How do i get rid of call __x86.get_pc_thunk.ax - c

I tried to compile and convert a very simple C program to assembly language.
I am using Ubuntu and the OS type is 64 bit.
This is the C Program.
void add();
int main() {
add();
return 0;
}
if i use gcc -S -m32 -fno-asynchronous-unwind-tables -o simple.S simple.c this is how my assembly source code File should look like:
.file "main1.c"
.text
.globl main
.type main, #function
main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
call add
movl $0, %eax
movl %ebp, %esp
popl %ebp
ret
.size main, .-main
.ident "GCC: (Debian 4.4.5-8) 4.4.5" // this part should say Ubuntu instead of Debian
.section .note.GNU-stack,"",#progbits
but instead it looks like this:
.file "main0.c"
.text
.globl main
.type main, #function
main:
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
pushl %ebp
movl %esp, %ebp
pushl %ebx
pushl %ecx
call __x86.get_pc_thunk.ax
addl $_GLOBAL_OFFSET_TABLE_, %eax
movl %eax, %ebx
call add#PLT
movl $0, %eax
popl %ecx
popl %ebx
popl %ebp
leal -4(%ecx), %esp
ret
.size main, .-main
.section
.text.__x86.get_pc_thunk.ax,"axG",#progbits,__x86.get_pc_thunk.ax,comdat
.globl __x86.get_pc_thunk.ax
.hidden __x86.get_pc_thunk.ax
.type __x86.get_pc_thunk.ax, #function
__x86.get_pc_thunk.ax:
movl (%esp), %eax
ret
.ident "GCC: (Ubuntu 6.3.0-12ubuntu2) 6.3.0 20170406"
.section .note.GNU-stack,"",#progbits
At my University they told me to use the Flag -m32 if I am using a 64 bit Linux version. Can somebody tell me what I am doing wrong?
Am I even using the correct Flag?
edit after -fno-pie
.file "main0.c"
.text
.globl main
.type main, #function
main:
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
pushl %ebp
movl %esp, %ebp
pushl %ecx
subl $4, %esp
call add
movl $0, %eax
addl $4, %esp
popl %ecx
popl %ebp
leal -4(%ecx), %esp
ret
.size main, .-main
.ident "GCC: (Ubuntu 6.3.0-12ubuntu2) 6.3.0 20170406"
.section .note.GNU-stack,"",#progbits
it looks better but it's not exactly the same.
for example what does leal mean?

As a general rule, you cannot expect two different compilers to generate the same assembly code for the same input, even if they have the same version number; they could have any number of extra "patches" to their code generation. As long as the observable behavior is the same, anything goes.
You should also know that GCC, in its default -O0 mode, generates intentionally bad code. It's tuned for ease of debugging and speed of compilation, not for either clarity or efficiency of the generated code. It is often easier to understand the code generated by gcc -O1 than the code generated by gcc -O0.
You should also know that the main function often needs to do extra setup and teardown that other functions do not need to do. The instruction leal 4(%esp),%ecx is part of that extra setup. If you only want to understand the machine code corresponding to the code you wrote, and not the nitty details of the ABI, name your test function something other than main.
(As pointed out in the comments, that setup code is not as tightly tuned as it could be, but it doesn't normally matter, because it's only executed once in the lifetime of the program.)
Now, to answer the question that was literally asked, the reason for the appearance of
call __x86.get_pc_thunk.ax
is because your compiler defaults to generating "position-independent" executables. Position-independent means the operating system can load the program's machine code at any address in (virtual) memory and it'll still work. This allows things like address space layout randomization, but to make it work, you have to take special steps to set up a "global pointer" at the beginning of every function that accesses global variables or calls another function (with some exceptions). It's actually easier to explain the code that's generated if you turn optimization on:
main:
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
pushl %ebp
movl %esp, %ebp
pushl %ebx
pushl %ecx
This is all just setting up main's stack frame and saving registers that need to be saved. You can ignore it.
call __x86.get_pc_thunk.bx
addl $_GLOBAL_OFFSET_TABLE_, %ebx
The special function __x86.get_pc_thunk.bx loads its return address -- which is the address of the addl instruction that immediately follows -- into the EBX register. Then we add to that address the value of the magic constant _GLOBAL_OFFSET_TABLE_, which, in position-independent code, is the difference between the address of the instruction that uses _GLOBAL_OFFSET_TABLE_ and the address of the global offset table. Thus, EBX now points to the global offset table.
call add#PLT
Now we call add#PLT, which means call add, but jump through the "procedure linkage table" to do it. The PLT takes care of the possibility that add is defined in a shared library rather than the main executable. The code in the PLT uses the global offset table and assumes that you have already set EBX to point to it, before calling an #PLT symbol.  That's why main has to set up EBX even though nothing appears to use it. If you had instead written something like
extern int number;
int main(void) { return number; }
then you would see a direct use of the GOT, something like
call __x86.get_pc_thunk.bx
addl $_GLOBAL_OFFSET_TABLE_, %ebx
movl number#GOT(%ebx), %eax
movl (%eax), %eax
We load up EBX with the address of the GOT, then we can load the address of the global variable number from the GOT, and then we actually dereference the address to get the value of number.
If you compile 64-bit code instead, you'll see something different and much simpler:
movl number(%rip), %eax
Instead of all this mucking around with the GOT, we can just load number from a fixed offset from the program counter. PC-relative addressing was added along with the 64-bit extensions to the x86 architecture. Similarly, your original program, in 64-bit position-independent mode, will just say
call add#PLT
without setting up EBX first. The call still has to go through the PLT, but the PLT uses PC-relative addressing itself and doesn't need any help from its caller.
The only difference between __x86.get_pc_thunk.bx and __x86.get_pc_thunk.ax is which register they store their return address in: EBX for .bx, EAX for .ax. I have also seen GCC generate .cx and .dx variants. It's just a matter of which register it wants to use for the global pointer -- it must be EBX if there are going to be calls through the PLT, but if there aren't any then it can use any register, so it tries to pick one that isn't needed for anything else.
Why does it call a function to get the return address? Older compilers would do this instead:
call 1f
1: pop %ebx
but that screws up return-address prediction, so nowadays the compiler goes to a little extra trouble to make sure every call is paired with a ret.

