro sizeof(char) to 1 and optimise out the imul anyway. I personally think sizeof(char) is a good thing to include, as it implies that the size of the type is relevant. That way, if you move to wchar_t or similar later to gain Unicode support, it's obvious that you need to change that calculation too.
    – Polynomial
    Aug 26, 2014 at 11:04
  • 1
    @Polynomial 1 is a most descriptive term than sizeof(char).
    – curiousguy
    Aug 28, 2014 at 19:26

2 Answers 2


This is extremely unsafe, to the point of being pointless:

  1. Your hash function is not a one-way function. One can instantly (with constant and low runtime) calculate an input producing any given hash if you allow arbitrary 4 character passwords as inputs by undoing the XOR with the initial hash value formed from the password length. With a little ingenuity, it is likely that such a preimage can be very effectively generated even if the input character set is restricted.

  2. Even if your hash function was a cryptographically strong one-way function, it would not be a good password hashing function because it is fast. Ideally, you want a key derivative function that, like PBKDF2 or scrypt, has a configurable work factor allowing you to adjust the difficulty level for your attacker (and the workload for your server) to new conditions over time.

  3. Your hash is so far on the short side that it is hard to call it secure even if everything else were fine. As Andrey pointed out in an older answer, any 32-bit hash can be guessed in (typ.) 0.5 * 2^32 attempts. To not be a security bottleneck, this number should always be greater than what it would take to guess the password. Yet 32 bits of entropy is so low that a similar number (28 bits) works to illustrate a weak password in the famous xkcd password cartoon.


32 bit hash function cannot be possibly safe for the purpose of password verification.

Problem here is that it is "easy" to find a colliding password, that is, a password, that hashes to the "correct" hash value despite being different from the original password. On average it will take 2^31 password trials to get such collision, which is considered very weak by today standard.

  • How do you conclude that on "average it will take 2^31 password trials to get such" preimages?
    – user49075
    Aug 22, 2014 at 14:38
  • 2
    If output of hash function is uniformly distributed wouldn't it take on average N/2 tries to find a preimage (N being a number of possible outputs)?
    – Andrey
    Aug 22, 2014 at 14:56
  • Yes, if it were uniformly distributed, or if the attacker was forced or opted to use a generic attack that would work even if it were, then N/2 attempts on average is correct. This hash function looks like it may take approximately one trial (since it looks, well, invertible).
    – user27909
    Aug 22, 2014 at 16:48
  • @pyramids Yes, I've seen your excellent answer and agree that collision can be found faster for this particular function. IMO, function isn't invertible in the sense that it won't allow recovery of the original input (simply because it's lossy), but it seems to allow for fast preimage computation.
    – Andrey
    Aug 22, 2014 at 16:56
  • @Andrey: No, that's if there were N possible inputs.
    – user49075
    Aug 22, 2014 at 16:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .