# CSE4303 Introduction to Computer Security (Lecture 18)

> Due to lack of my attention, this lecture note is generated by AI to create continuations of the previous lecture note. I kept this warning because the note was created by AI.

#### Software security

### Overview

#### Outline

- Context
- Prominent software vulnerabilities and exploits
- Buffer overflows
  - Background: C code, compilation, memory layout, execution
  - Baseline exploit
  - Challenges
  - Defenses, countermeasures, counter-countermeasures

### Buffer overflows

#### All programs are stored in memory

- The process's view of memory is that it owns all of it.
- For a `32-bit` process, the virtual address space runs from:
  - `0x00000000`
  - to `0xffffffff`
- In reality, these are virtual addresses.
  - The OS and CPU map them to physical addresses.

#### The instructions themselves are in memory

- Program text is also stored in memory.
- The slide shows instructions such as:

```asm
0x4c2 sub $0x224,%esp
0x4c1 push %ecx
0x4bf mov %esp,%ebp
0x4be push %ebp
```

- Important point:
  - code and data are both memory-resident
  - control flow therefore depends on values stored in memory

#### Data's location depends on how it's created

- Static initialized data example

```c
static const int y = 10;
```

- Static uninitialized data example

```c
static int x;
```

- Command-line arguments and environment are set when the process starts.
- Stack data appears when functions run.

```c
int f() {
    int x;
    ...
}
```

- Heap data appears at runtime.

```c
malloc(sizeof(long));
```

- Summary from the slide
  - Known at compile time
    - text
    - initialized data
    - uninitialized data
  - Set when process starts
    - command line and environment
  - Runtime
    - stack
    - heap

#### We are going to focus on runtime attacks

- Stack and heap grow in opposite directions.
- Compiler-generated instructions adjust the stack size at runtime.
- The stack pointer tracks the active top of the stack.
- Repeated `push` instructions place values onto the stack.
- The slides use the sequence:
  - `push 1`
  - `push 2`
  - `push 3`
  - `return`
- Heap allocation is apportioned by the OS and managed in-process by `malloc`.
- The lecture says: focusing on the stack for now.

```text
0x00000000                              0xffffffff
Heap  --------------------------------->     <--------------------------------- Stack
```

#### Stack layout when calling functions

Questions asked on the slide:

- What do we do when we call a function?
  - What data need to be stored?
  - Where do they go?
- How do we return from a function?
  - What data need to be restored?
  - Where do they come from?

Example used in the slide:

```c
void func(char *arg1, int arg2, int arg3)
{
    char loc1[4];
    int loc2;
    int loc3;
}
```

Important layout points:

- Arguments are pushed in reverse order of code.
- Local variables are pushed in the same order as they appear in the code.
- The slide then introduces two unknown slots between locals and arguments.

#### Accessing variables

Example:

```c
void func(char *arg1, int arg2, int arg3)
{
    char loc1[4];
    int loc2;
    int loc3;
    ...
    loc2++;
    ...
}
```

Question from the slide:
- Where is `loc2`?

Step-by-step answer developed in the slides:

- Its absolute address is undecidable at compile time.
- We do not know exactly where `loc2` is in absolute memory.
- We do not know how many arguments there are in general.
- But `loc2` is always a fixed offset before the frame metadata.
- This motivates the frame pointer.

Definitions from the slide:

- Stack frame
  - the current function call's region on the stack
- Frame pointer
  - `%ebp`
- Example answer
  - `loc2` is at `-8(%ebp)`

#### Notation

- `%ebp`
  - a memory address stored in the frame-pointer register
- `(%ebp)`
  - the value at memory address `%ebp`
  - like dereferencing a pointer

The slide sequence then shows:

```asm
pushl %ebp
movl %esp, %ebp
```

- Meaning:
  - first save the old frame pointer on the stack
  - then set the new frame pointer to the current stack pointer

#### Returning from functions

Example caller:

```c
int main()
{
    ...
    func("Hey", 10, -3);
    ...
}
```

Questions from the slides:

- How do we restore `%ebp`?
- How do we resume execution at the correct place?

Slide answers:

- Push `%ebp` before locals.
- Set `%ebp` to current `%esp`.
- Set `%ebp` to `(%ebp)` at return.
- Push next `%eip` before `call`.
- Set `%eip` to `4(%ebp)` at return.

