I wrote function in C that does work fine for me, but I need to write the same function in assembly. I generated assembly code of this function using `gcc -S rrd.c and I need some help in implementing it in asm. I'm using Linux, x86 assembly.
My C function:
double rrd (double a)
{
FILE *f=fopen("inital.txt","r");
if(f==NULL)
return 1;
double first, second;
do
{
fscanf(f, "%lf", &first);
fscanf(f, "%lf", &second);
}
while(a >= first);
close(f);
return second;
}
This is what I get:
http://pastebin.com/zyEBgTEC
And this is what I have so far:
.data
mode:
.string "r"
file:
.string "inital.txt"
format:
.string "%lf"
value:
.space 8
first:
.space 8
second:
.space 8
pointer:
.space 8
.text
.global re
re:
pushl %ebp
movl %esp, %ebp
fldl 8(%ebp)
fstpl value
pushl $mode
pushl $file
call fopen
movl %eax, pointer
cmpl $0, pointer
jne loop
jmp end
loop:
leal 16(%esp), %edx
pushl %edx
pushl $format
pushl pointer
call __isoc99_fscanf
leal 24(%esp), %edx
pushl %edx
pushl $format
pushl pointer
call __isoc99_fscanf
fldl 16(%esp)
fldl 32(%esp)
pushl pointer
call close
end:
leave
ret
Result after fldl instructions:
st0 -2.3534389814351579410727679528547297e-185 (raw 0xbd99ccccd5ffff9ac000)
st1 -1.9968719781118224876337308160145767 (raw 0xbfffff99804025016800)
My problem is that I don't exactly know where are the results of fscanf instructions and I don't know why there are those leal instructions.
Could anyone help me to rewrite this function?
So it should look like this:
wiecej:
leal pierwsza, %edx
pushl %edx
pushl $format
pushl wskaznik
call __isoc99_fscanf
leal druga, %edx
pushl %edx
pushl $format
pushl wskaznik
call __isoc99_fscanf
fldl pierwsza
fldl b
fucomip %st(1), %st
fstp %st(0)
setae %al
testb %al, %al
jne wiecej
fldl druga
Related
I wrote a single c program that prints input to std output. Then I converted it to assembly language. By the way I am using AT&T Syntax.
This is the simple C code.
#include <stdio.h>
int main()
{
int c;
while ((c = getchar ()) != EOF)
{
putchar(c);
}
return 0;
}
int c is a local variable.
Then I converted it to assembly language.
.file "question_1.c"
.text
.globl main
.type main, #function
//prolog
main:
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
pushl %ebp
movl %esp, %ebp
pushl %ecx
subl $20, %esp // we add 20 bytes to the stack
jmp .L2
.L3:
subl $12, %esp
pushl -12(%ebp)
call putchar
addl $16, %esp
.L2:
call getchar
movl %eax, -12(%ebp)
cmpl $-1, -12(%ebp)
jne .L3
//assumption this is the epilog
movl $0, %eax
movl -4(%ebp), %ecx
leave
leal -4(%ecx), %esp
ret
.size main, .-main
.ident "GCC: (Ubuntu 4.9.4-2ubuntu1) 4.9.4"
.section .note.GNU-stack,"",#progbits
normally in the epilog we are supposed to addl 20 because in the prolog we subl 20.
So the is the stack frame still there?
Or am I missing out a crucial point?
I also have a question regarding the main function. Normally functions are normally "called" but where does it happen in the assembly code?
Thank you in advance.
Just after the main label, leal 4(%esp), %ecx saves four plus the stack pointer in %ecx. At the end of the routine, leal -4(%ecx), %esp writes four less than the saved value to the stack pointer. This directly restores the original value, instead of doing it by adding the amount that was subtracted.
I don't understand why gcc -S -m32 produces these particular lines of code:
movl %eax, 28(%esp)
movl $desc, 4(%esp)
movl 28(%esp), %eax
movl %eax, (%esp)
call sort_gen_asm
My question is why %eax is pushed and then popped? And why movl used instead of pushl and popl respectively? Is it faster? Is there some coding convention I don't yet know? I've just started looking at asm-output closely, so I don't know much.
