Stack View when printf is called? - c

I was just learning about format string vulnerabilities that makes me ask this question
Consider the following simple program:
#include<stdio.h>
void main(int argc, char **argv)
{
char *s="SomeString";
printf(argv[1]);
}
Now clearly, this code is vulnerable to a format String vulnerability. I.e. when the command line argument is %s, then the value SomeString is printed since printf pops the stack once.
What I dont understand is the structure of the stack when printf is called
In my head I imagine the stack to be as follows:
grows from left to right ----->
main() ---> printf()-->
RET to libc_main | address of 's' | current registers| ret ptr to main | ptr to format string|
if this is the case, how does inputting %s to the program, cause the value of s to be popped ?
(OR) If I am totally wrong about the stack structure , please correct me

The stack contents depends a lot on the following:
the CPU
the compiler
the calling conventions (i.e. how parameters are passed in the registers and on the stack)
the code optimizations performed by the compiler
This is what I get by compiling your tiny program with x86 mingw using gcc stk.c -S -o stk.s:
.file "stk.c"
.def ___main; .scl 2; .type 32; .endef
.section .rdata,"dr"
LC0:
.ascii "SomeString\0"
.text
.globl _main
.def _main; .scl 2; .type 32; .endef
_main:
LFB6:
.cfi_startproc
pushl %ebp
.cfi_def_cfa_offset 8
.cfi_offset 5, -8
movl %esp, %ebp
.cfi_def_cfa_register 5
andl $-16, %esp
subl $32, %esp
call ___main
movl $LC0, 28(%esp)
movl 12(%ebp), %eax
addl $4, %eax
movl (%eax), %eax
movl %eax, (%esp)
call _printf
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
.cfi_endproc
LFE6:
.def _printf; .scl 2; .type 32; .endef
And this is what I get using gcc stk.c -S -O2 -o stk.s, that is, with optimizations enabled:
.file "stk.c"
.def ___main; .scl 2; .type 32; .endef
.section .text.startup,"x"
.p2align 2,,3
.globl _main
.def _main; .scl 2; .type 32; .endef
_main:
LFB7:
.cfi_startproc
pushl %ebp
.cfi_def_cfa_offset 8
.cfi_offset 5, -8
movl %esp, %ebp
.cfi_def_cfa_register 5
andl $-16, %esp
subl $16, %esp
call ___main
movl 12(%ebp), %eax
movl 4(%eax), %eax
movl %eax, (%esp)
call _printf
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
.cfi_endproc
LFE7:
.def _printf; .scl 2; .type 32; .endef
As you can see, in the latter case there's no pointer to "SomeString" on the stack. In fact, the string isn't even present in the compiled code.
In this simple code there are no registers saved on the stack because there aren't any variables allocated to registers that need to be preserved across the call to printf().
So, the only things you get on the stack here are the string pointer (optionally), unused space due to stack alignment (andl $-16, %esp + subl $32, %esp align the stack and allocate space for local variables, none here), the printf()'s parameter, the return address for returning from printf() back to main().
In the former case the pointer to "SomeString" and the printf()'s parameter (value of argv[1]) are quite far away from one another:
movl $LC0, 28(%esp) ; address of "SomeString" is at esp+28
movl 12(%ebp), %eax
addl $4, %eax
movl (%eax), %eax
movl %eax, (%esp) ; address of a copy of argv[1] is at esp
call _printf
To make the two addresses stored one right after the other on the stack, if that's what you want, you'd need to play with the code, compilation/optimization options or use a different compiler.
Or you could supply a format string in argv[1] such that printf() would reach it. You could, for example, include a number of fake parameters in the format string.
For example, if I compile this piece of code using gcc stk.c -o stk.exe and run it as stk.exe %u%u%u%u%u%u%s, I'll get the following output from it:
4200532268676042006264200532880015253SomeString
All of this is pretty hacky and it's not trivial to make it work right.

