Linux assembly x86 | trying to get stack value, syntax error - linux

I have been messing around with linux assembly on an x86 machine,
Basically my question is: I have pushed couple values into the stack moved the stack pointer into the base pointer and moved a value of 8 into a register to get a pushed value and in the end i wanted to get the value and put it into %ebx for the system call so i would get the value, but it seems to get an error. no clue why.
Error is: junk (%ebp) after register
Example:
.section .data
.section .text
.globl _start
_start:
pushl $50
pushl $20
movl %esp,%ebp
movl $8,%edx
movl %edx(%ebp),%ebx ## Supposed to be return value at system termination // PROBLEM HERE
movl $1,%eax ## System call
int $0x80 # Terminate program

I think part of the problem might be that in x86 the stack actually grows downwards, not up. You're adding to the base pointer, which is giving junk, where you have to subtract from it. I don't have an x86 machine so I can't test this, but have you tried something like movl -%edx(%ebp),%ebx?
Oops, I reversed the direction of the operands in my head. In this case, your stack looks like this:
1952 - ???
1948 - 20
1944 - 50 <- ebp <- esp
So when you take ebp+8, you aren't getting 20, you're getting address 1952, and you don't know what that contains.

Check out the links in https://stackoverflow.com/tags/x86/info. I updated them recently, and added the info about using gdb to single-step asm.
What do you mean "get an error"? Segmentation fault? Syntax error? (The normal syntax is (%ebp, %edx). Only numeric-constant displacements go outside the parens, e.g. -4(%ebp, %edx))
Also, if you're going to use stack frame pointers at all, do the mov %esp, %ebp after pushing any registers you want to preserve, but before pushing args to any functions you're going to call. However, there's no need to use %ebp that way at all, though. gcc defaults to -fomit-frame-pointer since 4.4 I think. It can make it easier to keep track of where your local variables are, if you're pushing/popping stuff.
You might want to just start with 64bit asm, instead of messing around with the obsolete x86 args-on-the-stack ABI.
This just made me think of what's probably wrong with your code. You're probably getting a segfault. (But you didn't say if it was that, syntax error, or something else.) Because you probably built your code in 64bit mode. Build a 32bit binary, or change your code to use %rsp.
You might want to just start with 64bit asm, instead of messing around with the obsolete x86 args-on-the-stack ABI.
This just made me think of what's probably wrong with your code. You're probably getting a segfault. (But you didn't say if it was that, syntax error, or something else.) Because you probably built your code in 64bit mode. Build a 32bit binary, or change your code to use %rsp.

Related

Global Offset Table: "Pointers to Pointers"? Is this handled by the loader?

This question is about Linux (Ubuntu) executables.
I'll detail things as I understand them to make it clearer if anything's off (so please correct me where applicable):
The GOT acts an extra level of indirection to enable accessing data from a text section which needs to be position-independent, for instance because the text section might be readonly and the actual addresses of the data may be unknown at (static) linking time.
The GOT then holds addresses to the actual data's location, which is known at loading time, and so the dynamic linker (which is invoked by the loader) is able to modify the appropriate GOT entries and make them point to the actual data.
The main thing that confuses me – not the only one at the moment, mind you :-) – is that this means the addresses in the text section now point to a value "of a different type":
If there was no GOT, I'd have expected this address (for instance in a RIP-relative addressing mode) to point to the actual value I'm after. With a GOT, though, I expect it to point to the appropriate GOT entry, which in turn holds the address to the value I'm after. In this case, there's an extra "dereferencing" required here.
Am I somehow misunderstanding this? If I use RIP-relative addressing, shouldn't the computed address (RIP+offset) be the actual address used in the instruction? So (in AT&T syntax):
mov $fun_data(%rip), %rax
To my understanding, without GOT, this should be "rax = *(rip + (fun_data - rip))", or in short: rax = *fun_data.
With GOT, however, I expect this to be equivalent to rax = **fun_data, since *fun_data is just the GOT entry to the real fun_data.
Am I wrong about this, or is it just that the loader somehow knows to access the real data if the pointer is into the GOT? (In other words: that in a PIE, I suppose, some pointers effectively become pointers-to-pointers?)
Am I wrong about this
No.
or is it just that the loader somehow knows to access the real data if the pointer is into the GOT?
The compiler knows that double dereference is required.
Compile this source with and without -fPIC and observe for yourself:
extern int dddd;
int fn() { return dddd; }
Without -fPIC, you get (expected):
movl dddd(%rip), %eax
With -fPIC you get "double dereference":
movq dddd#GOTPCREL(%rip), %rax # move pointer to dddd into RAX
movl (%rax), %eax # dereference it

On x64 Linux, what is the difference between syscall, int 0x80 and ret to exit a program?

