I have this ".s" file, written in AT&T assembly.
.globl interleave
interleave:
pushl %ebx
pushl %esi
pushl %edi
movl 16(%esp), %ebx #a
movl 20(%esp), %esi #b
movl 24(%esp), %edi #c
D: movb (%ebx), %cl
testb %cl, %cl
jz W
movb %cl, (%edi) #*c
incl %edi
incl %ebx
T: movb (%esi), %dl
testb %dl, %dl
jz W
movb %dl, (%edi) #*c
incl %edi
incl %esi
W: orb %cl,%dl
jz E
#movb $0, %al
jmp D
E: movb $0, (%edi)
popl %edi
popl %esi
popl %ebx
ret
I want to compile it on windows 10 with cygwin with the following main file, but it does not work.
void interleave(const char* a, const char* b, char* c) ;
int main(int argc, char const *argv[]){
const char* a = "car";
const char* b = "old";
char c[] = "";
interleave(a,b,c);
printf("%s (expected coalrd)\n", c);
return 0;}
With gcc i get es1B.s:3: Error: invalid instruction suffix for push
With gcc -m32 I get collect2: error: ld returned 1 exit status
I even tried to compile it in 32 bit with i686-w64-mingw32-gcc but I get undefined reference to interleave
I am able to compile it and run it on linux with gcc -m32 , but is there a way to make this work on windows?
Thanks
Solved by adding an underscore before the function name as suggested in the comments:
.globl _interleave
_interleave:
...
Compiling with i686-w64-mingw32-gcc now works.
Related
Why is that replacing
movl $84, 3(%rsi)
with
movl $3, %ecx
leal (%esi, %ecx, 1), %r11d
movl $84, (%r11d)
results in Segmentation fault (core dumped) and how can I solve it?
(I will be using %ecx as a counter later on to loop through the array)
As I understand it, movl $84, 3(%rsi) moves the literal value 84 (ASCII 'T') into result[2] (which works and correct, based on the printed output: 24 = 02aTa)
On the other hand,
movl $3, %ecx
leal (%esi, %ecx, 1), %r11d
movl $84, (%r11d)
line 1 moves integer value 3 into %ecx
line 2 calculates %esi + 1 * %ecx = %esi + %ecx which is the same as 3(%esi) (incorrect, as it does not work)
line 3 moves integer value 84 ('T') into the address above (which should be the same as movl $84, 3(%rsi))
Replacing %esi with %rsi and so on (including using leaq, movq) in the non-working part, I get
*** stack smashing detected ***: ./a.out terminated
Aborted (core dumped)
So my question is: How can I achieve the effect of movl $84, %ecx(%esi)?
Compile with gcc main.c helper.s
main.c
#include <stdio.h>
void helper(unsigned int value, char * result);
int main() {
char result[6];
helper(24, result);
printf("%5u = %s\n", 24, result);
return 0;
}
helper.s
.globl helper
helper:
# 0
movl $48, 0(%rsi)
# 2
movl $50, 1(%rsi)
# a
movl $97, 2(%rsi)
#ifdef DEBUG
# T
movl $84, 3(%rsi)
#ELSE
/*
movl $3, %ecx
leal (%esi, %ecx, 1), %r11d
# T
movl $84, (%r11d)
*/
#ENDIF
# a
movl $97, 4(%rsi)
movl $0x0000000000000000, 5(%rsi)
ret
Replacing all %rsi with %esi (and all 64-bit registers with 32-bit registers), I get Segmentation fault (core dumped). Why is this?
This is a homework task. I've got a C program that calls a function calc(int, float*, float*, float*, float*) implemented with NASM. I want to do floating-point division with the data passed from C, but first I wanted to check if I access the data correctly.