On x86 the stack on a function call could look something like:
: :
+--------------+
: alignment :
+--------------+
12(%ebp) | arg2 |
+--------------+
8(%ebp) | arg1 |
+--------------+
4(%ebp) | ret | -----> return address
+--------------+
(%ebp) | ebp | -----> previous ebp value
+--------------+
-4(%ebp) | local1 | -----> local vars, sometimes they can overflow ;-)
+--------------+
: alignment :
+--------------+
: :
If you used -fomit-frame-pointer ebp would not be saved on the stack. At different optimization levels some variables may disappear (be optimized out), ...
Other ABIs store function arguments on registers, instead of saving them on the stack. Later, before calling another function, live registers may spill into the stack.

Related

assembly language - c-language and mnemonics

I wrote a simple program in c-language, the classic helloworld. I wanted to know how it looked liked when the compiler translated it to assembly code.
I use MinGW and the command:
gcc -S hellow.c
When I opened this file I expected it would, AT THE LEAST, be somewhat similar to a hello-world program written directly in assembly, that is:
jmp 115
db 'Hello world!$' (db = define bytes)
-a 115
mov ah, 09 (09 for displaying strings ... ah = 'command register')
mov dx, 102 (adress of the string)
int 21
int 20
Instead it look like this:
.file "hellow.c"
.def ___main;
.scl 2;
.type 32;
.endef
.section
.rdata,"dr"
LC0:
.ascii "Hello world!\0"
.text
.globl _main
.def _main;
.scl 2;
.type 32;
.endef
_main:
LFB6:
.cfi_startproc
pushl %ebp
.cfi_def_cfa_offset 8
.cfi_offset 5, -8
movl %esp, %ebp
.cfi_def_cfa_register 5
andl $-16, %esp
subl $16, %esp
call ___main
movl $LC0, (%esp)
call _puts
movl $0, %eax
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
.cfi_endproc
LFE6:
.def _puts;
.scl 2;
.type 32;
.endef
I know litte about assembly language, but i DO recognice the so called mnemonics like ADD, POP, PUSH, MOV, JMP, INT etc. Could not see much of these in the code generated by the c-compiler.
What did I missunderstand?
This prepares the arguments to call a function __main that probably does all initial setup that is needed for a C program
andl $-16, %esp
subl $16, %esp
call ___main
This prepares the arguments and calls function _puts. LC0 is a symbol that contains the string to be printed.
movl $LC0, (%esp)
call _puts
This prepares the return value of main and returns
movl $0, %eax
leave
ret
Your example code uses Intel syntax, while the standard output from gcc is AT&T syntax. You can change that by using
gcc -S hellow.c -masm=intel
The resulting output should look more familiar.
However, if the compiler generates the source then it looks rather different, then what you would write by hand.
The int would be used if you compile for DOS, but even so, these calls would be wrapped in C standard functions, like puts in this case.

Understanding x86 Assembly Code from C code

C code:
#include <stdio.h>
main() {
int i;
for (i = 0; i < 10; i++) {
printf("%s\n", "hello");
}
}
ASM:
.file "simple_loop.c"
.section .rodata
.LC0:
.string "hello"
.text
.globl main
.type main, #function
main:
.LFB0:
.cfi_startproc
pushl %ebp # push ebp onto stack
.cfi_def_cfa_offset 8
.cfi_offset 5, -8
movl %esp, %ebp # setup base pointer or stack ?
.cfi_def_cfa_register 5
andl $-16, %esp # ?
subl $32, %esp # ?
movl $0, 28(%esp) # i = 0
jmp .L2
.L3:
movl $.LC0, (%esp) # point stack pointer to "hello" ?
call puts # print "hello"
addl $1, 28(%esp) # i++
.L2:
cmpl $9, 28(%esp) # if i < 9
jle .L3 # goto l3
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
So I am trying to improve my understanding of x86 assembly code. For the above code, I marked off what I believe I understand. As for the question marked content, could someone share some light? Also, if any of my comments are off, please let me know.
andl $-16, %esp # ?
subl $32, %esp # ?
This reserves some space on the stack. First, the andl instruction rounds the %esp register down to the next lowest multiple of 16 bytes (exercise: find out what the binary value of -16 is to see why). Then, the subl instruction moves the stack pointer down a bit further (32 bytes), reserving some more space (which it will use next). I suspect this rounding is done so that access through the %esp register is slightly more efficient (but you'd have to inspect your processor data sheets to figure out why).
movl $.LC0, (%esp) # point stack pointer to "hello" ?
This places the address of the string "hello" onto the stack (this instruction doesn't change the value of the %esp register itself). Apparently your compiler considers it more efficient to move data onto the stack directly, rather than to use the push instruction.