I decided yesterday to learn assembly (NASM syntax) after years of C++ and Python and I'm already confused about the way to exit a program. It's mostly about ret because it's the suggested instruction on SASM IDE.
I'm speaking for main obviously. I don't care about x86 backward compatibility. Only the x64 Linux best way. I'm curious.
If you use printf or other libc functions, it's best to ret from main or call exit. (Which are equivalent; main's caller will call the libc exit function.)
If not, if you were only making other raw system calls like write with syscall, it's also appropriate and consistent to exit that way, but either way, or call exit are 100% fine in main.
If you want to work without libc at all, e.g. put your code under _start: instead of main: and link with ld or gcc -static -nostdlib, then you can't use ret. Use mov eax, 231 (__NR_exit_group) / syscall.
main is a real & normal function like any other (called with a valid return address), but _start (the process entry point) isn't. On entry to _start, the stack holds argc and argv, so trying to ret would set RIP=argc, and then code-fetch would segfault on that unmapped address. Nasm segmentation fault on RET in _start
System call vs. ret-from-main
Exiting via a system call is like calling _exit() in C - skip atexit() and libc cleanup, notably not flushing any buffered stdout output (line buffered on a terminal, full-buffered otherwise).
This leads to symptoms such as Using printf in assembly leads to empty output when piping, but works on the terminal (or if your output doesn't end with \n, even on a terminal.)
main is a function, called (indirectly) from CRT startup code. (Assuming you link your program normally, like you would a C program.) Your hand-written main works exactly like a compiler-generate C main function would. Its caller (__libc_start_main) really does do something like int result = main(argc, argv); exit(result);,
e.g. call rax (pointer passed by _start) / mov edi, eax / call exit.
So returning from main is exactly1 like calling exit.
Syscall implementation of exit() for a comparison of the relevant C functions, exit vs. _exit vs. exit_group and the underlying asm system calls.
C question: What is the difference between exit and return? is primarily about exit() vs. return, although there is mention of calling _exit() directly, i.e. just making a system call. It's applicable because C main compiles to an asm main just like you'd write by hand.
Footnote 1: You can invent a hypothetical intentionally weird case where it's different. e.g. you used stack space in main as your stdio buffer with sub rsp, 1024 / mov rsi, rsp / ... / call setvbuf. Then returning from main would involve putting RSP above that buffer, and __libc_start_main's call to exit could overwrite some of that buffer with return addresses and locals before execution reached the fflush cleanup. This mistake is more obvious in asm than C because you need leave or mov rsp, rbp or add rsp, 1024 or something to point RSP at your return address.
In C++, return from main runs destructors for its locals (before global/static exit stuff), exit doesn't. But that just means the compiler makes asm that does more stuff before actually running the ret, so it's all manual in asm, like in C.
The other difference is of course the asm / calling-convention details: exit status in EAX (return value) or EDI (first arg), and of course to ret you have to have RSP pointing at your return address, like it was on function entry. With call exit you don't, and you can even do a conditional tailcall of exit like jne exit. Since it's a noreturn function, you don't really need RSP pointing at a valid return address. (RSP should be aligned by 16 before a call, though, or RSP%16 = 8 before a tailcall, matching the alignment after call pushes a return address. It's unlikely that exit / fflush cleanup will do any alignment-required stores/loads to the stack, but it's a good habit to get this right.)
(This whole footnote is about ret vs. call exit, not syscall, so it's a bit of a tangent from the rest of the answer. You can also run syscall without caring where the stack-pointer points.)
SYS_exit vs. SYS_exit_group raw system calls
The raw SYS_exit system call is for exiting the current thread, like pthread_exit().
(eax=60 / syscall, or eax=1 / int 0x80).
SYS_exit_group is for exiting the whole program, like _exit.
(eax=231 / syscall, or eax=252 / int 0x80).