#### Stack and functions: Summary

- Calling function
  - push arguments onto the stack in reverse order
  - push the return address
    - the address of the instruction that should run after control returns
  - jump to the function's address
- Called function
  - push old frame pointer `%ebp` onto the stack
  - set frame pointer `%ebp` to current `%esp`
  - push local variables onto the stack
  - access locals as offsets from `%ebp`
- Returning function
  - reset previous stack frame
    - `%ebp = (%ebp)`
  - jump back to return address
    - `%eip = 4(%ebp)`

#### Quick overview (again)

- Buffer
  - contiguous set of a given data type
  - common in C
    - all strings are buffers of `char`
- Overflow
  - put more into the buffer than it can hold
- Question
  - where does the extra data go?
- Slide answer
  - now that we know memory layouts, we can reason about where the overwrite lands

#### A buffer overflow example

Example 1 from the slide:

```c
void func(char *arg1)
{
    char buffer[4];
    strcpy(buffer, arg1);
    ...
}

int main()
{
    char *mystr = "AuthMe!";
    func(mystr);
    ...
}
```

Step-by-step effect shown in the slides:

- Initial stack region includes:
  - `buffer`
  - saved `%ebp`
  - saved `%eip`
  - `&arg1`
- First 4 bytes copied:
  - `A u t h`
- Remaining bytes continue writing:
  - `M e ! \0`
- Because `strcpy` keeps copying until it sees `\0`, bytes go past the end of the buffer.
- In the example, upon return:
  - `%ebp` becomes `0x0021654d`
- Result:
  - segmentation fault
  - shown as `SEGFAULT (0x00216551)` in the slide sequence

#### A buffer overflow example: changing control data vs. changing program data

Example 2 from the slide:

```c
void func(char *arg1)
{
    int authenticated = 0;
    char buffer[4];
    strcpy(buffer, arg1);
    if (authenticated) { ... }
}

int main()
{
    char *mystr = "AuthMe!";
    func(mystr);
    ...
}
```

Step-by-step effect shown in the slides:

- Initial stack contains:
  - `buffer`
  - `authenticated`
  - saved `%ebp`
  - saved `%eip`
  - `&arg1`
- Overflow writes:
  - `A u t h` into `buffer`
  - `M e ! \0` into `authenticated`
- Result:
  - code still runs
  - user now appears "authenticated"

Important lesson:
- A buffer overflow does not need to crash.
- It may silently change program data or logic.

#### `gets` vs `fgets`

Unsafe function shown in the slide:

```c
void vulnerable()
{
    char buf[80];
    gets(buf);
}
```

Safer version shown in the slide:

```c
void safe()
{
    char buf[80];
    fgets(buf, 64, stdin);
}
```

Even safer pattern from the next slide:

```c
void safer()
{
    char buf[80];
    fgets(buf, sizeof(buf), stdin);
}
```

Reference from slide:
- [List of vulnerable C functions](https://security.web.cern.ch/security/recommendations/en/codetools/c.shtml)

#### User-supplied strings

- In the toy examples, the strings are constant.
- In reality they come from users in many ways:
  - text input
  - packets
  - environment variables
  - file input
- Validating assumptions about user input is extremely important.

#### What's the worst that could happen?

Using:

```c
char buffer[4];
strcpy(buffer, arg1);
```

- `strcpy` will let you write as much as you want until a `\0`.
- If attacker-controlled input is long enough, the memory past the buffer becomes "all ours" from the attacker's perspective.
- That raises the key question from the slide:
  - what could you write to memory to wreak havoc?

#### Code injection

- Title-only transition slide.
- It introduces the move from accidental overwrite to deliberate attacker payloads.

#### High-level idea

Example used in the slide:

```c
void func(char *arg1)
{
    char buffer[4];
    sprintf(buffer, arg1);
    ...
}
```

Two-step plan shown in the slides:

- 1. Load my own code into memory.
- 2. Somehow get `%eip` to point to it.