The extra junk you're seeing is due to your version of GCC special-casing main to compensate for possibly-broken entry point code starting it with a misaligned stack. I'm not sure how to disable this or if it's even possible, but renaming the function to something other than main will suppress it for the sake of your reading.
After renaming to xmain I get:
xmain:
pushl %ebp
movl %esp, %ebp
subl $8, %esp
call add
movl $0, %eax
leave
ret

Related

Can/do C compilers optimize out adress-of in inline functions?

Let's say I have following code:
int f() {
int foo = 0;
int bar = 0;
foo++;
bar++;
// many more repeated operations in actual code
foo++;
bar++;
return foo+bar;
}
Abstracting repeated code into a separate functions, we get
static void change_locals(int *foo_p, int *bar_p) {
*foo_p++;
*bar_p++;
}
int f() {
int foo = 0;
int bar = 0;
change_locals(&foo, &bar);
change_locals(&foo, &bar);
return foo+bar;
}
I'd expect the compiler to inline the change_locals function, and optimize things like *(&foo)++ in the resulting code to foo++.
If I remember correctly, taking address of a local variable usually prevents some optimizations (e.g. it can't be stored in registers), but does this apply when no pointer arithmetic is done on the address and it doesn't escape from the function? With a larger change_locals, would it make a difference if it was declared inline (__inline in MSVC)?
I am particularly interested in behavior of GCC and MSVC compilers.
inline (and all its cousins _inline, __inline...) are ignored by gcc. It might inline anything it decides is an advantage, except at lower optimization levels.
The code procedure by gcc -O3 for x86 is:
.text
.p2align 4,,15
.globl f
.type f, #function
f:
pushl %ebp
xorl %eax, %eax
movl %esp, %ebp
popl %ebp
ret
.ident "GCC: (GNU) 4.4.4 20100630 (Red Hat 4.4.4-10)"
It returns zero because *ptr++ doesn't do what you think. Correcting the increments to:
(*foo_p)++;
(*bar_p)++;
results in
.text
.p2align 4,,15
.globl f
.type f, #function
f:
pushl %ebp
movl $4, %eax
movl %esp, %ebp
popl %ebp
ret
So it directly returns 4. Not only did it inline them, but it optimized the calculations away.
Vc++ from vs 2005 provides similar code, but it also created unreachable code for change_locals(). I used the command line
/O2 /FD /EHsc /MD /FA /c /TP
If I remember correctly, taking
address of a local variable usually
prevents some optimizations (e.g. it
can't be stored in registers), but
does this apply when no pointer
arithmetic is done on the address and
it doesn't escape from the function?
The general answer is that if the compiler can ensure that no one else will change a value behind its back, it can safely be placed in a register.
Think of this as though the compiler first performs inlining, then transforms all those *&foo (which results from the inlining) to simply foo before deciding if they should be placed in registers on in memory on the stack.
With a larger change_locals, would it
make a difference if it was declared
inline (__inline in MSVC)?
Again, generally speaking, whether or not a compiler decides to inline something is done using heuristics. If you explicitly specify that you want something to be inlines, the compiler will probably weight this into its decision process.
I've tested gcc 4.5, MSC and IntelC using this:
#include <stdio.h>
void change_locals(int *foo_p, int *bar_p) {
(*foo_p)++;
(*bar_p)++;
}
int main() {
int foo = printf("");
int bar = printf("");
change_locals(&foo, &bar);
change_locals(&foo, &bar);
printf( "%i\n", foo+bar );
}
And all of them did inline/optimize the foo+bar value, but also did
generate the code for change_locals() (but didn't use it).
Unfortunately, there's still no guarantee that they'd do the same for
any kind of such a "local function".
gcc:
__Z13change_localsPiS_:
pushl %ebp
movl %esp, %ebp
movl 8(%ebp), %edx
movl 12(%ebp), %eax
incl (%edx)
incl (%eax)
leave
ret
_main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
pushl %ebx
subl $28, %esp
call ___main
movl $LC0, (%esp)
call _printf
movl %eax, %ebx
movl $LC0, (%esp)
call _printf
leal 4(%ebx,%eax), %eax
movl %eax, 4(%esp)
movl $LC1, (%esp)
call _printf
xorl %eax, %eax
addl $28, %esp
popl %ebx
leave
ret

