59

I'm looking at writing a JIT compiler for a hobby virtual machine I've been working on recently. I know a bit of assembly, (I'm mainly a C programmer. I can read most assembly with reference for opcodes I don't understand, and write some simple programs.) but I'm having a hard time understanding the few examples of self-modifying code I've found online.

This is one such example: http://asm.sourceforge.net/articles/smc.html

The example program provided does about four different modifications when run, none of which are clearly explained. Linux kernel interrupts are used several times, and aren't explained or detailed. (The author moved data into several registers before calling the interrupts. I assume he was passing arguments, but these arguments aren't explained at all, leaving the reader to guess.)

What I'm looking for is the simplest, most straightforward example in code of a self-modifying program. Something that I can look at, and use to understand how self-modifying code in x86 assembly has to be written, and how it works. Are there any resources you can point me to, or any examples you can give that would adequately demonstrate this?

I'm using NASM as my assembler.

EDIT: I'm also running this code on Linux.

Seki
  • 11,135
  • 7
  • 46
  • 70
jakogut
  • 4,409
  • 6
  • 29
  • 41
  • 1
    http://linux.die.net/man/2/mprotect should explain what the arguments for mprotect are. The function ID to call is passed in EAX and the next arguments are passed in EBX ECX and EDX. – KitsuneYMG Jan 27 '11 at 05:11
  • Related: [How to get c code to execute hex machine code?](https://stackoverflow.com/q/9960721) shows copying machine-code bytes into a page with write+exec permission, and calling a function in them. Including the necessary GNU C `__builtin___clear_cache` on the range (actually just syncs I-cache however is required on the target ISA.) – Peter Cordes Jan 10 '22 at 01:30

7 Answers7

62

wow, this turned out to be a lot more painful than I expected. 100% of the pain was linux protecting the program from being overwritten and/or executing data.

Two solutions shown below. And a lot of googling was involved so the somewhat simple put some instruction bytes and execute them was mine, the mprotect and aligning on page size was culled from google searches, stuff I had to learn for this example.

The self modifying code is straight forward, if you take the program or at least just the two simple functions, compile and then disassemble you will get the opcodes for those instructions. or use nasm to compile blocks of assembler, etc. From this I determined the opcode to load an immediate into eax then return.

Ideally you simply put those bytes in some ram and execute that ram. To get linux to do that you have to change the protection, which means you have to send it a pointer that is aligned on a mmap page. So allocate more than you need, find the aligned address within that allocation that is on a page boundary and mprotect from that address and use that memory to put your opcodes and then execute.

the second example takes an existing function compiled into the program, again because of the protection mechanism you cannot simply point at it and change bytes, you have to unprotect it from writes. So you have to back up to the prior page boundary call mprotect with that address and enough bytes to cover the code to be modified. Then you can change the bytes/opcodes for that function in any way you want (so long as you don't spill over into any function you want to continue to use) and execute it. In this case you can see that fun() works, then I change it to simply return a value, call it again and now it has been modified.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/mman.h>

unsigned char * testfun;

unsigned int fun(unsigned int a) {
    return (a + 13);
}

unsigned int fun2(void) {
    return (13);
}

int main(void) {
    unsigned int ra;
    unsigned int pagesize;
    unsigned char * ptr;
    unsigned int offset;

    pagesize = getpagesize();
    testfun = malloc(1023 + pagesize + 1);
    if (testfun == NULL) return (1);
    //need to align the address on a page boundary
    printf("%p\n", testfun);
    testfun = (unsigned char * )(((long) testfun + pagesize - 1) & ~(pagesize - 1));
    printf("%p\n", testfun);

    if (mprotect(testfun, 1024, PROT_READ | PROT_EXEC | PROT_WRITE)) {
        printf("mprotect failed\n");
        return (1);
    }

