Reverse engineering

How to Break Simple Software Protections

Soufiane Tahiri
September 17, 2012 by
Soufiane Tahiri

Some software developers are really lazy when it comes to protecting their products, and in some cases, the protection they implement (just like most "infamous" softwares) is really easy and simple to break.

In this article I'll explain to you how easy it is to break these kinds of protections. Generally, when a program starts, it tests if it's registered (purchased) or not, and depending on the result (0 or 1) it shows you a navigation screen or activates limitations … or not!

What really happens is that our software makes a CALL to the routine which makes this test (TEST, CMP…) then after this CALL it puts a static value in a register that is usually EAX, then the software checks the result of the concerned register (in this case AL).


[plain]EAX (32bits) = AX (16bits) = AL (8bits) + AH (8bits)[/plain]

After putting 1 or 0 in this register, either a JE or a JNE determines if the software is registered or not. These Jumps (like others except JMP) are directly influenced by the Zero Flag which is modified according to the value of EAX.

Well, now we know how this works, so even a newbie cracker will not find difficulties to break this kind of protection, and he even has several ways to do it, the simplest one is to find the CALL, then force the register EAX or AL into getting the right value (0 or 1).

You have to know that a "Call" is ALWAYS ended by a RET.

[plain]Pattern of a disassembled protection:</pre>


:00408740 E87B5B0000 Call 0040E2C0 ; calls serial check routine..

:00408748 85C0 test eax, eax ; checks if sn is good

:0040874B 7527 Jne 00408774 ; jump if ZF = 1

:0040874C ---><span style="color: red;">"Invalid Registration code."

</span>*********More Instructions**********

:00408774 ---> <span style="color: #00b050;">"Thank you for Support!!"[/plain]

Assuming that our software asked us for a serial key, if we input an invalid one, the CALL begins the check routine from the address 40E2C0, till a RET then it sends a result to EAX. If EAX contains the correct value, JNE shows us that our software is correctly registered.

Some may think to change the conditional jump JNE at 40874B to an unconditional one, so this way we can force the software to say that we are registered, except that we have to input a serial number each time we close and start the software.

But at this stage we know that everything is around EAX's value, and by forcing its value to have either 1 or 0 in the beginning of the routine started by the CALL (in our case at address 40E2C0), we will force our software to be registered or not.

In Assembly language, MOV is the instruction that tells an x86/IA-32 processor to move an immediate value into a register. So to force EAX to have 0, we can put mov eax, 0 Or mov al, 0.

Which means, all that the routine under the CALL must do is to move/put 0 into EAX; in other words, it tells the software that the serial number given is correct without filling it, and then sends the correct value to EAX.

In our example, the CALL at address 408740 calls this serial check routine:

[plain]:0040E2C0 81EC340400 sub esp, 434

:0040E2xx 8B0D605F4500 Mov ecx, dword ptr ds:[455F60]


******** Cutting more instructions ***********


:0040Exxx C3 RET ' Note that every routine called by a CALL is ended by a RET (C3 hex)[/plain]

By changing sub esp, 434 and mov ecx, dword ptr ds:[455F60] to mov eax, 1 and RET

[plain]0040E2C0 B801000000 mov, eax,1

:0040E2xx C3 RET

:0040E2xx 0D ?

:0040E2xx 60 ?

:0040E2xx 5F ?

:0040E2xx 45 ?

:0040E2xx 00 ?[/plain]

We get fully registered software without knowing how a correct serial number is calculated and without even having a correct one.

We can fill instructions after RET with NOPs (90 hex), but it's not necessary since they will not be executed.

More details about Mov picked from Wikipedia:

The binary code for this instruction is 10110 followed by a 3-bit identifier for which register to use. The identifier for the AL register is 000, so the following machine code loads the AL register with the data 01100001 what gives : 10110000 01100001

This binary computer code can be made more human-readable by expressing it in hexadecimal as follows: B0 61

Here, B0 means 'Move a copy of the following value into AL', and 61 is a hexadecimal representation of the value 01100001, which is 97 in decimal. Intel assembly language provides the mnemonic MOV (an abbreviation of move) for instructions such as this, so the machine code above can be written as follows in assembly language, complete with an explanatory comment if required, after the semicolon. This is much easier to read and to remember:

[plain]MOV AL, 61h ; Load AL with 97 decimal (61 hex)[/plain]


Soufiane Tahiri
Soufiane Tahiri

Soufiane Tahiri is is an InfoSec Institute contributor and computer security researcher, specializing in reverse code engineering and software security. He is also founder of and practiced reversing for more then 8 years. Dynamic and very involved, Soufiane is ready to catch any serious opportunity to be part of a workgroup.

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