By using the objdump command I figured that the address 0x02a8 in memory contains start the path /lib64/ld-linux-x86-64.so.2, and this path ends with a 0x00 byte, due to the C standard.
So I tried to write a simple C program that will print this line (I used a sample from the book "RE for beginners" by Denis Yurichev - page 24):
#include <stdio.h>
int main(){
printf(0x02a8);
return 0;
}
But I was disappointed to get a segmentation fault instead of the expected /lib64/ld-linux-x86-64.so.2 output.
I find it strange to use such a "fast" call of printf without specifiers or at least pointer cast, so I tried to make the code more natural:
#include <stdio.h>
int main(){
char *p = (char*)0x02a8;
printf(p);
printf("\n");
return 0;
}
And after running this I still got a segmentation fault.
I don't believe this is happening because of restricted memory areas, because in the book it all goes well at the 1st try. I am not sure, maybe there is something more that wasn't mentioned in that book.
So need some clear explanation of why the segmentation faults keep happening every time I try running the program.
I'm using the latest fully-upgraded Kali Linux release.
Disappointing to see that your "RE for beginners" book does not go into the basics first, and spits out this nonsense. Nonetheless, what you are doing is obviously wrong, let me explain why.
Normally on Linux, GCC produces ELF executables that are position independent. This is done for security purposes. When the program is run, the operating system is able to place it anywhere in memory (at any address), and the program will work just fine. This technique is called Address Space Layout Randomization, and is a feature of the operating system that nowdays is enabled by default.
Normally, an ELF program would have a "base address", and would be loaded exactly at that address in order to work. However, in case of a position independent ELF, the "base address" is set to 0x0, and the operating system and the interpreter decide where to put the program at runtime.
When using objdump on a position independent executable, every address that you see is not a real address, but rather, an offset from the base of the program (that will only be known at runtime). Therefore it is not possible to know the position of such a string (or any other variable) at runtime.
If you want the above to work, you will have to compile an ELF that is not position independent. You can do so like this:
gcc -no-pie -fno-pie prog.c -o prog
It no longer works like that. The 64-bit Linux executables that you're likely using are position-independent and they're loaded into memory at an arbitrary address. In that case ELF file does not contain any fixed base address.
While you could make a position-dependent executable as instructed by Marco Bonelli it is not how things work for arbitrary executables on modern 64-bit linuxen, so it is more worthwhile to learn to do this with position-independent ones, but it is a bit trickier.
This worked for me to print ELF i.e. the elf header magic, and the interpreter string. This is dirty in that it probably only works for a small executable anyway.
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
int main(){
// convert main to uintptr_t
uintptr_t main_addr = (uintptr_t)main;
// clear bottom 12 bits so that it points to the beginning of page
main_addr &= ~0xFFFLLU;
// subtract one page so that we're in the elf headers...
main_addr -= 0x1000;
// elf magic
puts((char *)main_addr);
// interpreter string, offset from hexdump!
puts((char *)main_addr + 0x318);
}
There is another trick to find the beginning of the ELF executable in memory: the so-called auxiliary vector and getauxval:
The getauxval() function retrieves values from the auxiliary vector,
a mechanism that the kernel's ELF binary loader uses to pass certain
information to user space when a program is executed.
The location of the ELF program headers in memory will be
#include <sys/auxv.h>
char *program_headers = (char*)getauxval(AT_PHDR);
The actual ELF header is 64 bytes long, and the program headers start at byte 64 so if you subtract 64 from this you will get a pointer to the magic string again, therefore our code can be simplified to
#include <stdio.h>
#include <inttypes.h>
#include <sys/auxv.h>
int main(){
char *elf_header = (char *)getauxval(AT_PHDR) - 0x40;
puts(elf_header + 0x318); // or whatever the offset was in your executable
}
And finally, an executable that figures out the interpreter position from the ELF headers alone, provided that you've got a 64-bit ELF, magic numbers from Wikipedia...
#include <stdio.h>
#include <inttypes.h>
#include <sys/auxv.h>
int main() {
// get pointer to the first program header
char *ph = (char *)getauxval(AT_PHDR);
// elf header at this position
char *elfh = ph - 0x40;
// segment type 0x3 is the interpreter;
// program header item length 0x38 in 64-bit executables
while (*(uint32_t *)ph != 3) ph += 0x38;
// the offset is 64 bits at 0x8 from the beginning of the
// executable
uint64_t offset = *(uint64_t *)(ph + 0x8);
// print the interpreter path...
puts(elfh + offset);
}
I guess it segfaults because of the way you use printf: you dont use the format parameter how it is designed to be.
When you want to use the printf function to read data the first argument it takes is a string that will format how the display will work int printf(char *fmt , ...) "the ... represent the data you want to display accordingly to the format string parameter
so if you want to print a string
//format as text
printf("%s\n", pointer_to_beginning_of_string);
//
If this does not work cause it probably will it is because you are trying to read memory that you are not supposed to access.
try adding extra flags " -Werror -Wextra -Wall -pedantic " with your compiler and show us the errors please.
I've been writing an OS using this tutorial. I am at the part where
the boot loader is completed and C is used for programming (and then linked together ...). But that just as a note, I believe the problem I have is related to gcc.
I build an i386-elf cross compiler for the OS. And everything works fine, I can execute my code everything works. Except that all global variables are initialized zero, although I provided a default value.
int test_var = 1234;
// yes, void main() is correct (the boot-loader will call this)
void main() {}
If I debug this code with GDB, I get: (gcc-7.1.0, target: i328-elf)
(gdb) b main
Breakpoint 1 at 0x1554: file src/kernel/main.c, line 11.
(gdb) c
Continuing.
Breakpoint 1, main () at src/kernel/main.c:11
11 void main() {
(gdb) p test_var
$1 = 0
If i run the same code on my local machine (gcc-6.3.0, target: x86_64), it prints 1234.
My question is: Did I misconfigure gcc, is this a mistake in my OS, is this a known problem? I couldn't find anything about it.
My entire source-code: link
I use the following commands to compile my stuff:
# ...
i386-elf-gcc -g -ffreestanding -Iinclude/ -c src/kernel/main.c -o out/kernel/main.o
# ...
i386-elf-ld -e 0x1000 -Ttext 0x1000 -o out/kernel.elf out/kernel_entry.o out/kernel/main.o # some other stuff ...
i386-elf-objcopy -O binary out/kernel.elf out/kernel.bin
cat out/boot.bin out/kernel.bin > out/os.bin
qemu-system-i386 -drive "format=raw,file=out/os.bin"
EDIT: As #EugeneSh. suggested here some logic to make sure, that it's not removed:
#include <cpu/types.h>
#include <cpu/isr.h>
#include <kernel/print.h>
#include <driver/vga.h>
int test_var = 1234;
void main() {
vga_text_init();
switch (test_var) {
case 1234: print("That's correct"); break;
case 0: print("It's zero"); break;
// I don't have a method like atoi() in place, I would use
// GDB to get the value
default: print("It's something else");
}
}
Sadly it prints It's zero
Compiler never clears uninitialized global variables to zero, its logic in built inside loader,
So when you allocate memory for data segment then it size contains bss section also. So you have to check bss section offset, alignment & size withing data segment and memset() them to '0'.
As you are writing your OS so may be all the library routines are not available so better write memset() function using assembly.
I wrote the simple C program (test.c) below:-
#include<stdio.h>
int main()
{
return 0;
}
and executed the follwing to understand size changes in .bss segment.
gcc test.c -o test
size test
The output came out as:-
text data bss dec hex filename
1115 552 8 1675 68b test
I didn't declare anything globally or of static scope. So please explain why the bss segment size is of 8 bytes.
I made the following change:-
#include<stdio.h>
int x; //declared global variable
int main()
{
return 0;
}
But to my surprise, the output was same as previous:-
text data bss dec hex filename
1115 552 8 1675 68b test
Please explain.
I then initialized the global:-
#include<stdio.h>
int x=67; //initialized global variable
int main()
{
return 0;
}
The data segment size increased as expected, but I didn't expect the size of bss segment to reduce to 4 (on the contrary to 8 when nothing was declared). Please explain.
text data bss dec hex filename
1115 556 4 1675 68b test
I also tried the comands objdump, and nm, but they too showed variable x occupying .bss (in 2nd case). However, no change in bss size is shown upon size command.
I followed the procedure according to:
http://codingfox.com/10-7-memory-segments-code-data-bss/
where the outputs are coming perfectly as expected.
When you compile a simple main program you are also linking startup code.
This code is responsible, among other things, to init bss.
That code is the code that "uses" 8 bytes you are seeing in .bss section.