    //400687: b8 0d 00 00 00          mov    $0xd,%eax
    //40068d: c3                      retq

    testfun[0] = 0xb8;
    testfun[1] = 0x0d;
    testfun[2] = 0x00;
    testfun[3] = 0x00;
    testfun[4] = 0x00;
    testfun[5] = 0xc3;

    ra = ((unsigned int( * )()) testfun)();
    printf("0x%02X\n", ra);

    testfun[0] = 0xb8;
    testfun[1] = 0x20;
    testfun[2] = 0x00;
    testfun[3] = 0x00;
    testfun[4] = 0x00;
    testfun[5] = 0xc3;

    ra = ((unsigned int( * )()) testfun)();
    printf("0x%02X\n", ra);

    printf("%p\n", fun);
    offset = (unsigned int)(((long) fun) & (pagesize - 1));
    ptr = (unsigned char * )((long) fun & (~(pagesize - 1)));

    printf("%p 0x%X\n", ptr, offset);

    if (mprotect(ptr, pagesize, PROT_READ | PROT_EXEC | PROT_WRITE)) {
        printf("mprotect failed\n");
        return (1);
    }

    //for(ra=0;ra&lt;20;ra++) printf("0x%02X,",ptr[offset+ra]); printf("\n");

    ra = 4;
    ra = fun(ra);
    printf("0x%02X\n", ra);

    ptr[offset + 0] = 0xb8;
    ptr[offset + 1] = 0x22;
    ptr[offset + 2] = 0x00;
    ptr[offset + 3] = 0x00;
    ptr[offset + 4] = 0x00;
    ptr[offset + 5] = 0xc3;

    ra = 4;
    ra = fun(ra);
    printf("0x%02X\n", ra);

    return (0);
}
Jared Burrows
  • 54,294
  • 25
  • 151
  • 185
old_timer
  • 69,149
  • 8
  • 89
  • 168
  • 2
    not only Linux but most modern OSes also protect writable memory from executing – phuclv Nov 28 '13 at 11:35
  • 2
    Can this be done in Windows, i.e. unprotecting a page of RAM, or would we be stuck with blue screens of death? I want to use this method to create a self-modifying encryption system. – tentimes Dec 15 '13 at 09:27
  • The code worked fine on 32-bit Arch Linux, but failed on 64-bit RHEL (both the 64-bit ELF, of course, but also when using the 32-bit ELF). Don't know if this has to do with additional memory protection on RHEL or something else. The output was: ``` 0x9a00008 0x9a01000 mprotect failed ``` – Alexander Mar 13 '14 at 14:10
  • 1
    This isn't *self*-modifying code, it's just normal JIT into a buffer. Those `mov`-immediate instructions don't modify their own instruction bytes. – Peter Cordes Aug 31 '18 at 05:07
  • The technical problem here was not "SELF" modifying code, the technical problem here was protection, once you overcome protection then you can self modify to your hearts content. Wasnt meant to be a "SELF" modifying code answer. – old_timer Aug 31 '18 at 12:30
  • It is self modifying in the sense that this program modified its own memory space at run time. Call this JIT if you want, could easily argue your statement is JIT as well. Again the problem is protection not modification. – old_timer Aug 31 '18 at 12:32
  • 3
    @Alexander mprotect fails because the calculation for page alignment is broken for 64 bit. You need `long pagesize`. Otherwise addresses are 64 bit values and the bitmasks used are 32 bit. – MemAllox Jun 10 '19 at 11:56
12

Since you're writing a JIT compiler, you probably don't want self-modifying code, you want to generate executable code at runtime. These are two different things. Self-modifying code is code that is modified after it has already started running. Self-modifying code has a large performance penalty on modern processors, and therefore would be undesirable for a JIT compiler.

Generating executable code at runtime should be a simple matter of mmap()ing some memory with PROT_EXEC and PROT_WRITE permissions. You could also call mprotect() on some memory you allocated yourself, as dwelch did above.