What exactly happens when you dereference a static variable in C?

So lets say I have this code
int my_static_int = 4;
func(&my_static_int);
I passed the function a pointer to my_static_int, obviously. But what happens when the code is compiled? Avenue I've considered:
1) When you declare a non-pointer variable, C automatically creates its pointer and does something internally like typedefs my_static_int to be *(internal_reference)
Anyway, I hope that my question is descriptive enough
Pointers are just a term to help us humans understand what's going on.
The & operator when used with a variable simply means address of. No "pointer" is created at runtime, you are simply passing in the address of the variable into the function.
If you have:
int x = 3;
int* p = &x;
Then p is a variable which holds a memory address. Inside that memory address is an int.
If you really want to know how the code looks under the covers, you have to get the compiler to generate the assembler code (gcc can do this with the -S option).
When you truly grok C and pointers at their deepest level, you'll realise that it's just the address of the variable being passed in rather than the value of the variable. There's no need for creating extra memory to hold a pointer since the pointer is moved directly from the code to the stack (the address will probably have been set either at link time or load time, not run time).
There's also no need for internal type creation since the compiled code already knows the type and how to manipulate it.
Keeping in mind that this is implementation-specific, consider the following code:
int my_static_int = 4;
static void func (int *x) {
*x = *x + 7;
}
int main (void) {
func(&my_static_int);
return 0;
}
which, when compiled with gcc -S to get the assembler, produces:
.file "qq.c"
.globl _my_static_int
.data
.align 4
_my_static_int:
.long 4
.text
.def _func; .scl 3; .type 32; .endef
_func:
pushl %ebp
movl %esp, %ebp
movl 8(%ebp), %eax
movl 8(%ebp), %edx
movl (%edx), %edx
addl $7, %edx
movl %edx, (%eax)
popl %ebp
ret
.def ___main; .scl 2; .type 32; .endef
.globl _main
.def _main; .scl 2; .type 32; .endef
_main:
pushl %ebp
movl %esp, %ebp
subl $8, %esp
andl $-16, %esp
movl $0, %eax
addl $15, %eax
addl $15, %eax
shrl $4, %eax
sall $4, %eax
movl %eax, -4(%ebp)
movl -4(%ebp), %eax
call __alloca
call ___main
movl $_my_static_int, (%esp)
call _func
movl $0, %eax
leave
ret
The important bit is these sections:
movl $_my_static_int, (%esp) ; load address of variable onto stack.
call _func ; call the function.
:
movl 8(%ebp), %eax ; get passed parameter (the address of the var) into eax
movl 8(%ebp), %edx ; and also into edx.
movl (%edx), %edx ; get the value from the address (dereference).
addl $7, %edx ; add 7 to it.
movl %edx, (%eax) ; and put it back into the same address.
Hence the address is passed, and used to get at the variable.
When the code is compiled, function func receives the address of your my_static_int variable as parameter. Nothing else.
There no need to create any implicit pointers when you declare a non-pointer variable. It is not clear from your question how you came to this weird idea.
Why not look at the assembly output? You can do this with gcc using the -S option, or (if your system uses the GNU toolchain) using the objdump -d command on the resulting object file or executable file.
The simple answer is that the object code generates a reference to the symbol where my_static_int is allocated (which is typically in the static data segment of your object module).
So the address of the variable is resolved at load time (when it is assigned a real physical address), and the loader fixes up the reference to the variable, filling it in with its address.

How do C compilers implement functions that return large structures?