You can strip that code using -nostartfiles gcc option:
-nostartfiles
Do not use the standard system startup files when linking. The standard system libraries are used normally, unless -nostdlib or -nodefaultlibs is used
To make a test use the following code
#include<stdio.h>
int _start()
{
return 0;
}
and compile it with
gcc -nostartfiles test.c
Youll see .bss set to 0
text data bss dec hex filename
206 224 0 430 1ae test
Your first two snippets are identical since you aren't using the variable x.
Try this
#include<stdio.h>
volatile int x;
int main()
{
x = 1;
return 0;
}
and you should see a change in .bss size.
Please note that those 4/8 bytes are something inside the start-up code. What it is and why it varies in size isn't possible to tell without digging into all the details of mentioned start-up code.
I don't understand why the address of the variables changes based on how I run the program. I have this simple code in C:
#include <stdlib.h>
#include <stdio.h>
int main(){
char buffer[100];
gets(buffer);
printf("buf: %p\n", (void*)&buffer);
printf(buffer);
return 0;
}
before compiling I used this command to disable the randomization:
echo 0 | sudo tee /proc/sys/kernel/randomize_va_space
but this is what happens:
image
in the first execution the address of buffer is bffff084 in the second is bffff024
This happens because of Address Space Layout Randomization. Each time program starts, random number is added to stack address so it would be harder to predict absolute addresses of buffers and write injectable shell code.
Since your program is vulnerable to buffer overflow (due to usage of gets and statically allocated buffer), it does its work.
I'm trying to figure out how to execute machine code stored in memory.
I have the following code:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char* argv[])
{
FILE* f = fopen(argv[1], "rb");
fseek(f, 0, SEEK_END);
unsigned int len = ftell(f);
fseek(f, 0, SEEK_SET);
char* bin = (char*)malloc(len);
fread(bin, 1, len, f);
fclose(f);
return ((int (*)(int, char *)) bin)(argc-1, argv[1]);
}
The code above compiles fine in GCC, but when I try and execute the program from the command line like this:
./my_prog /bin/echo hello
The program segfaults. I've figured out the problem is on the last line, as commenting it out stops the segfault.
I don't think I'm doing it quite right, as I'm still getting my head around function pointers.
Is the problem a faulty cast, or something else?
You need a page with write execute permissions. See mmap(2) and mprotect(2) if you are under unix. You shouldn't do it using malloc.
Also, read what the others said, you can only run raw machine code using your loader. If you try to run an ELF header it will probably segfault all the same.
Regarding the content of replies and downmods:
1- OP said he was trying to run machine code, so I replied on that rather than executing an executable file.
2- See why you don't mix malloc and mman functions:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <sys/mman.h>
int main()
{
char *a=malloc(10);
char *b=malloc(10);
char *c=malloc(10);
memset (a,'a',4095);
memset (b,'b',4095);
memset (c,'c',4095);
puts (a);
memset (c,0xc3,10); /* return */
/* c is not alligned to page boundary so this is NOOP.
Many implementations include a header to malloc'ed data so it's always NOOP. */
mprotect(c,10,PROT_READ|PROT_EXEC);
b[0]='H'; /* oops it is still writeable. If you provided an alligned
address it would segfault */
char *d=mmap(0,4096,PROT_READ|PROT_WRITE|PROT_EXEC,MAP_PRIVATE|MAP_ANON,-1,0);
memset (d,0xc3,4096);
((void(*)(void))d)();
((void(*)(void))c)(); /* oops it isn't executable */
return 0;
}
It displays exactly this behavior on Linux x86_64 other ugly behavior sure to arise on other implementations.
Using malloc works fine.
OK this is my final answer, please note I used the orignal poster's code.