This is an excerpt from the C program:
printf("read.c: F data1[0]=%f\n", data1[0]);
printf("read.c: X data1[0]=%X\n", *(int*)(&data1[0]));
calc(nlines, data1, data2, result1, result2);
For testing, I wanted to print out exactly the same from the assembler code, but whatever I tried, it wouldn't give me the right results. To be precise, outputting the %X format gives the same result, but the %f format gives some incredibly huge number.
global calc
extern printf
; -----------------------------------------------------------------------
; extern void calc(int nlines, float* data1, float* data2,
; float* result1, float* result2)
; -----------------------------------------------------------------------
calc:
section .data
.strf db "calc.asm: F data1[0]=%f", 10, 0
.strx db "calc.asm: X data1[0]=%X", 10, 0
section .text
enter 0, 0
; Move the value of float* data1 into ecx.
mov ecx, [esp + 12]
; Move the contents of data1[0] into esi.
mov esi, [ecx]
push esi
push .strf
call printf
add esp, 8
push esi
push .strx
call printf
add esp, 8
leave
ret
Outputs
read.c: F data1[0]=20.961977
read.c: X data1[0]=41A7B221
calc.asm: F data1[0]=-8796958457989122902187458235483374032941932827208012972482327255932202912296419757153331437662235555722313731094096197990916443553479942683040096290755684437514827018615169352974748429901549205109479495668937369584705401541113350145698235773041651907978442730240007381959397006695721667307435228446926569472.000000
calc.asm: X data1[0]=41A7B221
I've also looked into fld, but I couldn't find out how I can push the loaded value on stack. This didnt work:
; Move float* data1 into ecx
mov ecx, [esp + 12]
; Load the floating point number into esi.
fld dword [ecx]
fst esi
How to do it right?
I've stripped down read.c to this code
#include <stdio.h>
#include <stdlib.h>
#define MAXLINES 1024
extern void calc(int, float*, float*, float*, float*);
int main(int argc, char** argv)
{
int nlines;
float* data1 = malloc(sizeof(float)*MAXLINES);
float*data2, *results1, *results2;
printf("read.c: F data1[0]=%f\n", data1[0]);
printf("read.c: X data1[0]=%X\n", *(int*)(&data1[0]));
calc(nlines, data1, data2, results1, results2);
return 0;
}
and this is the assembler output:
.file "test.c"
.section .rodata
.LC0:
.string "read.c: F data1[0]=%f\n"
.LC1:
.string "read.c: X data1[0]=%X\n"
.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 $64, %esp
movl $4096, (%esp)
call malloc
movl %eax, 44(%esp)
movl 44(%esp), %eax
flds (%eax)
fstpl 4(%esp)
movl $.LC0, (%esp)
call printf
movl 44(%esp), %eax
movl (%eax), %eax
movl %eax, 4(%esp)
movl $.LC1, (%esp)
call printf
movl 60(%esp), %eax
movl %eax, 16(%esp)
movl 56(%esp), %eax
movl %eax, 12(%esp)
movl 52(%esp), %eax
movl %eax, 8(%esp)
movl 44(%esp), %eax
movl %eax, 4(%esp)
movl 48(%esp), %eax
movl %eax, (%esp)
call calc
movl $0, %eax
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
.cfi_endproc
.LFE2:
.size main, .-main
.ident "GCC: (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4"
.section .note.GNU-stack,"",#progbits
.LC1:
.string "read.c: F data1[0]=%f\n"
.LC2:
.string "read.c: X data1[0]=%X\n"
.text
.globl main
.type main, #function
main:
.LFB4:
.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 $64, %esp
movl 44(%esp), %eax
flds (%eax)
fstpl 4(%esp)
movl $.LC1, (%esp)
call printf
movl 44(%esp), %eax
movl (%eax), %eax
movl %eax, 4(%esp)
movl $.LC2, (%esp)
call printf
movl 60(%esp), %eax
movl %eax, 16(%esp)
movl 56(%esp), %eax
movl %eax, 12(%esp)
movl 52(%esp), %eax
movl %eax, 8(%esp)
movl 44(%esp), %eax
movl %eax, 4(%esp)
movl 48(%esp), %eax
movl %eax, (%esp)
call calc
movl $0, %eax
leave
.cfi_restore 5
.cfi_def_cfa 4, 4
ret
.cfi_endproc
.LFE4:
.size main, .-main
.ident "GCC: (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4"
.section .note.GNU-stack,"",#progbits
Ok, I've now had a chance to test this and verify that what I suggested in my comment works. Here's my modified version of the assembly code, with some comments to explain the things I've added/changed:
global _calc
extern _printf
; -----------------------------------------------------------------------
; extern void calc(int nlines, float* data1, float* data2,
; float* result1, float* result2)
; -----------------------------------------------------------------------
_calc:
section .data
.strf db "calc.asm: F data1[0]=%f", 10, 0
.strx db "calc.asm: X data1[0]=%X", 10, 0
section .text
enter 0, 0
; Move the value of float* data1 into ecx.