The return value of a function is usually stored on the stack or in a register. But for a large structure, it has to be on the stack. How much copying has to happen in a real compiler for this code? Or is it optimized away?
For example:
struct Data {
unsigned values[256];
};
Data createData()
{
Data data;
// initialize data values...
return data;
}
(Assuming the function cannot be inlined..)
None; no copies are done.
The address of the caller's Data return value is actually passed as a hidden argument to the function, and the createData function simply writes into the caller's stack frame.
This is known as the named return value optimisation. Also see the c++ faq on this topic.
commercial-grade C++ compilers implement return-by-value in a way that lets them eliminate the overhead, at least in simple cases
...
When yourCode() calls rbv(), the compiler secretly passes a pointer to the location where rbv() is supposed to construct the "returned" object.
You can demonstrate that this has been done by adding a destructor with a printf to your struct. The destructor should only be called once if this return-by-value optimisation is in operation, otherwise twice.
Also you can check the assembly to see that this happens:
Data createData()
{
Data data;
// initialize data values...
data.values[5] = 6;
return data;
}
here's the assembly:
__Z10createDatav:
LFB2:
pushl %ebp
LCFI0:
movl %esp, %ebp
LCFI1:
subl $1032, %esp
LCFI2:
movl 8(%ebp), %eax
movl $6, 20(%eax)
leave
ret $4
LFE2:
Curiously, it allocated enough space on the stack for the data item subl $1032, %esp, but note that it takes the first argument on the stack 8(%ebp) as the base address of the object, and then initialises element 6 of that item. Since we didn't specify any arguments to createData, this is curious until you realise this is the secret hidden pointer to the parent's version of Data.
But for a large structure, it has to be on the heap stack.
Indeed so! A large structure declared as a local variable is allocated on the stack. Glad to have that cleared up.
As for avoiding copying, as others have noted:
Most calling conventions deal with "function returning struct" by passing an additional parameter that points the location in the caller's stack frame in which the struct should be placed. This is definitely a matter for the calling convention and not the language.
With this calling convention, it becomes possible for even a relatively simple compiler to notice when a code path is definitely going to return a struct, and for it to fix assignments to that struct's members so that they go directly into the caller's frame and don't have to be copied. The key is for the compiler to notice that all terminating code paths through the function return the same struct variable. If that's the case, the compiler can safely use the space in the caller's frame, eliminating the need for a copy at the point of return.
There are many examples given, but basically
This question does not have any definite answer. it will depend on the compiler.
C does not specify how large structs are returned from a function.
Here's some tests for one particular compiler, gcc 4.1.2 on x86 RHEL 5.4
gcc trivial case, no copying
[00:05:21 1 ~] $ gcc -O2 -S -c t.c
[00:05:23 1 ~] $ cat t.s
.file "t.c"
.text
.p2align 4,,15
.globl createData
.type createData, #function
createData:
pushl %ebp
movl %esp, %ebp
movl 8(%ebp), %eax
movl $1, 24(%eax)
popl %ebp
ret $4
.size createData, .-createData
.ident "GCC: (GNU) 4.1.2 20080704 (Red Hat 4.1.2-46)"
.section .note.GNU-stack,"",#progbits
gcc more realistic case , allocate on stack, memcpy to caller
#include <stdlib.h>
struct Data {
unsigned values[256];
};
struct Data createData()
{
struct Data data;
int i;
for(i = 0; i < 256 ; i++)
data.values[i] = rand();
return data;
}
[00:06:08 1 ~] $ gcc -O2 -S -c t.c
[00:06:10 1 ~] $ cat t.s
.file "t.c"
.text
.p2align 4,,15
.globl createData
.type createData, #function
createData:
pushl %ebp
movl %esp, %ebp
pushl %edi
pushl %esi
pushl %ebx
movl $1, %ebx
subl $1036, %esp
movl 8(%ebp), %edi
leal -1036(%ebp), %esi
.p2align 4,,7
.L2:
call rand
movl %eax, -4(%esi,%ebx,4)
addl $1, %ebx
cmpl $257, %ebx
jne .L2
movl %esi, 4(%esp)
movl %edi, (%esp)
movl $1024, 8(%esp)
call memcpy
addl $1036, %esp
movl %edi, %eax
popl %ebx
popl %esi
popl %edi
popl %ebp
ret $4
.size createData, .-createData
.ident "GCC: (GNU) 4.1.2 20080704 (Red Hat 4.1.2-46)"
.section .note.GNU-stack,"",#progbits
gcc 4.4.2### has grown a lot, and does not copy for the above non-trivial case.
.file "t.c"
.text
.p2align 4,,15
.globl createData
.type createData, #function
createData:
pushl %ebp
movl %esp, %ebp
pushl %edi
pushl %esi
pushl %ebx
movl $1, %ebx
subl $1036, %esp
movl 8(%ebp), %edi
leal -1036(%ebp), %esi
.p2align 4,,7
.L2:
call rand
movl %eax, -4(%esi,%ebx,4)
addl $1, %ebx
cmpl $257, %ebx
jne .L2
movl %esi, 4(%esp)
movl %edi, (%esp)
movl $1024, 8(%esp)
call memcpy
addl $1036, %esp
movl %edi, %eax
popl %ebx
popl %esi
popl %edi
popl %ebp
ret $4
.size createData, .-createData
.ident "GCC: (GNU) 4.1.2 20080704 (Red Hat 4.1.2-46)"
.section .note.GNU-stack,"",#progbits
In addition, VS2008 (compiled the above as C) will reserve struct Data on the stack of createData() and do a rep movsd loop to copy it back to the caller in Debug mode, in Release mode it will move the return value of rand() (%eax) directly back to the caller
typedef struct {
unsigned value[256];
} Data;
Data createData(void) {
Data r;
calcualte(&r);
return r;
}
Data d = createData();
msvc(6,8,9) and gcc mingw(3.4.5,4.4.0) will generate code like the following pseudocode
void createData(Data* r) {
calculate(&r)
}
Data d;
createData(&d);
gcc on linux will issue a memcpy() to copy the struct back on the stack of the caller. If the function has internal linkage, more optimizations become available though.