The C code:
void print_array(int *data, size_t sz);
void sort_gen_asm(array_t*, comparer_t);
int main(int argc, char *argv[]) {
FILE *file;
array_t *array;
file = fopen("test", "rb");
if (file == NULL) {
err(EXIT_FAILURE, NULL);
}
array = array_get(file);
sort_gen_asm(array, desc);
print_array(array->data, array->sz);
array_destroy(array);
fclose(file);
return 0;
}
It gives this output:
.file "main.c"
.section .rodata
.LC0:
.string "rb"
.LC1:
.string "test"
.text
.globl main
.type main, #function
main:
.LFB2:
.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
movl $.LC0, 4(%esp)
movl $.LC1, (%esp)
call fopen
movl %eax, 24(%esp)
cmpl $0, 24(%esp)
jne .L2
movl $0, 4(%esp)
movl $1, (%esp)
call err
.L2:
movl 24(%esp), %eax
movl %eax, (%esp)
call array_get
movl %eax, 28(%esp)
movl $desc, 4(%esp)
movl 28(%esp), %eax
movl %eax, (%esp)
call sort_gen_asm
movl 28(%esp), %eax
movl 4(%eax), %edx
movl 28(%esp), %eax
movl (%eax), %eax
movl %edx, 4(%esp)
movl %eax, (%esp)
call print_array
movl 28(%esp), %eax
movl %eax, (%esp)
call array_destroy
movl 24(%esp), %eax
movl %eax, (%esp)
call fclose
movl $0, %eax
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
.cfi_endproc
.LFE2:
.size main, .-main
.ident "GCC: (Ubuntu/Linaro 4.8.1-10ubuntu8) 4.8.1"
.section .note.GNU-stack,"",#progbits
The save / load of eax is because you did not compile with optimizations. So any read/write of a variable will emit a read/write of a memory address.
Actually, for (almost) any line of code you will be able to identify the exact piece of assembler code resulting from it (let me advise you to compile with gcc -g -c -O0 and then objdump -S file.o):
#array = array_get(file);
call array_get
movl %eax, 28(%esp) #write array
#sort_gen_asm(array, desc);
movl 28(%esp), %eax #read array
movl %eax, (%esp)
...
About not pushing/poping, it is a standard zero-cost optimization. Instead of push/pop every time you want to call a function you just substract the maximum needed space to esp at the beginning of the function and then save your function arguments at the bottom of the empty space. There are a lot of advantages: faster code (no changing esp), it doesn't need to compute the argument in any particular order, and the esp will need to be substracted anyway for the local variables space.
Some things have to do with calling conventions. Others with optimisations.
sort_gen_asm seems to use cdecl calling convention which requires it's arguments to be pushed onto the stack in reverse order. thus:
movl $desc, 4(%esp)
movl %eax, (%esp)
The other moves are partially unoptimised compiler routines:
movl %eax, 28(%esp) # save contents of %eax on the stack before calling
movl 28(%esp), %eax # retrieve saved 28(%esp) in order to prepare it as an argument
# Unoptimised compiler seems to have forgotten that it's
# still in the register
I'm trying to understand the assembly code of the C function. I could not understand why andl -16 is done at the main. Is it for allocating space for the local variables. If so why subl 32 is done for main.