mov ecx, [esp + 12]
; Move the contents of data1[0] into esi.
mov esi, [ecx]
fld dword [ecx] ; Load a single-precision float onto the FP stack.
sub esp,8 ; Make room for a double on the stack.
fstp qword [esp] ; Store the top of the FP stack on the regular stack as
; a double, and pop it off the FP stack.
push .strf
call _printf
add esp, 12 ; 12 == sizeof(char*) + sizeof(double)
push esi
push .strx
call _printf
add esp, 8
leave
ret
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
x86 assembly registers — Why do they work the way they do?
I've compiled the following program:
#include <stdio.h>
int square(int x) {
return x * x;
}
int main() {
int y = square(9);
printf("%d\n", y);
return 0;
}
two times with different options on OSX with GCC 4.2.1:
gcc foo.c -o foo_32.s -S -fverbose-asm -m32 -O1
gcc foo.c -o foo_64.s -S -fverbose-asm -m64 -O1
The result for 32 bit:
_square: ## #square
## BB#0: ## %entry
pushl %ebp
movl %esp, %ebp
movl 8(%ebp), %eax
imull %eax, %eax
popl %ebp
ret
And for 64-bit:
_square: ## #square
Leh_func_begin1:
## BB#0: ## %entry
pushq %rbp
Ltmp0:
movq %rsp, %rbp
Ltmp1:
movl %edi, %eax
imull %eax, %eax
popq %rbp
ret
As is evident, the 32-bit version retrieves the parameter from the stack, which is what one would expect with cdecl. The 64-bit version however uses the EDI register to pass the parameter.
Doesn't this violate the System V AMD64 ABI, which specifies that the RDI, RSI, RDX, RCX, R8, R9, XMM0–7 registers should be used? Or is this only the case for true 64-bit values like long?
EDI is simply the lower half of RDI so the compiler is passing the argument in RDI, but the argument is only 32-bits long, so it only takes up half of the register.
I was trying to come up with inline assembly for gcc to get both division and modulus using single divl instruction. Unfortunately, I am not that good at assembly. Could someone please help me on this? Thank you.
You're looking for something like this:
__asm__("divl %2\n"
: "=d" (remainder), "=a" (quotient)
: "g" (modulus), "d" (high), "a" (low));
Although I agree with the other commenters that usually GCC will do this for you and you should avoid inline assembly when possible, sometimes you need this construct.
For instance, if the high word is less than the modulus, then it is safe to perform the division like this. However, GCC isn't smart enough to realize this, because in the general case dividing a 64 bit number by a 32 bit number can lead to overflow, and so it calls to a library routine to do extra work. (Replace with 128 bit/64 bit for 64 bit ISAs.)
You shouldn't try to optimize this yourself. GCC already does this.
volatile int some_a = 18, some_b = 7;
int main(int argc, char *argv[]) {
int a = some_a, b = some_b;
printf("%d %d\n", a / b, a % b);
return 0;
}
Running
gcc -S test.c -O
yields
main:
.LFB11:
.cfi_startproc
subq $8, %rsp
.cfi_def_cfa_offset 16
movl some_a(%rip), %esi
movl some_b(%rip), %ecx
movl %esi, %eax
movl %esi, %edx
sarl $31, %edx
idivl %ecx
movl %eax, %esi
movl $.LC0, %edi
movl $0, %eax
call printf
movl $0, %eax
addq $8, %rsp
.cfi_def_cfa_offset 8
ret
Notice that the remainder, %edx, is not moved because it is also the third argument passed to printf.