Understanding empty main()'s translation into assembly

Could somebody please explain what GCC is doing for this piece of code? What is it initializing? The original code is:
#include <stdio.h>
int main()
{
}
And it was translated to:
.file "test1.c"
.def ___main; .scl 2; .type 32; .endef
.text
.globl _main
.def _main; .scl 2; .type 32; .endef
_main:
pushl %ebp
movl %esp, %ebp
subl $8, %esp
andl $-16, %esp
movl $0, %eax
addl $15, %eax
addl $15, %eax
shrl $4, %eax
sall $4, %eax
movl %eax, -4(%ebp)
movl -4(%ebp), %eax
call __alloca
call ___main
leave
ret
I would be grateful if a compiler/assembly guru got me started by explaining the stack, register and the section initializations. I cant make head or tail out of the code.
EDIT:
I am using gcc 3.4.5. and the command line argument is gcc -S test1.c
Thank You,
kunjaan.
I should preface all my comments by saying, I am still learning assembly.
I will ignore the section initialization. A explanation for the section initialization and basically everything else I cover can be found here:
http://en.wikibooks.org/wiki/X86_Assembly/GAS_Syntax
The ebp register is the stack frame base pointer, hence the BP. It stores a pointer to the beginning of the current stack.
The esp register is the stack pointer. It holds the memory location of the top of the stack. Each time we push something on the stack esp is updated so that it always points to an address the top of the stack.
So ebp points to the base and esp points to the top. So the stack looks like:
esp -----> 000a3 fa
000a4 21
000a5 66
000a6 23
ebp -----> 000a7 54
If you push e4 on the stack this is what happens:
esp -----> 000a2 e4
000a3 fa
000a4 21
000a5 66
000a6 23
ebp -----> 000a7 54
Notice that the stack grows towards lower addresses, this fact will be important below.
The first two steps are known as the procedure prolog or more commonly as the function prolog. They prepare the stack for use by local variables (See procedure prolog quote at the bottom).
In step 1 we save the pointer to the old stack frame on the stack by calling
pushl %ebp. Since main is the first function called, I have no idea what the previous value of %ebp points too.
Step 2, We are entering a new stack frame because we are entering a new function (main). Therefore, we must set a new stack frame base pointer. We use the value in esp to be the beginning of our stack frame.
Step 3. Allocates 8 bytes of space on the stack. As we mentioned above, the stack grows toward lower addresses thus, subtracting by 8, moves the top of the stack by 8 bytes.
Step 4; Aligns the stack, I've found different opinions on this. I'm not really sure exactly what this is done. I suspect it is done to allow large instructions (SIMD) to be allocated on the stack,
http://gcc.gnu.org/ml/gcc/2008-01/msg00282.html
This code "and"s ESP with 0xFFFF0000,
aligning the stack with the next
lowest 16-byte boundary. An
examination of Mingw's source code
reveals that this may be for SIMD
instructions appearing in the "_main"
routine, which operate only on aligned
addresses. Since our routine doesn't
contain SIMD instructions, this line
is unnecessary.
http://en.wikibooks.org/wiki/X86_Assembly/GAS_Syntax
Steps 5 through 11 seem to have no purpose to me. I couldn't find any explanation on google. Could someone who really knows this stuff provide a deeper understanding. I've heard rumors that this stuff is used for C's exception handling.
Step 5, stores the return value of main 0, in eax.
Step 6 and 7 we add 15 in hex to eax for unknown reason. eax = 01111 + 01111 = 11110
Step 8 we shift the bits of eax 4 bits to the right. eax = 00001 because the last bits are shift off the end 00001 | 111.
Step 9 we shift the bits of eax 4 bits to the left, eax = 10000.
Steps 10 and 11 moves the value in the first 4 allocated bytes on the stack into eax and then moves it from eax back.