In a single-threaded program you can use either, but conceptually exit_group makes more sense to me if you're going to use raw system calls. glibc's _exit() wrapper function actually uses the exit_group system call (since glibc 2.3). See Syscall implementation of exit() for more details.
However, nearly all the hand-written asm you'll ever see uses SYS_exit1. It's not "wrong", and SYS_exit is perfectly acceptable for a program that didn't start more threads. Especially if you're trying to save code size with xor eax,eax / inc eax (3 bytes in 32-bit mode) or push 60 / pop rax (3 bytes in 64-bit mode), while push 231/pop rax would be even larger than mov eax,231 because it doesn't fit in a signed imm8.
Note 1: (Usually actually hard-coding the number, not using __NR_... constants from asm/unistd.h or their SYS_... names from sys/syscall.h)
And historically, it's all there was. Note that in unistd_32.h, __NR_exit has call number 1, but __NR_exit_group = 252 wasn't added until years later when the kernel gained support for tasks that share virtual address space with their parent, aka threads started by clone(2). This is when SYS_exit conceptually became "exit current thread". (But one could easily and convincingly argue that in a single-threaded program, SYS_exit does still mean exit the whole program, because it only differs from exit_group if there are multiple threads.)
To be honest, I've never used eax=252 / int 0x80 in anything, only ever eax=1. It's only in 64-bit code where I often use mov eax,231 instead of mov eax,60 because neither number is "simple" or memorable the way 1 is, so might as well be a cool guy and use the "modern" exit_group way in my single-threaded toy program / experiment / microbenchmark / SO answer. :P (If I didn't enjoy tilting at windmills, I wouldn't spend so much time on assembly, especially on SO.)
And BTW, I usually use NASM for one-off experiments so it's inconvenient to use pre-defined symbolic constants for call numbers; with GCC to preprocess a .S before running GAS you can make your code self-documenting with #include <sys/syscall.h> so you can use mov $SYS_exit_group, %eax (or $__NR_exit_group), or mov eax, __NR_exit_group with .intel_syntax noprefix.
Don't use the 32-bit int 0x80 ABI in 64-bit code:
What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? explains what happens if you use the COMPAT_IA32_EMULATION int 0x80 ABI in 64-bit code.
It's totally fine for just exiting, as long as your kernel has that support compiled in, otherwise it will segfault just like any other random int number like int 0x7f. (e.g. on WSL1, or people that built custom kernels and disabled that support.)
But the only reason you'd do it that way in asm would be so you could build the same source file with nasm -felf32 or nasm -felf64. (You can't use syscall in 32-bit code, except on some AMD CPUs which have a 32-bit version of syscall. And the 32-bit ABI uses different call numbers anyway so this wouldn't let the same source be useful for both modes.)
Related:
Why am I allowed to exit main using ret? (CRT startup code calls main, you're not returning directly to the kernel.)
Nasm segmentation fault on RET in _start - you can't ret from _start
Using printf in assembly leads to empty output when piping, but works on the terminal stdout buffer (not) flushing with raw system call exit
Syscall implementation of exit() call exit vs. mov eax,60/syscall (_exit) vs. mov eax,231/syscall (exit_group).
Can't call C standard library function on 64-bit Linux from assembly (yasm) code - modern Linux distros config GCC in a way that call exit or call puts won't link with nasm -felf64 foo.asm && gcc foo.o.
Is main() really start of a C++ program? - Ciro's answer is a deep dive into how glibc + its CRT startup code actually call main (including x86-64 asm disassembly in GDB), and shows the glibc source code for __libc_start_main.
Linux x86 Program Start Up
or - How the heck do we get to main()? 32-bit asm, and more detail than you'll probably want until you're a lot more comfortable with asm, but if you've ever wondered why CRT runs so much code before getting to main, that covers what's happening at a level that's a couple steps up from using GDB with starti (stop at the process entry point, e.g. in the dynamic linker's _start) and stepi until you get to your own _start or main.
https://stackoverflow.com/tags/x86/info lots of good links about this and everything else.

