loop through all possible combination of 8 bytes-CodePudding

I want to do something like this:

for i in range('\xff'*8):
  hash = b'challenge'   i
  inp = hashlib.sha256(hash).digest()

I don't know how i can get all the possible combinations of 8 bytes with bitwise operations.

CodePudding user response：

You can do this, but using struct would be preferred, I've also added the zero check (increase 2 to 26...):

start = b'\0' * 2
import hashlib
for i in range(256 ** 8):
    byte = b'challenge'   bytes.fromhex("6x" % i)
    inp = hashlib.sha256(byte).digest()
    if inp.startswith(start):
        print(byte, inp)
        break

Output:

b'challenge\x00\x00\x00\x00\x00\x00*\xde' b'\x00\x00\x9a\x84H\xbc\x07\xd7\xdc\x07#\xb8\x08A\xb1&\xdcD\xb0 \x84\\9y#\xc9\xcf\xaa\xff\xb7\xbf\x9a'
...

I mean you're already using hashlib, so you could as well use another standard library:

start = b'\0' * 2
import hashlib, struct
if struct.pack("Q", 1) == struct.pack(">Q", 1):
    print("Using faster method:")
    for i in range(256 ** 8):
        byte = b'challenge'   struct.pack("Q", i)
        inp = hashlib.sha256(byte).digest()
        if inp.startswith(start):
            print(byte, inp)
            break
else:
    for i in range(256 ** 8):
        byte = b'challenge'   struct.pack(">Q", i)
        inp = hashlib.sha256(byte).digest()
        if inp.startswith(start):
            print(byte, inp)
            break