Question

I’m trying to convert some Ruby code into Python. I’m having troubles with this line:

Digest::MD5.digest(message).unpack('L*')

I think I should use the struct module and the hashlib one, but if I do:

struct.unpack('L', hashlib.md5(message).digest())

I get this error:

struct.error: unpack requires a bytes object of length 4

What should I do?
Thank you,

rubik

P.S. The output should be a list of 4 x 32 bit ints:

irb(main):039:0> Digest::MD5.digest('Hash').unpack('L*')
=> [631892218, 1967199614, 3683860954, 4130231798]

Answer 1

There is no support for arbitrary length unpacks (the * operator). You’ll have to specify a repeater manually.

Luckily the struct module does let you specify a fixed length, and the hashlib module tells you how many bytes to expect. By putting an integer number before the L you specify the number of times to apply the pattern. And a specific hash in the hashlib library has a .digest_size attribute that tells you how many bytes long a specific hash is.

To combine these:

structspec = '%iL' % (hashlib.hash('md5').digest_size / 4)
struct.unpack(structspec, hashlib.md5(message).digest())

If you don’t even want to hardcode the 4 there, you can ask struct.calcsize for the size of an L unsigned long as well:

structspec = '%iL' % (hashlib.hash('md5').digest_size / struct.calcsize('L'))
struct.unpack(structspec, hashlib.md5(message).digest())

In fact, on 64 bit platforms like my Mac, L is 8 bytes, so the latter calculation matters enormously if you are deploying across different architectures.