# Reverse engineering¶

Reverse engineering is the process of analysing low level code and extracting higher level information(eg: code in higher level language, design information etc). Specifically, here it is about analysing assembly code in executables and understand their behaviour.

In this section, we will learn some reverse engineering but it is best learnt by a lot of practice: we can help you by giving some general guidelines and directions but without effort from your side, it is not possible to learn effectively. With that said, let's begin reversing some very simple Linux programs. Just like before, we will assume you are using a recent version of Ubuntu GNU/Linux(any LTS version >= 14.04).

# Tool setup¶

Reverse engineering requires use of several tools. The two most popular kinds of tools we will use are:

1. Disassemblers: These allow us to view the assembly code.
2. Debuggers: These allow us to analyse a program when it is running to inspect memory values, register values and more.

## Baby steps in reverse engineering¶

Alright let's get started with reverse engineering using the many tools you installed above. We will start with a very simple crackme challenge. A crackme is a binary which expects a specific input as answer and that input should be discovered by analysing the binary. Click here to download the crackme binary and let's get started.

As a first step, try running the binary without any arguments to understand something about it's behaviour. Open a terminal, navigate to the folder where you downloaded the binary and give the binary execute permission by running the following command:

 1 $chmod +x ./simplecrackme0x00  After the above step, run the binary using the following command and give it some inputs:  1 2 3 $ ./simplecrackme00 Enter answer: test Incorrect! Try again! 

Play around with the binary giving different inputs to gain some understanding of the binary for sometime and then return here.

From the several executions of the binary, we can conclude the following:

1. The binary takes some input from user.
2. The binary performs some kind of validation on the input.
3. Depending on the result of the validation, the binary concludes that the input is valid or not.

By multiple executions of the binary, we were able to infer useful information about the binary! This is an important step: executing the binary can reveal useful information. You can do this step as long as you can to infer as much useful information as possible. However, in majority of the cases, the amount of useful information you can derive from this kind of analysis is limited.

So now what should be our next step? That's right - we have to analyse the validation performed on the input to understand it and hopefully find out the correct input. There are several ways. We will use IDA Pro to see how helpful(powerful) it can be. When you run IDA Pro, you will get a screen as below: Figure: IDA startup screen.

Click on the "New" button, choose the binary you downloaded and click "Open" to load the binary into IDA Pro. On loading the binary, IDA Pro does a lot of preprocessing and finally displays a window similar to below: Figure: IDA screen after loading executable(click to expand).

The graph displayed is a nice representation of the execution of the assembly instructions in main. Such a graph is called as the Control Flow Graph(CFG) of the function: it represents the control flow execution as a graph. As you can see, there are 4 nodes in the graph. Each node is called a basic block. The arrows connecting the nodes indicate the flow of execution between the basic blocks. There are 3 kinds of arrows in the graph displayed in the CFG:

1. Blue arrows: These indicate that control always along this path.
2. Green arrows: These indicate that the control flows along this path only if a condition is true.
3. Red arrows: These indicate that the control flows along this path only if a condition is false.

On analysing the 4 basic blocks in the function main and using some lessons we learnt in the assembly programing section, we arrive at following conclusions:

1. At address 0x80484B5, the string "Enter answer:" is displayed by the printf function.
2. At address 0x80484C9, the user input is read by the scanf function.
3. The basic block 3 is executed if the input we supplied is wrong because the string "Incorrect! Try again!" is printed in that basic block. From the initial analysis by executing the binary, we found that "Incorrect! Try again!" is printed only if wrong input is supplied.
4. The basic block 2 is probably executed if we give the right input because the string "Correct!" is printed by instructions in this block.

Thus, we can further refine our earlier understanding of the binary as follows:

1. The program reads input at address 0x80484C9 and validates it few instructions later.
2. The program makes some condition checking at or near 0x80484D6 and prints "Correct" or "Incorrect" depending on the result of the condition check.
3. We want the message "Correct" to be displayed and so the condition check at 0x80484D6 should evaluate to false since only then is the message displayed.

Now with the above knowledge and your experience with assembly programming and using a debugger, can you analyse the binary and find out the correct answer? Attempt solving it on your own before looking at the solution. It's good practice to try: even if you can't solve, you will have learnt a lot by attempting.