Understanding assembly language _start label in a C program

I had written a simple c program and was trying to do use GDB to debug the program. I understand the use of following in main function:
On entry
push %ebp
mov %esp,%ebp
On exit
leave
ret
Then I tried gdb on _start and I got the following
xor %ebp,%ebp
pop %esi
mov %esp,%ecx
and $0xfffffff0,%esp
push %eax
push %esp
push %edx
push $0x80484d0
push $0x8048470
push %ecx
push %esi
push $0x8048414
call 0x8048328 <__libc_start_main#plt>
hlt
nop
nop
nop
nop
I am unable to understand these lines, and the logic behind this.
Can someone provide any guidance to help explain the code of _start?
Here is the well commented assembly source of the code you posted.
Summarized, it does the following things:
establish a sentinel stack frame with ebp = 0 so code that walks the stack can find its end easily
Pop the number of command line arguments into esi so we can pass them to __libc_start_main
Align the stack pointer to a multiple of 16 bits in order to comply with the ABI. This is not guaranteed to be the case in some versions of Linux so it has to be done manually just in case.
The addresses of __libc_csu_fini, __libc_csu_init, the argument vector, the number of arguments and the address of main are pushed as arguments to __libc_start_main
__libc_start_main is called. This function (source code here) sets up some glibc-internal variables and eventually calls main. It never returns.
If for any reason __libc_start_main should return, a hlt instruction is placed afterwards. This instruction is not allowed in user code and should cause the program to crash (hopefully).
The final series of nop instructions is padding inserted by the assembler so the next function starts at a multiple of 16 bytes for better performance. It is never reached in normal execution.
for gnu tools the _start label is the entry point of the program. for the C language to work you need to have a stack you need to have some memory/variables zeroed and some set to the values you chose:
int x = 5;
int y;
int fun ( void )
{
static int z;
}
all three of these variables x,y,z are essentially global, one is a local global. since we wrote it that way we assume that when our program starts x contains the value 5 and it is assumed that y is zero. in order for those things to happen, some bootstrap code is required and that is what happens (and more) between _start and main().
Other toolchains may choose to use a different label to define the entry/start point, but gnu tools use _start. there may be other things your tools require before main() is called C++ for example requires more than C.

Can anybody explain some simple assembly code?

I have just started to learn assembly. This is the dump from gdb for a simple program which prints hello ranjit.
Dump of assembler code for function main:
0x080483b4 <+0>: push %ebp
0x080483b5 <+1>: mov %esp,%ebp
0x080483b7 <+3>: sub $0x4,%esp
=> 0x080483ba <+6>: movl $0x8048490,(%esp)
0x080483c1 <+13>: call 0x80482f0 <puts#plt>
0x080483c6 <+18>: leave
0x080483c7 <+19>: ret
My questions are :
Why every time ebp is pushed on to stack at start of the program? What is in the ebp which is necessary to run this program?
In second line why is ebp copied to esp?
I can't get the third line at all. what I know about SUB syntax is "sub dest,source", but here how can esp be subtracted from 4 and stored in 4?
What is this value "$0x8048490"? Why it is moved to esp, and why this time is esp closed in brackets? Does it denote something different than esp without brackets?
Next line is the call to function but what is this "0x80482f0"?
What is leave and ret (maybe ret means returning to lib c.)?
operating system : ubuntu 10, compiler : gcc
ebp is used as a frame pointer in Intel processors (assuming you're using a calling convention that uses frames).
It provides a known point of reference for locating passed-in parameters (on one side) and local variables (on the other) no matter what you do with the stack pointer while your function is active.
The sequence:
push %ebp ; save callers frame pointer
mov %esp,%ebp ; create a new frame pointer
sub $N,%esp ; make space for locals
saves the frame pointer for the previous stack frame (the caller), loads up a new frame pointer, then adjusts the stack to hold things for the current "stack level".
Since parameters would have been pushed before setting up the frame, they can be accessed with [bp+N] where N is a suitable offset.
Similarly, because locals are created "under" the frame pointer, they can be accessed with [bp-N].
The leave instruction is a single one which undoes that stack frame. You used to have to do it manually but Intel introduced a faster way of getting it done. It's functionally equivalent to:
mov %ebp, %esp ; restore the old stack pointer
pop %ebp ; and frame pointer
(the old, manual way).
Answering the questions one by one in case I've missed something:
To start a new frame. See above.
It isn't. esp is copied to ebp. This is AT&T notation (the %reg is a dead giveaway) where (among other thing) source and destination operands are swapped relative to Intel notation.
See answer to (2) above. You're subtracting 4 from esp, not the other way around.
It's a parameter being passed to the function at 0x80482f0. It's not being loaded into esp but into the memory pointed at by esp. In other words, it's being pushed on the stack. Since the function being called is puts (see (5) below), it will be the address of the string you want putsed.
The function name in the <> after the address. It's calling the puts function (probably the one in the standard library though that's not guaranteed). For a description of what the PLT is, see here.
I've already explained leave above as unwinding the current stack frame before exiting. The ret simply returns from the current function. If the current functtion is main, it's going back to the C startup code.
In my career I learned several assembly languages, you didn't mention which but it appears Intel x86 (segmented memory model as PaxDiablo pointed out). However, I have not used assembly since last century (lucky me!). Here are some of your answers:
The EBP register is pushed onto the stack at the beginning because we need it further along in other operations of the routine. You don't want to just discard its original value thus corrupting the integrity of the rest of the application.
If I remember correctly (I may be wrong, long time) it is the other way around, we are moving %esp INTO %ebp, remember we saved it in the previous line? now we are storing some new value without destroying the original one.
Actually they are SUBstracting the value of four (4) FROM the contents of the %esp register. The resulting value is not stored on "four" but on %esp. If %esp had 0xFFF8 after the SUB it will contain 0xFFF4. I think this is called "Immediate" if my memory serves me. What is happening here (I reckon) is the computation of a memory address (4 bytes less).
The value $0x8048490 I don't know. However, it is NOT being moved INTO %esp but rather INTO THE ADDRESS POINTED TO BY THE CONTENTS OF %esp. That is why the notation is (%esp) rather than %esp. This is kind of a common notation in all assembly languages I came about in my career. If on the other hand the right operand was simply %esp, then the value would have been moved INTO the %esp register. Basically the %esp register's contents are being used for addressing.
It is a fixed value and the string on the right makes me think that this value is actually the address of the puts() (Put String) compiler library routine.
"leave" is an instrution that is the equivalent of "pop %ebp". Remember we saved the contents of %ebp at the beginning, now that we are done with the routine we are restoring it back into the register so that the caller gets back to its context. The "ret" instruction is the final instruction of the routine, it "returns" to the caller.