Steps 12 and 13 setup the c library.
We have reached the function epilogue. That is, the part of the function which returns the stack pointers, esp and ebp to the state they were in before this function was called.
Step 14, leave sets esp to the value of ebp, moving the top of stack to the address it was before main was called. Then it sets ebp to point to the address we saved on the top of the stack during step 1.
Leave can just be replaced with the following instructions:
mov %ebp, %esp
pop %ebp
Step 15, returns and exits the function.
1. pushl %ebp
2. movl %esp, %ebp
3. subl $8, %esp
4. andl $-16, %esp
5. movl $0, %eax
6. addl $15, %eax
7. addl $15, %eax
8. shrl $4, %eax
9. sall $4, %eax
10. movl %eax, -4(%ebp)
11. movl -4(%ebp), %eax
12. call __alloca
13. call ___main
14. leave
15. ret
Procedure Prolog:
The first thing a function has to do
is called the procedure prolog. It
first saves the current base pointer
(ebp) with the instruction pushl %ebp
(remember ebp is the register used for
accessing function parameters and
local variables). Now it copies the
stack pointer (esp) to the base
pointer (ebp) with the instruction
movl %esp, %ebp. This allows you to
access the function parameters as
indexes from the base pointer. Local
variables are always a subtraction
from ebp, such as -4(%ebp) or
(%ebp)-4 for the first local variable,
the return value is always at 4(%ebp)
or (%ebp)+4, each parameter or
argument is at N*4+4(%ebp) such as
8(%ebp) for the first argument while
the old ebp is at (%ebp).
http://www.milw0rm.com/papers/52
A really great stack overflow thread exists which answers much of this question.
Why are there extra instructions in my gcc output?
A good reference on x86 machine code instructions can be found here:
http://programminggroundup.blogspot.com/2007/01/appendix-b-common-x86-instructions.html
This a lecture which contains some of the ideas used below:
http://csc.colstate.edu/bosworth/cpsc5155/Y2006_TheFall/MySlides/CPSC5155_L23.htm
Here is another take on answering your question:
http://www.phiral.net/linuxasmone.htm
None of these sources explain everything.
Here's a good step-by step breakdown of a simple main() function as compiled by GCC, with lots of detailed info: GAS Syntax (Wikipedia)
For the code you pasted, the instructions break down as follows:
First four instructions (pushl through andl): set up a new stack frame
Next five instructions (movl through sall): generating a weird value for eax, which will become the return value (I have no idea how it decided to do this)
Next two instructions (both movl): store the computed return value in a temporary variable on the stack
Next two instructions (both call): invoke the C library init functions
leave instruction: tears down the stack frame
ret instruction: returns to caller (the outer runtime function, or perhaps the kernel function that invoked your program)
Well, dont know much about GAS, and i'm a little rusty on Intel assembly, but it looks like its initializing main's stack frame.
if you take a look, __main is some kind of macro, must be executing initializations.
Then, as main's body is empty, it calls leave instruction, to return to the function that called main.
From http://en.wikibooks.org/wiki/X86_Assembly/GAS_Syntax#.22hello.s.22_line-by-line:
This line declares the "_main" label, marking the place that is called from the startup code.
pushl %ebp
movl %esp, %ebp
subl $8, %esp
These lines save the value of EBP on the stack, then move the value of ESP into EBP, then subtract 8 from ESP. The "l" on the end of each opcode indicates that we want to use the version of the opcode that works with "long" (32-bit) operands;
andl $-16, %esp
This code "and"s ESP with 0xFFFF0000, aligning the stack with the next lowest 16-byte boundary. (neccesary when using simd instructions, not useful here)