When should I omit the frame pointer?

Is there any substantial optimization when omitting the frame pointer?
If I have understood correctly by reading this page, -fomit-frame-pointer is used when we want to avoid saving, setting up and restoring frame pointers.
Is this done only for each function call and if so, is it really worth to avoid a few instructions for every function?
Isn't it trivial for an optimization.
What are the actual implications of using this option apart from the debugging limitations?
I compiled the following C code with and without this option
int main(void)
{
int i;
i = myf(1, 2);
}
int myf(int a, int b)
{
return a + b;
}
,
# gcc -S -fomit-frame-pointer code.c -o withoutfp.s
# gcc -S code.c -o withfp.s
.
diff -u 'ing the two files revealed the following assembly code:
--- withfp.s 2009-12-22 00:03:59.000000000 +0000
+++ withoutfp.s 2009-12-22 00:04:17.000000000 +0000
## -7,17 +7,14 ##
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
- pushl %ebp
- movl %esp, %ebp
pushl %ecx
- subl $36, %esp
+ subl $24, %esp
movl $2, 4(%esp)
movl $1, (%esp)
call myf
- movl %eax, -8(%ebp)
- addl $36, %esp
+ movl %eax, 20(%esp)
+ addl $24, %esp
popl %ecx
- popl %ebp
leal -4(%ecx), %esp
ret
.size main, .-main
## -25,11 +22,8 ##
.globl myf
.type myf, #function
myf:
- pushl %ebp
- movl %esp, %ebp
- movl 12(%ebp), %eax
- addl 8(%ebp), %eax
- popl %ebp
+ movl 8(%esp), %eax
+ addl 4(%esp), %eax
ret
.size myf, .-myf
.ident "GCC: (GNU) 4.2.1 20070719
Could someone please shed light on the key points of the above code where -fomit-frame-pointer did actually make the difference?
Edit: objdump's output replaced with gcc -S's
-fomit-frame-pointer allows one extra register to be available for general-purpose use. I would assume this is really only a big deal on 32-bit x86, which is a bit starved for registers.*
One would expect to see EBP no longer saved and adjusted on every function call, and probably some additional use of EBP in normal code, and fewer stack operations on occasions where EBP gets used as a general-purpose register.
Your code is far too simple to see any benefit from this sort of optimization-- you're not using enough registers. Also, you haven't turned on the optimizer, which might be necessary to see some of these effects.
* ISA registers, not micro-architecture registers.
The only downside of omitting it is that debugging is much more difficult.
The major upside is that there is one extra general purpose register which can make a big difference on performance. Obviously this extra register is used only when needed (probably in your very simple function it isn't); in some functions it makes more difference than in others.
You can often get more meaningful assembly code from GCC by using the -S argument to output the assembly:
$ gcc code.c -S -o withfp.s
$ gcc code.c -S -o withoutfp.s -fomit-frame-pointer
$ diff -u withfp.s withoutfp.s
GCC doesn't care about the address, so we can compare the actual instructions generated directly. For your leaf function, this gives:
myf:
- pushl %ebp
- movl %esp, %ebp
- movl 12(%ebp), %eax
- addl 8(%ebp), %eax
- popl %ebp
+ movl 8(%esp), %eax
+ addl 4(%esp), %eax
ret
GCC doesn't generate the code to push the frame pointer onto the stack, and this changes the relative address of the arguments passed to the function on the stack.
Profile your program to see if there is a significant difference.
Next, profile your development process. Is debugging easier or more difficult? Do you spend more time developing or less?
Optimizations without profiling are a waste of time and money.

