Z3方程求解器-位掩码操作

Question

After reverse engineering a program, I found the constraints.

I have to find two unsigned numbers such that:

x + y = 0xC0ED91BD

and

x * y = Z

In the original program, the multiplication of the numbers is performed using imul instruction.

Z should be such that when only the lower 32-bits of Z are checked, it should be less than 60.

Z & 0x00000000ffffffff < 60

However, when I add the second equation as a constraint, z3 gives me an error.

Here's my code:

#! /usr/bin/python

from z3 import *
import sys

s = Solver()
result = int(sys.argv[1], 16)

x = Int('x')
y = Int('y')

s.add(x + y == result)
s.add(x > 0, y > 0)
s.add((x * y) & 0x00000000ffffffff < 60)

while s.check() == sat:
    print s.model()
    s.add(Or(x != s.model()[x], y != s.model()[y]))

Updated:

Here is the code based on the recommended solution:

#! /usr/bin/python

from z3 import *
import sys

s = Solver()
x = BitVec('x', 32)
y = BitVec('y', 32)

result = BitVecVal(int(sys.argv[1], 16), 32)

s.add(x + y == result)
s.add(x > 0, y > 0)
s.add(x * y < 10)

print s.check()
print s.model()

The binary is 64-bit.

However, the multiplication operation is performed using 32-bit integers as shown below:

mov     edx, [rbp+var_14]
mov     eax, [rbp+var_10]
imul    eax, edx

So if eax = 0x425a95e5 and edx = 0x7e92fbd8

Then after multiplication using imul instruction, eax will store: 0x00000038.

Both, the Carry Flag and Overflow Flag bits in the EFLAGS register will be set after it.

Answer 1

The issue here is that you have declared x and y to be arbitrarily wide integers, as opposed to bit-vectors of a fixed length. Simply change your declarations to match what the underlying bit-size should be. Assuming you want your arithmetic to be done in 32 bits, you'd say:

x = BitVec('x', 32)
y = BitVec('y', 32)

instead. You should also declare result similarly:

result = BitVecVal(int(sys.argv[1], 16), 32)

Once you make these changes, your program should work just fine.

Note that in this case masking with 0x00000000ffffffff isn't really necessary; since the numbers are already 32-bits wide. You only need to keep it if your x and y are larger; say 64 bits.

With the above changes, I get the following output when I run your program and call it with 0xC0ED91BD :

[y = 2123561944, x = 1113232869]
[y = 1440310864, x = 1796483949]
[y = 1171875408, x = 2064919405]
... many other lines ..

This might look confusing, as the numbers seem to be larger than what you think they should be. But remember that arithmetic over bitvectors is done modulo 2^n where n is the bit-size, so the results are actually correct:

>>> 2123561944 + 1113232869 % 2**32 == 0xC0ED91BD
True
>>> 2123561944 * 1113232869 % 2**32 < 60
True

etc.