movl $0, %eax
movl %eax, -4(%ebp)
movl -4(%ebp), %eax
This code moves zero into EAX, then moves EAX into the memory location EBP-4, which is in the temporary space we reserved on the stack at the beginning of the procedure. Then it moves the memory location EBP-4 back into EAX; clearly, this is not optimized code.
call __alloca
call ___main
These functions are part of the C library setup. Since we are calling functions in the C library, we probably need these. The exact operations they perform vary depending on the platform and the version of the GNU tools that are installed.
Here's a useful link.
http://unixwiz.net/techtips/win32-callconv-asm.html
It would really help to know what gcc version you are using and what libc. It looks like you have a very old gcc version or a strange platform or both. What's going on is some strangeness with calling conventions. I can tell you a few things:
Save the frame pointer on the stack according to convention:
pushl %ebp
movl %esp, %ebp
Make room for stuff at the old end of the frame, and round the stack pointer down to a multiple of 4 (why this is needed I don't know):
subl $8, %esp
andl $-16, %esp
Through an insane song and dance, get ready to return 1 from main:
movl $0, %eax
addl $15, %eax
addl $15, %eax
shrl $4, %eax
sall $4, %eax
movl %eax, -4(%ebp)
movl -4(%ebp), %eax
Recover any memory allocated with alloca (GNU-ism):
call __alloca
Announce to libc that main is exiting (more GNU-ism):
call ___main
Restore the frame and stack pointers:
leave
Return:
ret
Here's what happens when I compile the very same source code with gcc 4.3 on Debian Linux:
.file "main.c"
.text
.p2align 4,,15
.globl main
.type main, #function
main:
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
pushl %ebp
movl %esp, %ebp
pushl %ecx
popl %ecx
popl %ebp
leal -4(%ecx), %esp
ret
.size main, .-main
.ident "GCC: (Debian 4.3.2-1.1) 4.3.2"
.section .note.GNU-stack,"",#progbits
And I break it down this way:
Tell the debugger and other tools the source file:
.file "main.c"
Code goes in the text section:
.text
Beats me:
.p2align 4,,15
main is an exported function:
.globl main
.type main, #function
main's entry point:
main:
Grab the return address, align the stack on a 4-byte address, and save the return address again (why I can't say):
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
Save frame pointer using standard convention:
pushl %ebp
movl %esp, %ebp
Inscrutable madness:
pushl %ecx
popl %ecx
Restore the frame pointer and the stack pointer:
popl %ebp
leal -4(%ecx), %esp
Return:
ret
More info for the debugger?:
.size main, .-main
.ident "GCC: (Debian 4.3.2-1.1) 4.3.2"
.section .note.GNU-stack,"",#progbits
By the way, main is special and magical; when I compile
int f(void) {
return 17;
}
I get something slightly more sane:
.file "f.c"
.text
.p2align 4,,15
.globl f
.type f, #function
f:
pushl %ebp
movl $17, %eax
movl %esp, %ebp
popl %ebp
ret
.size f, .-f
.ident "GCC: (Debian 4.3.2-1.1) 4.3.2"
.section .note.GNU-stack,"",#progbits
There's still a ton of decoration, and we're still saving the frame pointer, moving it, and restoring it, which is utterly pointless, but the rest of the code make sense.
It looks like GCC is acting like it is ok to edit main() to include CRT initialization code. I just confirmed that I get the exact same assembly listing from MinGW GCC 3.4.5 here, with your source text.
The command line I used is:
gcc -S emptymain.c
Interestingly, if I change the name of the function to qqq() instead of main(), I get the following assembly:
.file "emptymain.c"
.text
.globl _qqq
.def _qqq; .scl 2; .type 32; .endef
_qqq:
pushl %ebp
movl %esp, %ebp
popl %ebp
ret
which makes much more sense for an empty function with no optimizations turned on.

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