*This is a text version of a talk that I gave at PyCon AU 2019.*

Let’s say you’ve got this program:

```
pie_price = 3.14
num_pies = int(input("How many pies?"))
pie_owing = pie_price * num_pies
if pie_owing > 10:
print("You're over the pie budget")
```

How do you test that the line that prints “you’re over the pie budget” can run? One way is to just run the program, type in a large number, and verify that you see it.

But what if you couldn’t ask for input? For example, maybe this part of the code is buried deep within a larger process, and reaching it is tricky; maybe the code under test is operating in a continuous integration environment, and no user input is available. What do you do then to ensure that this line is reachable?

Why, *producing a formal proof*, of course. It’s the only sensible way.

In this post, we’ll walk through the theory and practice of using *symbolic execution*, a static analysis technique, for bug discovery. In particular, we’ll focus on a specific type of bug: how can we prove that a line of code is, or is not, reachable?

## How to solve it

Let’s start by reframing the question into something more formal:

Or, to put it another way:

Let’s work the problem by doing it by hand. Here’s the code again:

```
pie_price = 3.14
num_pies = int(input("How many pies?"))
pie_owing = pie_price * num_pies
if pie_owing > 10:
print("You're over the pie budget")
```

We know from the first line that `pie_price`

is 3.14. However, we don’t know the value of num_pies, because it depends upon user input. In order for any of the rest of the code to work, though, we need to have a label for the value stored in num_pies.

This is where the *symbol * in *symbolic execution* comes in: we’ll introduce a symbolic value – let’s call it 🥧 – and declare that the variable num_pies contains 🥧. We don’t know *anything* about what’s stored in 🥧, but we do know some facts *about* it.

Specifically, we know a single fact about it right now: 🥧 is an integer, which means that it supports any operation that other integers support: addition, multiplication, comparison, and so on.

Our next line, `pie_owing = pie_price * num_pies`

, has a similar problem: we can’t know the value of `pie_owing`

, because it’s the result of a multiplication between a known (or *concrete*) value and the symbolic value 🥧. So, what do we store in `pie_owing`

? We’ll store the entire *expression* `pie_price * 🥧`

in there.

The final line of code before the `print`

statement is a conditional: `if pie_owing > 10`

. If we proceed on to the next line, then it follows that the value of `pie_owing`

– whatever it is – is greater than 10.

We now have enough information to put together a collection of logical assertions that must be true in order to reach the `print`

statement. They are:

`pie_price`

= 3.14`num_pies`

= 🥧`pie_owing`

=`pie_price`

×`num_pies`

`pie_owing`

> 10

Great. Our next question is: can we demonstrate that these equations can all be true at the same time?

Could we even do it… *with a computer*?

## Proving it with Z3

The Z3 Theorem Prover is a library from Microsoft that’s capable of, among many other things, answering this problem. It also has bindings to lots of popular languages, including Python.

To answer our question, we’ll construct several equations that represent the constraints on the input that are in place when the `print`

line is reached, and feed them into a solver; we can then ask the solver to check to see if they can be true at the same time.

```
from z3 import *
# Create the solver
s = Solver()
# Declare our variables: "pie_price", which we know the
# value of, "num_pies", which we don't, and "pies_owing", which depends upon the values of the other two
pie_price = Real('pie_price')
num_pies = Int('num_pies')
pies_owing = pie_price * num_pies
# Assert that pie_price is equal to 3.14
s.add(pie_price == 3.14)
# Assert that pies_owing is greater than 10
s.add(pies_owing > 10)
# Ask if these these can be true at the same time
s.check() # returns "sat" - they can be!!
```

We’ve now demonstrated that in order to reach the line, `print("You're over the pie budget")`

, a set of equations must be true at the same time; additionally, Z3 indicates that they can indeed be. Therefore, we’ve proved that the line is reachable, and we never needed to ask the user for input.

Incidentally, we can ask Z3 to produce a *model* of its solution, which means it will produce a value for all of the variable in question, *including* num_pies – the value we’d ordinarily get from the user. That is, Z3 can produce a value for num_pies that would result in the print statement to run.

```
s.model()[num_pies] # 4
```

## Generating the Equations Automatically

In the previous example, we had to read through the code and manually produced the equations that are in place. Wouldn’t it be nicer, though, if we could have a system do this for us?

To do this, we’ll take advantage of the fact that Python is very easy to decompile into byte code. Using the `dis`

module, we can take any Python function, and produce the byte code that represents it. Converting the code to byte code is important, because byte code is significantly simpler, and easier to analyse.

Once we have the byte code, we need to find a way to determine the possible paths through the code that execution can take, depending on the inputs given the program. We then need to determine the constraints on the variables that affect the path; if at any point the constraints are not compatible with each other, the path is impossible. If *all* paths that reach a line of code are impossible, then the line of code is unreachable under any circumstance.

For example, consider this snippet of code:

```
i = 1
if i == 0:
print("Whoa!")
```

There is *theoretical*ly a path of execution that goes from line 1, through line 2, and ends at line 3, but if you think about it, it would require the variable `i`

to be equal to 0 and also to 1. This is impossible, and as a result, the path is impossible; because this is the *only* path that reaches line 3, that line is unreachable.

This means that our next problem is: given a block of code, how do we calculate the possible paths through that code?

## Basic Blocks and Control Flow Graphs

As before, let’s start with a chunk of code, which we’ll use as our example.

```
def test(a):
x = 0
if a > 0 and a < 5:
x = 1
b = a + 1
if x == 1 and b > 6:
print("Hello!")
```

Our question for this code is: can the final line of code, `print("Hello")`

, ever be reached? And can the process of discovering this be automated?