C - what is the return value of a semicolon?

im just curious about the following example
#include<stdio.h>
int test();
int test(){
// int a = 5;
// int b = a+1;
return ;
}
int main(){
printf("%u\n",test());
return 0;
}
i compiled it with 'gcc -Wall -o semicolon semicolon.c' to create an executable
and 'gcc -Wall -S semicolon.c' to get the assembler code which is:
.file "semicolon.c"
.text
.globl test
.type test, #function
test:
pushl %ebp
movl %esp, %ebp
subl $4, %esp
leave
ret
.size test, .-test
.section .rodata
.LC0:
.string "%u\n"
.text
.globl main
.type main, #function
main:
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
pushl %ebp
movl %esp, %ebp
pushl %ecx
subl $20, %esp
call test
movl %eax, 4(%esp)
movl $.LC0, (%esp)
call printf
movl $0, %eax
addl $20, %esp
popl %ecx
popl %ebp
leal -4(%ecx), %esp
ret
.size main, .-main
.ident "GCC: (Ubuntu 4.3.3-5ubuntu4) 4.3.3"
.section .note.GNU-stack,"",#progbits
since im not such an assembler pro, i only know that printf prints what is in eax
but i dont fully understand what 'movl %eax, 4(%esp)' means which i assume fills eax before calling test
but what is the value then? what means 4(%esp) and what does the value of esp mean?
if i uncomment the lines in test() printf prints 6 - which is written in eax ^^
Your assembly language annotated:
test:
pushl %ebp # Save the frame pointer
movl %esp, %ebp # Get the new frame pointer.
subl $4, %esp # Allocate some local space on the stack.
leave # Restore the old frame pointer/stack
ret
Note that nothing in test touches eax.
.size test, .-test
.section .rodata
.LC0:
.string "%u\n"
.text
.globl main
.type main, #function
main:
leal 4(%esp), %ecx # Point past the return address.
andl $-16, %esp # Align the stack.
pushl -4(%ecx) # Push the return address.
pushl %ebp # Save the frame pointer
movl %esp, %ebp # Get the new frame pointer.
pushl %ecx # save the old top of stack.
subl $20, %esp # Allocate some local space (for printf parameters and ?).
call test # Call test.
Note that at this point, nothing has modified eax. Whatever came into main is still here.
movl %eax, 4(%esp) # Save eax as a printf argument.
movl $.LC0, (%esp) # Send the format string.
call printf # Duh.
movl $0, %eax # Return zero from main.
addl $20, %esp # Deallocate local space.
popl %ecx # Restore the old top of stack.
popl %ebp # And the old frame pointer.
leal -4(%ecx), %esp # Fix the stack pointer,
ret
So, what gets printed out is whatever came in to main. As others have pointed out it is undefined: It depends on what the startup code (or the OS) has done to eax previously.
The semicolon has no return value, what you have there is an "empty return", like the one used to return from void functions - so the function doesn't return anything.
This actually causes a warning when compiling:
warning: `return' with no value, in function returning non-void
And I don't see anything placed in eax before calling test.
About 4(%esp), this means taking the value from the stack pointer (esp) + 4. I.e. the one-before-last word on the stack.
The return value of an int function is passed in the EAX register. The test function does not set the EAX register because no return value is given. The result is therefore undefined.
A semicolon indeed has no value.
I think the correct answer is that a return <nothing> for an int function is an error, or at least has undefined behavor. That's why compiling this with -Wall yields
semi.c: In function ‘test’:
semi.c:6: warning: ‘return’ with no value, in function returning non-void
As for what the %4,esp holds... it's a location on the stack where nothing was (intentionally) stored, so it will likely return whatever junk is found at that location. This could be the last expression evaluated to variables in the function (as in your example) or something completely different. This is what "undefined" is all about. :)

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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