EDIT: The 32-bit version is less confusing. Passing -m32 yields
main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
subl $16, %esp
movl some_a, %eax
movl some_b, %ecx
movl %eax, %edx
sarl $31, %edx
idivl %ecx
movl %edx, 8(%esp)
movl %eax, 4(%esp)
movl $.LC0, (%esp)
call printf
movl $0, %eax
leave
ret
Fortunately, you don't have to resort to inline assembly to achieve this. gcc will do this automatically when it can.
$ cat divmod.c
struct sdiv { unsigned long quot; unsigned long rem; };
struct sdiv divide( unsigned long num, unsigned long divisor )
{
struct sdiv x = { num / divisor, num % divisor };
return x;
}
$ gcc -O3 -std=c99 -Wall -Wextra -pedantic -S divmod.c -o -
.file "divmod.c"
.text
.p2align 4,,15
.globl divide
.type divide, #function
divide:
.LFB0:
.cfi_startproc
movq %rdi, %rax
xorl %edx, %edx
divq %rsi
ret
.cfi_endproc
.LFE0:
.size divide, .-divide
.ident "GCC: (GNU) 4.4.4 20100630 (Red Hat 4.4.4-10)"
.section .note.GNU-stack,"",#progbits
Yes -- a divl will produce the quotient in eax and the remainder in edx. Using Intel syntax, for example:
mov eax, 17
mov ebx, 3
xor edx, edx
div ebx
; eax = 5
; edx = 2
Here is an example in linux kernel code about divl
/*
* do_div() is NOT a C function. It wants to return
* two values (the quotient and the remainder), but
* since that doesn't work very well in C, what it
* does is:
*
* - modifies the 64-bit dividend _in_place_
* - returns the 32-bit remainder
*
* This ends up being the most efficient "calling
* convention" on x86.
*/
#define do_div(n, base) \
({ \
unsigned long __upper, __low, __high, __mod, __base; \
__base = (base); \
if (__builtin_constant_p(__base) && is_power_of_2(__base)) { \
__mod = n & (__base - 1); \
n >>= ilog2(__base); \
} else { \
asm("" : "=a" (__low), "=d" (__high) : "A" (n));\
__upper = __high; \
if (__high) { \
__upper = __high % (__base); \
__high = __high / (__base); \
} \
asm("divl %2" : "=a" (__low), "=d" (__mod) \
: "rm" (__base), "0" (__low), "1" (__upper)); \
asm("" : "=A" (n) : "a" (__low), "d" (__high)); \
} \
__mod; \
})
I came across this inline asm. I am not sure how it should look without this syntax... Could someone show it to me?
__asm__ volatile ("lock\n\tincl %0"
:"=m"(llvm_cbe_tmp__29)
:"m"(*(llvm_cbe_tmp__29))"cc");
lock
incl llvm_cbe_tmp__29
However, because the operand is specified abstractly, the compiler will generate the code needed to reference it, even if that means a load and store. As a result it is possible that more than two instructions or an addressing mode will be added.
Using gcc -S on this:
int main()
{
int *p;
asm volatile ("lock\n\tincl %0":"=m"(p):"m"(*(p)):"cc");
}
gives
.type main, #function
main:
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
pushl %ebp
movl %esp, %ebp
pushl %ecx
subl $20, %esp
movl -8(%ebp), %eax
#APP
# 4 "asm.c" 1
lock
incl -8(%ebp)
# 0 "" 2
#NO_APP
addl $20, %esp
popl %ecx
popl %ebp
leal -4(%ecx), %esp
ret