I could not understand the disassembly of the func1. As read the stack grows from higher order address to low order address for 8086 processors. So here why is the access on positive side of the ebp(for parameters offset) and why not in the negative side of ebp. The local variables inside the func1 is 3 + return address + saved registers - So it has to be 20, but why is it 24? (subl $24,esp)
#include<stdio.h>
int add(int a, int b){
int res = 0;
res = a + b;
return res;
}
int func1(int a){
int s1,s2,s3;
s1 = add(a,a);
s2 = add(s1,a);
s3 = add(s1,s2);
return s3;
}
int main(){
int a,b;
a = 1;b = 2;
b = func1(a);
printf("\n a : %d b : %d \n",a,b);
return 0;
}
assembly code :
.file "sample.c"
.text
.globl add
.type add, #function
add:
pushl %ebp
movl %esp, %ebp
subl $16, %esp
movl $0, -4(%ebp)
movl 12(%ebp), %eax
movl 8(%ebp), %edx
leal (%edx,%eax), %eax
movl %eax, -4(%ebp)
movl -4(%ebp), %eax
leave
ret
.size add, .-add
.globl func1
.type func1, #function
func1:
pushl %ebp
movl %esp, %ebp
subl $24, %esp
movl 8(%ebp), %eax
movl %eax, 4(%esp)
movl 8(%ebp), %eax
movl %eax, (%esp)
call add
movl %eax, -4(%ebp)
movl 8(%ebp), %eax
movl %eax, 4(%esp)
movl -4(%ebp), %eax
movl %eax, (%esp)
call add
movl %eax, -8(%ebp)
movl -8(%ebp), %eax
movl %eax, 4(%esp)
movl -4(%ebp), %eax
movl %eax, (%esp)
call add
movl %eax, -12(%ebp)
movl -12(%ebp), %eax
leave
ret
.size func1, .-func1
.section .rodata
.LC0:
.string "\n a : %d b : %d \n"
.text
.globl main
.type main, #function
main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
subl $32, %esp
movl $1, 28(%esp)
movl $2, 24(%esp)
movl 28(%esp), %eax
movl %eax, (%esp)
call func1
movl %eax, 24(%esp)
movl $.LC0, %eax
movl 24(%esp), %edx
movl %edx, 8(%esp)
movl 28(%esp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
movl $0, %eax
leave
ret
.size main, .-main
.ident "GCC: (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5"
.section .note.GNU-stack,"",#progbits
The andl $-16, %esp aligns the stack pointer to a multiple of 16 bytes, by clearing the low four bits.
The only places where positive offsets are used with (%ebp) are parameter accesses.
You did not state what your target platform is or what switches you used to compile with. The assembly code shows some Ubuntu identifier has been inserted, but I am not familiar with the ABI it uses, beyond that it is probably similar to ABIs generally used with the Intel x86 architecture. So I am going to guess that the ABI requires 8-byte alignment at routine calls, and so the compiler makes the stack frame of func1 24 bytes instead of 20 so that 8-byte alignment is maintained.
I will further guess that the compiler aligned the stack to 16 bytes at the start of main as a sort of “preference” in the compiler, in case it uses SSE instructions that prefer 16-byte alignment, or other operations that prefer 16-byte alignment.
So, we have:
In main, the andl $-16, %esp aligns the stack to a multiple of 16 bytes as a compiler preference. Inside main, 28(%esp) and 24(%esp) refer to temporary values the compiler saves on the stack, while 8(%esp), 4(%esp), and (%esp) are used to pass parameters to func1 and printf. We see from the fact that the assembly code calls printf but it is commented out in your code that you have pasted C source code that is different from the C source code used to generate the assembly code: This is not the correct assembly code generated from the C source code.
In func1, 24 bytes are allocated on the stack instead of 20 to maintain 8-byte alignment. Inside func1, parameters are accessed through 8(%ebp) and 4(%ebp). Locations from -12(%ebp) to -4(%ebp) are used to hold values of your variables. 4(%esp) and (%esp) are used to pass parameters to add.
Here is the stack frame of func1:
- 4(%ebp) = 20(%esp): s1.
- 8(%ebp) = 16(%esp): s2.
-12(%ebp) = 12(%esp): s3.
-16(%ebp) = 8(%esp): Unused padding.
-20(%ebp) = 4(%esp): Passes second parameter of add.
-24(%ebp) = 0(%esp): Passes first parameter of add.
I would suggest working through this with the output of objdump -S which will give you interlisting with the C source.
I'm trying to understand the assembly code during a recursive function call.
#include<stdio.h>
int recursive(int no){
if(no > 1){
no--;
recursive(no);
printf("\n %d \n",no);
}
else if(no == 1){
return 1;
}
}
int main(){
int a = 10;
recursive(a);
return 0;
}
disassembly :
.file "sample2.c"
.section .rodata
.LC0:
.string "\n %d \n"
.text
.globl recursive
.type recursive, #function
recursive:
pushl %ebp
movl %esp, %ebp
subl $24, %esp
cmpl $1, 8(%ebp)
jle .L2
subl $1, 8(%ebp)
movl 8(%ebp), %eax
movl %eax, (%esp)
call recursive
movl $.LC0, %eax
movl 8(%ebp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
jmp .L5
.L2:
cmpl $1, 8(%ebp)
jne .L5
movl $1, %eax
movl %eax, %edx
movl %edx, %eax
jmp .L4
.L5:
.L4:
leave
ret
.size recursive, .-recursive
.globl main
.type main, #function
main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
subl $32, %esp
movl $10, 28(%esp)
movl 28(%esp), %eax
movl %eax, (%esp)
call recursive
movl $0, %eax
leave
ret
.size main, .-main
.ident "GCC: (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5"
.section .note.GNU-stack,"",#progbits
I could understand .LC0 always holds the string literals. But I dont know what it really means. Would like to understand the code during the function call recursion was made.