Let’s start by asking `dis`

for the byte code.

```
import dis
dis.dis(test)
```

This produces something like this (I’ve truncated it to the first few lines):

```
2 0 LOAD_CONST 1 (0)
2 STORE_FAST 1 (x)
3 4 LOAD_FAST 0 (a)
6 LOAD_CONST 1 (0)
8 COMPARE_OP 4 (>)
10 POP_JUMP_IF_FALSE 24
12 LOAD_FAST 0 (a)
14 LOAD_CONST 2 (5)
16 COMPARE_OP 0 (<)
18 POP_JUMP_IF_FALSE 24
4 20 LOAD_CONST 3 (1)
22 STORE_FAST 1 (x)
5 >> 24 LOAD_FAST 0 (a)
26 LOAD_CONST 3 (1)
```

Each one of these lines is a low-level instruction to the Python virtual machine. The Python VM is a stack machine, which means that the instructions work by pushing and popping values on a stack. For example, the `LOAD_CONST`

and `LOAD_FAST`

operations push values onto the stack (either a constant value or a value stored in a variable), while the `COMPARE_OP`

operation pops two values off the stack, compares them, and pushes the result back onto the stack. Additionally, certain instructions are responsible for controlling the flow of execution: the `POP_JUMP_IF_FALSE`

instruction pops a value off the stack, and if it evaluates to `False`

, jumps to a numbered instruction; if it evaluates to `True`

, it proceeds to the next instruction instead.

How, then, can we find the possible paths through the code? One popular approach is to decompose the stream of instructions into basic blocks: runs of instructions that are only ever entered at the start, and only ever exit at the end (that is, it is impossible for the program to jump to a point that’s in the middle of a basic block).

To determine these basic blocks, the instructions are scanned, and certain instructions are marked as *leaders*:

- The first instruction is a leader.
- Instructions that are the destination of a jump are leaders.
- Instructions following a conditional jump are leaders.
- Instructions following a ‘stop’ instruction are leaders.

Once you know the leaders, you can then group up the instructions according to the most recent leader.

Next up, you form the connections between the blocks. Blocks have successors (blocks they lead to), and predecessors (blocks that lead to them.)

- Blocks that end in an unconditional jump have one successor – the target of the jump.
- Blocks that end in a
*conditional*jump have two successors – the target of the jump, and the next instruction. - Blocks that end in a ‘stop’ instruction have no successors.
- All other blocks have a single successor: the following instruction’s block.

With these rules in mind, we can take the byte code for our example function, and figure out the blocks.

Given these blocks and the way they link together, we can generate the *control flow graph* of the program. This graph shows how the blocks connect, and allows us to find the paths that execution can take through the program.

We’re now ready to start testing the paths that lead to the `print("Hello")`

function call, which is the second-to-last basic block (it’s the blue block, second from the right of the above image.) For the purposes of this article, we’ll select one of them arbitrarily, and prove that the path is valid or not; the same steps apply for testing any path.

## Finding impossible paths

In order to perform normal execution of the code, Python steps through each instruction, and performs them as normal – loading data into memory, requesting that the system get input, and so on. However, this only works when we’re running the *entire* program, which includes all of the work done to decide what parameters to use when calling the function `test`

. When we perform symbolic execution, and are examining only *portions* of the program, we no longer have the ability to know concrete values for every variable.

Let’s take a closer look at the first basic block:

```
2 0 LOAD_CONST 1 (0)
2 STORE_FAST 1 (x)
3 4 LOAD_FAST 0 (a)
6 LOAD_CONST 1 (0)
8 COMPARE_OP 4 (>)
10 POP_JUMP_IF_FALSE 24
```

The third instruction in this disassembly loads the contents of the variable `a`

, and pushes it onto the stack. However, `a`

is a parameter to the function, which means it’s not possible to get a concrete value for the variable when considering the code in isolation.

This means that when we encounter the instruction `LOAD_FAST`

a, we need to introduce a new symbolic value. That’s not the only symbolic value we need to track, though: on lines 1 and 2, we load the number 0, and store it in the variable `x`

. This means that we need to *declare* to Z3 that the variable `x`

exists, and assert that it is equal to 0.

Additionally, if we’re testing a specific path through the code, we already know whether the `POP_JUMP_IF_FALSE`

will jump or not. In the case of our selected path, if we’re proceeding from the first block to the second, it means that we’re taking the path in which the value on the stack is `True`

. This mean that we also assert that the result of comparing if `a`

is greater than 1 is `True`

.

In effect, setting a variable now means creating and recording an assertion that the variable contains a certain value, and when encountering a conditional jump, we assert that its condition is true (if we’re taking the true path), or false (if we’re not).

We continue this execution, creating additional constraints on the values as we encounter instructions that interact with them, and at the end of each block, we feed them into Z3 and ask if the assertions are compatible with each other. If they’re not, then the path is impossible, and we try again with a different path. If all of the paths that reach a block are impossible, then that block is unreachable under any circumstances.

In the specific case of our example, the line `print("Hello")`

is unreachable. For it to be reached, it would require either the value of `a`

to be both greater than 5 *and* less than 5 at the same time.

## Conclusions

Symbolic execution is really fun and useful, but it isn’t without its drawbacks. In particular:

- It’s vulnerable to an explosions in the number of paths, especially when looping (and
*especially*if the code can potentially loop infinitely) - If the same region of memory is referred to by two variables, it can be challenging for the analyser to detect this condition
- Elements in a collection require special handling; do you treat the collection itself as a value, or the values in the collection as individual values?
- It’s a lot more challenging in dynamically typed situations, where you don’t necessarily know the operations that can be performed on the values that you receive.

Nonetheless, it’s a fascinating field to play in, and can be tremendously rewarding. The video of the talk that I gave at PyCon AU 2019 is embedded below.