I'm loading from disk, the compiled version of this code to a heap allocated area "bin", just as the orignal code did (the name is fixed not using argv, and the value 0x674 is from;
objdump -F -D foo|grep -i hoho
08048674 <hohoho> (File Offset: 0x674):
This can be looked up at run time with the BFD (Binary File Descriptor library) or something else, you can call other binaries (not just yourself) so long as they are statically linked to the same set of lib's.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/mman.h>
unsigned char *charp;
unsigned char *bin;
void hohoho()
{
printf("merry mas\n");
fflush(stdout);
}
int main(int argc, char **argv)
{
int what;
charp = malloc(10101);
memset(charp, 0xc3, 10101);
mprotect(charp, 10101, PROT_EXEC | PROT_READ | PROT_WRITE);
__asm__("leal charp, %eax");
__asm__("call (%eax)" );
printf("am I alive?\n");
char *more = strdup("more heap operations");
printf("%s\n", more);
FILE* f = fopen("foo", "rb");
fseek(f, 0, SEEK_END);
unsigned int len = ftell(f);
fseek(f, 0, SEEK_SET);
bin = (char*)malloc(len);
printf("read in %d\n", fread(bin, 1, len, f));
printf("%p\n", bin);
fclose(f);
mprotect(&bin, 10101, PROT_EXEC | PROT_READ | PROT_WRITE);
asm volatile ("movl %0, %%eax"::"g"(bin));
__asm__("addl $0x674, %eax");
__asm__("call %eax" );
fflush(stdout);
return 0;
}
running...
co tmp # ./foo
am I alive?
more heap operations
read in 30180
0x804d910
merry mas
You can use UPX to manage the load/modify/exec of a file.
P.S. sorry for the previous broken link :|
It seems to me you're loading an ELF image and then trying to jump straight into the ELF header? http://en.wikipedia.org/wiki/Executable_and_Linkable_Format
If you're trying to execute another binary, why don't you use the process creation functions for whichever platform you're using?
An typical executable file has:
a header
entry code that is called before main(int, char **)
The first means that you can't generally expect byte 0 of the file to be executable; intead, the information in the header describes how to load the rest of the file in memory and where to start executing it.
The second means that when you have found the entry point, you can't expect to treat it like a C function taking arguments (int, char **). It may, perhaps, be usable as a function taking no paramters (and hence requiring nothing to be pushed prior to calling it). But you do need to populate the environment that will in turn be used by the entry code to construct the command line strings passed to main.
Doing this by hand under a given OS would go into some depth which is beyond me; but I'm sure there is a much nicer way of doing what you're trying to do. Are you trying to execute an external file as a on-off operation, or load an external binary and treat its functions as part of your program? Both are catered for by the C libraries in Unix.
It is more likely that that it is the code that is jumped to by the call through function-pointer that is causing the segfault rather than the call itself. There is no way from the code you have posted to determine that that code loaded into bin is valid. Your best bet is to use a debugger, switch to assembler view, break on the return statement and step into the function call to determine that the code you expect to run is indeed running, and that it is valid.
Note also that in order to run at all the code will need to be position independent and fully resolved.
Moreover if your processor/OS enables data execution prevention, then the attempt is probably doomed. It is at best ill-advised in any case, loading code is what the OS is for.
What you are trying to do is something akin to what interpreters do. Except that an interpreter reads a program written in an interpreted language like Python, compiles that code on the fly, puts executable code in memory and then executes it.
You may want to read more about just-in-time compilation too:
Just in time compilation
Java HotSpot JIT runtime
There are libraries available for JIT code generation such as the GNU lightning and libJIT, if you are interested. You'd have to do a lot more than just reading from file and trying to execute code, though. An example usage scenario will be:
Read a program written in a scripting-language (maybe
your own).
Parse and compile the source into an
intermediate language understood by
the JIT library.
Use the JIT library to generate code
for this intermediate
representation, for your target platform's CPU.
Execute the JIT generated code.
And for executing the code you'd have to use techniques such as using mmap() to map the executable code into the process's address space, marking that page executable and jumping to that piece of memory. It's more complicated than this, but its a good start in order to understand what's going on beneath all those interpreters of scripting languages such as Python, Ruby etc.
The online version of the book "Linkers and Loaders" will give you more information about object file formats, what goes on behind the scenes when you execute a program, the roles of the linkers and loaders and so on. It's a very good read.
You can dlopen() a file, look up the symbol "main" and call it with 0, 1, 2 or 3 arguments (all of type char*) via a cast to pointer-to-function-returning-int-taking-0,1,2,or3-char*
Use the operating system for loading and executing programs.
On unix, the exec calls can do this.
Your snippet in the question could be rewritten:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main(int argc, char* argv[])
{
return execv(argv[1],argv+2);
}
Executable files contain much more than just code. Header, code, data, more data, this stuff is separated and loaded into different areas of memory by the OS and its libraries. You can't load a program file into a single chunk of memory and expect to jump to it's first byte.
If you are trying to execute your own arbitrary code, you need to look into dynamic libraries because that is exactly what they're for.