The slide sequence draws this as:
- vulnerable buffer on stack
- attacker-controlled bytes placed in memory
- `%eip` redirected toward those bytes

#### This is nontrivial

- Pulling off this attack requires getting a few things really right, and some things only sorta right.
- The lecture says to think about what is tricky about the attack.
- Main security idea:
  - the key to defending it is to make the hard parts really hard

#### Challenge 1: Loading code into memory

- The attacker payload must be machine-code instructions.
  - already compiled
  - ready to run
- We have to be careful in how we construct it.
  - It cannot contain all-zero bytes.
    - otherwise `sprintf`, `gets`, `scanf`, and similar routines stop copying
  - It cannot make use of the loader.
    - because we are injecting the bytes directly
  - It cannot use the stack.
    - because we are in the process of smashing it
- The lecture then gives the name:
  - shellcode

#### What kind of code would we want to run?

- Goal: full-purpose shell
  - code to launch a shell is called shellcode
  - it is nontrivial to write shellcode that works as injected code
    - no zeroes
    - cannot use the stack
    - no loader dependence
  - there are many shellcodes already written
  - there are even competitions for writing the smallest shellcode
- Goal: privilege escalation
  - ideally, attacker goes from guest or non-user to root

#### Shellcode

High-level C version shown in the slides:

```c
#include <stdio.h>
int main() {
    char *name[2];
    name[0] = "/bin/sh";
    name[1] = NULL;
    execve(name[0], name, NULL);
}
```

Assembly version shown in the slides:

```asm
xorl %eax, %eax
pushl %eax
pushl $0x68732f2f
pushl $0x6e69622f
movl %esp, %ebx
pushl %eax
...
```

Machine-code bytes shown in the slides:

```text
"\x31\xc0"
"\x50"
"\x68""//sh"
"\x68""/bin"
"\x89\xe3"
"\x50"
...
```

Important point from the slide:
- those machine-code bytes can become part of the attacker's input

#### Challenge 2: Getting our injected code to run

- We cannot insert a fresh "jump into my code" instruction.
- We must use whatever code is already running.

#### Hijacking the saved `%eip`

- Strategy:
  - overwrite the saved return address
  - make it point into the injected bytes
- Core idea:
  - when the function returns, the CPU loads the overwritten return address into `%eip`

Question raised by the slides:
- But how do we know the address?

Failure mode shown in the slide sequence:
- if the guessed address is wrong, the CPU tries to execute data bytes
- this is most likely not valid code
- result:
  - invalid instruction
  - CPU "panic" / crash

#### Challenge 3: Finding the return address

- If we do not have the code, we may not know how far the buffer is from the saved `%ebp`.
- One approach:
  - try many different values
- Worst case:
  - `2^32` possible addresses on `32-bit`
  - `2^64` possible addresses on `64-bit`
- But without address randomization:
  - the stack always starts from the same fixed address
  - the stack grows, but usually not very deeply unless heavily recursive

#### Improving our chances: nop sleds

- `nop` is a single-byte instruction.
- Definition:
  - it does nothing except move execution to the next instruction
- NOP sled idea:
  - put a long sequence of `nop` bytes before the real malicious code
  - now jumping anywhere in that region still works
  - execution slides down into the payload

Why this helps:
- it increases the chance that an approximate address guess still succeeds
- the slides explicitly state:
  - now we improve our chances of guessing by a factor of `#nops`

```text
[padding][saved return address guess][nop nop nop ...][malicious code]
```

#### Putting it all together

- Payload components shown in the slides:
  - padding
  - guessed return address
  - NOP sled
  - malicious code
- Constraint noted by the lecture:
  - input has to start wherever the vulnerable `gets` / similar function begins writing

#### Buffer overflow defense #1: use secure bounds-checking functions

- User-level protection
- Replace unbounded routines with bounded ones.
- Prefer secure languages where possible:
  - Java
  - Rust
  - etc.

#### Buffer overflow defense #2: Address Space Layout Randomization (ASLR)

- Randomize starting address of program regions.
- Goal:
  - prevent attacker from guessing / finding the correct address to put in the return-address slot
- OS-level protection

#### Buffer overflow counter-technique: NOP sled

- Counter-technique against uncertain addresses
- By jumping somewhere into a wide sled, exact address knowledge becomes less necessary

#### Buffer overflow defense #3: Canary

- Put a guard value between vulnerable local data and control-flow data.
- If overflow changes the canary, the program can detect corruption before returning.
- OS-level / compiler-assisted protection in the lecture framing

#### Buffer overflow defense #4: No-execute bits (NX)

- Mark the stack as not executable.
- Requires hardware support.
- OS / hardware-level protection

#### Buffer overflow counter-technique: ret-to-libc and ROP

- Code in the C library is already stored at consistent addresses.
- Attacker can find code in the C library that has the desired effect.
  - possibly heavily fragmented
- Then return to the necessary address or addresses in the proper order.
- This is the motivation behind:
  - `ret-to-libc`
  - Return-Oriented Programming (ROP)

We will continue from defenses / exploitation follow-ups in the next lecture.