Loading an exe from an exe

I am exporting a function [using _declspec(dllexport)] from a C++ exe. The function works fine when invoked by the exe itself. I am loading this exe (lets call this exe1) from another exe [the test project's exe - I'll call this exe2] using static linking i.e I use exe1's .lib file while compiling exe2 and exe2 loads it into memory on startup just like any dll. This causes the function to fail in execution.
The exact problem is revealed in the disassembly for a switch case statement within the function.
Assembly code when exe1 invokes the function
switch (dwType)
0040FF84 mov eax,dword ptr [dwType]
0040FF87 mov dword ptr [ebp-4],eax
0040FF8A cmp dword ptr [ebp-4],0Bh
0040FF8E ja $LN2+7 (40FFD2h)
0040FF90 mov ecx,dword ptr [ebp-4]
0040FF93 jmp dword ptr (40FFE0h)[ecx*4]
Consider the final two instructions. The mov moves the passed in argument into ecx. At 40EFF0h we have addresses to the various instructions for the respective case statements. Thus, the jmp would take us to the relevant case instructions
Assembly code when exe2 invokes the function
switch (dwType)
0037FF84 mov eax,dword ptr [dwType]
0037FF87 mov dword ptr [ebp-4],eax
0037FF8A cmp dword ptr [ebp-4],0Bh
0037FF8E ja $LN2+7 (37FFD2h)
0037FF90 mov ecx,dword ptr [ebp-4]
0037FF93 jmp dword ptr [ecx*4+40FFE0h]
Spot whats going wrong? The instruction addresses. The code has now been loaded into a different spot in memory. When exe1 was compiled, the compiler assumed that we will always be launching it and hence it would always be loaded at 0x0040000 [as is the case with all windows exes]. So it hard-coded a few values like 40FFE0h into the instructions. Only in the second case 40FFE0 is as good as junk memory since the instruction address table we are looking for is not located there.
How can I solve this without converting exe1 to a dll?
just don't do it. It doesn't worth the bother.
I've tried doing what you're trying a while ago. You can possibly solve the non-relocatable exe problem by changing the option in the properties window under "Linker->Advenced->Fixed base address" but then you'll have other problems.
The thing that finally made me realize its a waste of time is realizing that the EXE doesn't have a DllMain() function. This means that the CRT library is not getting initialized and that all sorts of stuff don't work the way you expect it to.
Here's the question I posted about this a while back
Have you considered another way of doing this? Such as making the 2nd .exe into a .dll and invoking it with rundll32 when you want to use it as an executable?
Otherwise:
The generated assembly is fine. The problem is that Win32 portable executables have a base address (0x0040000 in this case) and a section that contain details locations of addresses so that they can be rebased when required.
So one for two things is happening:
- Either the compiler isn't including the IMAGE_BASE_RELOCATION records when it builds the .exe.
- Or the runtime isn't performing the base relocations when it dynamiclaly loads the .exe
- (possibly both)
If the .exe does contain the relocation records, you can read them and perform the base relocation yourself. You'll have to jump through hoops like making sure you have write access to the memory (VirtualAlloc etc.) but it's conceptually quite simple.
If the .exe doesn't contain the relocation records you're stuffed - either find a compiler option to force their inclusion, or find another way to do what you're doing.
Edit: As shoosh points out, you may run into other problems once you fix this one.

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