I could not understand what this piece of assembly code does,
subl $24, %esp
cmpl $1, 8(%ebp)
jle .L2
subl $1, 8(%ebp)
movl 8(%ebp), %eax
movl %eax, (%esp)
call recursive
movl $.LC0, %eax
movl 8(%ebp), %edx
movl %edx, 4(%esp)
movl %eax, (%esp)
call printf
jmp .L5
.L2:
cmpl $1, 8(%ebp)
jne .L5
movl $1, %eax
movl %eax, %edx
movl %edx, %eax
jmp .L4
Q1:
The recursive function contains 1 parameter. so after the padding alignment, it has to be 8. why is it 24.
Also in .L2 ,
movl $1, %eax
movl %eax, %edx
movl %edx, %eax
jmp .L4
Q2:
we have moved '1' to the accumulater, why are we moving again to data register and then back to the accumulator.
Q3:
Are we popping out of stack. If leave is used for popping out of stack, are we not popping the rest of the 8 stack frames ?
To answer the only thing in your post that matches your title:
Why are we not popping out from the stack and only push instruction in the assembly.
Because leave is equivalent to:
movl %ebp, %esp
popl %ebp
i'm a new one to learn assembly. i write a c file:
#include <stdlib.h>
int max( int c )
{
int d;
d = c + 1;
return d;
}
int main( void )
{
int a = 0;
int b;
b = max( a );
return 0;
}
and i use gcc -S as01.c and create a assembly file.
.file "as01.c"
.text
.globl max
.type max, #function
max:
pushl %ebp
movl %esp, %ebp
subl $32, %esp
movl $0, -4(%ebp)
movl $1, -24(%ebp)
movl $2, -20(%ebp)
movl $3, -16(%ebp)
movl $4, -12(%ebp)
movl $6, -8(%ebp)
movl 8(%ebp), %eax
addl $1, %eax
movl %eax, -4(%ebp)
movl -4(%ebp), %eax
leave
ret
.size max, .-max
.globl main
.type main, #function
main:
pushl %ebp
movl %esp, %ebp
subl $20, %esp
movl $0, -4(%ebp)
movl -4(%ebp), %eax
movl %eax, (%esp)
call max
movl %eax, -8(%ebp)
"as01.s" 38L, 638C
i' confused, beacuse movl %eax, -4(%ebp) movl -4(%ebp), %eax in max(),
i know that %eax is used for returning the value of any function.
I think %eax is a temporarily register for store the c + 1.
This is right?
thank you for your answer.
You don't have optimisation turned on, so the compiler is generating really bad code. The primary storage for all your values is in the stack frame, and values are loaded into registers only long enough to do the calculations.
The code actually breaks down into:
pushl %ebp
movl %esp, %ebp
subl $32, %esp
Standard function prologue, setting up a new stack frame, and reserving 50 bytes for the stack frame.
movl $0, -4(%ebp)
movl $1, -24(%ebp)
movl $2, -20(%ebp)
movl $3, -16(%ebp)
movl $4, -12(%ebp)
movl $6, -8(%ebp)
Fill the stack frame with dummy values (presumably as a debugging aid).
movl 8(%ebp), %eax
addl $1, %eax
movl %eax, -4(%ebp)
Read the parameter c out of the stack frame, add one to it, store it into a (different) stack slot.
movl -4(%ebp), %eax
leave
ret
Read the value back out of the stack slot and return it.
If you compile this with optimisation, you'll see most of the code vanish. If you use -fomit-frame-pointer -Os, you should end up with this:
max:
movl 4(%esp), %eax
incl %eax
ret
movl %eax, -4(%ebp)
Here the value computed for d (now stored in eax) is saved in d's memory cell.
movl -4(%ebp), %eax
While here the return value (d's) gets loaded into eax, because, as you know, eax holds functions' return value.
As #David said, you're compiling without optimization, so gcc generates easy-to-debug code, which is quite inefficient and repetitive sometimes.