Josh Haberman
  • 4,170
  • 1
  • 22
  • 43
  • Self modifying code doesn't always had performance penalties on modern processors. You have to be careful about what you change, and make sure the CPU cache is in sync and branch protection isn't altered. Changing those will tank your performance. – Beachhouse Dec 05 '14 at 18:41
  • 2
    if the self-modification happens relatively infrequently and/or on parts of the code that are not _currently_ being executed, is the temporary performance hit negligible? – Erik Kaplun Dec 14 '16 at 12:30
  • 3
    @ErikAllik: yeah, it's a one-time hit that costs a pipeline flush, maybe similar in cost to a memory-order mis-speculation. Maybe hundreds of cycles, so pretty easy to amortize of repeated use of the updated code. – Peter Cordes Aug 31 '18 at 05:10
4

A little bit simpler example based on the example above. Thanks to dwelch helped a lot.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <sys/mman.h>

char buffer [0x2000];
void* bufferp;

char* hola_mundo = "Hola mundo!";
void (*_printf)(const char*,...);

void hola()
{ 
    _printf(hola_mundo);
}

int main ( void )
{
    //Compute the start of the page
    bufferp = (void*)( ((unsigned long)buffer+0x1000) & 0xfffff000 );
    if(mprotect(bufferp, 1024, PROT_READ|PROT_EXEC|PROT_WRITE))
    {
        printf("mprotect failed\n");
        return(1);
    }
    //The printf function has to be called by an exact address
    _printf = printf;

    //Copy the function hola into buffer
    memcpy(bufferp,(void*)hola,60 //Arbitrary size);


    ((void (*)())bufferp)();  

    return(0);
}
felknight
  • 1,383
  • 1
  • 9
  • 25
4

This is written in AT&T assembly. As you can see from the execution of the program, output has changed because of self-modifying code.

Compilation: gcc -m32 modify.s modify.c

the -m32 option is used because the example works on 32 bit machines

Aessembly:

.globl f4
.data     

f4:
    pushl %ebp       #standard function start
    movl %esp,%ebp

f:
    movl $1,%eax # moving one to %eax
    movl $0,f+1  # overwriting operand in mov instuction over
                 # the new immediate value is now 0. f+1 is the place
                 # in the program for the first operand.

    popl %ebp    # standard end
    ret

C test-program:

 #include <stdio.h>

 // assembly function f4
 extern int f4();
 int main(void) {
 int i;
 for(i=0;i<6;++i) {
 printf("%d\n",f4());
 }
 return 0;
 }

Output:

1
0
0
0
0
0
4

I'm working on a self-modifying game to teach x86 assembly, and had to solve this exact problem. I used the following three libraries:

AsmJit + AsmTk for assembling: https://github.com/asmjit/asmjit + https://github.com/asmjit/asmtk UDIS86 for disassembling: https://github.com/vmt/udis86

Instructions are read with Udis86, the user can edit them as a string, and then AsmJit/AsmTk is used to assemble the new bytes. These can be written back to memory, and as other users have pointed out, the write-back requires using VirtualProtect on Windows or mprotect on Unix to fix the memory page permissions.

The code samples are a just a little long for StackOverflow, so I'll refer you to an article I wrote with code samples:

https://medium.com/squallygame/how-we-wrote-a-self-hacking-game-in-c-d8b9f97bfa99

A functioning repo is here (very light-weight):

https://github.com/Squalr/SelfHackingApp

Zachary Canann
  • 1,131
  • 2
  • 13
  • 23
3

You can also look at projects like GNU lightning. You give it code for a simplified RISC-type machine, and it generates correct machine dynamically.

A very real problem you should think about is interfacing with foreign libraries. You will probably need to support at least some system-level calls/operations for your VM to be useful. Kitsune's advice is a good start to get you thinking about system-level calls. You would probably use mprotect to ensure that the memory you have modified becomes legally executable. (@KitsuneYMG)

Some FFI allowing calls to dynamic libraries written in C should be sufficient to hide a lot of the OS specific details. All these issues can impact your design quite a bit, so it is best to start thinking about them early.

Kevin A. Naudé
  • 3,992
  • 19
  • 20
0

I've never written self-modifying code, although I have a basic understanding about how it works. Basically you write on memory the instructions you want to execute then jump there. The processor interpret those bytes you've written an instructions and (tries) to execute them. For example, viruses and anti-copy programs may use this technique.
Regarding the system calls, you were right, arguments are passed via registers. For a reference of linux system calls and their argument just check here.

BlackBear
  • 22,411
  • 10
  • 48
  • 86