1. Find out where is the input stored after being read by scanf.
2. We saw that some kind of validation of the input is performed. Try to find out where is the validation performed. After that, try to understand what validation is performed.

# Baby steps in reverse engineering detailed walkthrough¶

Let's now perform a detailed analysis of the given executable.

Warning

If you have not attempted finding the answer on your own, we recommend you do so before reading ahead. It is alright if you don't solve it on your own: you will learn a lot by attempting to solve it.

Note

This is one of the many approaches possible: this one uses only IDA Pro to find the answer. This can be solved using both IDA Pro and gdb or using gdb alone. In most reversing exercises, you will need to use both a disassembler and a debugger also to solve the challenges.

If we analyse main, we can see that at address 0x80484A6, the value 0x12345678 is stored somewhere in memory. If we carefully look at the instruction, we see that ESP register is used to compute the address and thus, this value is stored somewhere on the stack. More specifically, it is a local variable because it is stored in the stack frame of the function main (if this is not clear, go back to assembly programming and refresh this part). Let's convert the memory location [esp + 0x1c] into a variable. To do so, click on the instruction at 0x80484A6 and press the key 'k'. You will see that a new local variable has been added with the name "var_4". You can rename this variable to something more meaningful, say "unknown_number". There are two ways to do so:

1. Click on "var_4" and press 'n' key.
2. Right click on "var_4" and choose "rename".

The next 2 instructions at addresses 0x80484AE and 0x80484B5 are storing some value on the stack and calling the function puts. From our assembly programming sections, we recall that as per the function calling convention on Linux, the arguments are stored on the stack before the function is called. Here, we see that the function puts is called and as per the man page documentation for printf, the first argument is a pointer to a string as argument. IDA has very cleverly dereferenced the pointer and displays the string beside the copy instruction. Thus, we can conclude that puts call at address 0x80484B5 displays the string "Enter answer: ".

Let us know look at the next 4 instructions from address 0x80484BA to 0x80484C9. At 0x80484C9, we see that scanf is invoked and we know from our previous experience(and also from man page for scanf) that is accepts multiple arguments: the first argument is a format string and all the other arguments are addresses of variables for storing the input. Here, we see that only 2 arguments are passed to scanf and hence, we can conclude that scanf is reading 1 input from user. This is also what we observed from when executed the binary before analysing it in IDA Pro: it accepted one input and then validated it. Additionally, at address 0x80484C2, we see that the first argument is set and it happens to be the string "%ld". Based on this input format, we can conclude that scanf is reading a long integer from the user.

Exercise

Try to draw the stack frame just before scanf is called. Show which the arguments are present on stack, in what order and the type of each argument on the stack.

To summarize what we have discovered till now by analysing the binary in IDA:

1. The binary stores the value 0x12345678 in unknown_number.
2. The binary prints the string "Enter number: " using puts.
3. The binary accepts a long integer as input using scanf and stores the input value in user_input.

Warning

Be sure that you are clear with all 3 discoveries listed above. If you are not, go back and read the solution again till you are clear with all of them and only then continue below.

Now, let's look at the instructions from address 0x80484CE to 0x80484D6. In the instructions at address 0x80484CE and 0x80484D2, we see two local variables being accessed. Convert them to local variables by clicking on the instruction and pressing the key 'k'. You will see that IDA identifies as the two local variables we created earlier: user_input and unknown_number. In instruction at 0x804849E, we can see that the user_input is stored in the EAX register and in the following instruction, we see that the value in the EAX register is compared with unknown_number using cmp instruction. From the assembly programming section, recall that cmp instruction compares two values and sets few flags depending on the result of the compare operation. After the comparison between user_input and unknown_number, we see the jnz instruction to 0x80484D6. From assembly programming, recall that

1. jnz will jump if the zero flag is set.
2. The cmp instruction sets the zero flag if the 2 values being compared are equal.

To summarize the instructions 0x80484CE to 0x80484D6:

1. The values user_input and unknown_number are compared.
2. If they are equal, the "Correct!" is printed else "Incorrect!" is printed.

Thus, we have found what is the expected input: 0x12345678(305419896 in decimal), the value of unknown_number. Let's try giving the number as input:

 1 2 3 \$ ./simplecrackme0x00 Enter answer: 305419896 Correct! 

Great work! You just solved the crackme!