I want to hard code a map in Haskell. I can see at least three ways to do it:
Using multiple equations:
message 200 = "OK"
message 404 = "Not found"
...
Using a case expression:
message s = case s of
200 -> "OK"
404 -> "Not found"
Actually using a Map.
Which is the most efficient way do it? Is one solution faster than the others, and why?
Are the first two solutions equivalent? (Will the compiler generates the same code?)
What is the recommended way (easier to read)?
(Note that I use Int in my example, but that is not essential. The keys might be Strings as well so I'm interested in both cases.)
Pattern matching on Ints happens in O(log(n)) time, like a map lookup.
Consider the following code, compiled to x86 assembly by ghc -S
module F (
f
) where
f :: Int -> String
f 0 = "Zero"
f 1 = "One"
f 2 = "Two"
f 3 = "Three"
f 4 = "Four"
f 5 = "Five"
f 6 = "Six"
f 7 = "Seven"
f _ = "Undefined"
The compiled assembly code is
.text
.align 4,0x90
.long _F_f_srt-(_sl8_info)+0
.long 0
.long 65568
_sl8_info:
.Lcma:
movl 3(%esi),%eax
cmpl $4,%eax
jl .Lcmq
cmpl $6,%eax
jl .Lcmi
cmpl $7,%eax
jl .Lcme
cmpl $7,%eax
jne .Lcmc
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_cm7_str,0(%ebp)
jmp _stg_ap_n_fast
.Lcmc:
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_clB_str,0(%ebp)
jmp _stg_ap_n_fast
.Lcme:
cmpl $6,%eax
jne .Lcmc
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_cm3_str,0(%ebp)
jmp _stg_ap_n_fast
.Lcmg:
cmpl $4,%eax
jne .Lcmc
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_clV_str,0(%ebp)
jmp _stg_ap_n_fast
.Lcmi:
cmpl $5,%eax
jl .Lcmg
cmpl $5,%eax
jne .Lcmc
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_clZ_str,0(%ebp)
jmp _stg_ap_n_fast
.Lcmk:
cmpl $2,%eax
jne .Lcmc
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_clN_str,0(%ebp)
jmp _stg_ap_n_fast
.Lcmm:
testl %eax,%eax
jne .Lcmc
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_clF_str,0(%ebp)
jmp _stg_ap_n_fast
.Lcmo:
cmpl $1,%eax
jl .Lcmm
cmpl $1,%eax
jne .Lcmc
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_clJ_str,0(%ebp)
jmp _stg_ap_n_fast
.Lcmq:
cmpl $2,%eax
jl .Lcmo
cmpl $3,%eax
jl .Lcmk
cmpl $3,%eax
jne .Lcmc
movl $_ghczmprim_GHCziCString_unpackCStringzh_closure,%esi
movl $_clR_str,0(%ebp)
jmp _stg_ap_n_fast
.text
.align 4,0x90
.long _F_f_srt-(_F_f_info)+0
.long 65541
.long 0
.long 65551
.globl _F_f_info
_F_f_info:
.Lcmu:
movl 0(%ebp),%esi
movl $_sl8_info,0(%ebp)
testl $3,%esi
jne .Lcmx
jmp *(%esi)
.Lcmx:
jmp _sl8_info
This is doing a binary search on the integer arguments. .Lcma is branching on <4 then <6 then <7. The first comparsion goes to .Lcmq which is branching on <2 then <3. The first comparison from that goes to .Lcmo, which branches on <1.
With ghc -O2 -S, instead we get this, where we can see the same pattern:
.text
.align 4,0x90
.long _F_zdwf_srt-(_F_zdwf_info)+0
.long 65540
.long 0
.long 33488911
.globl _F_zdwf_info
_F_zdwf_info:
.LcqO:
movl 0(%ebp),%eax
cmpl $4,%eax
jl .Lcr6
cmpl $6,%eax
jl .LcqY
cmpl $7,%eax
jl .LcqU
cmpl $7,%eax
jne .LcqS
movl $_F_f1_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.LcqS:
movl $_F_f9_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.LcqU:
cmpl $6,%eax
jne .LcqS
movl $_F_f2_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.LcqW:
cmpl $4,%eax
jne .LcqS
movl $_F_f4_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.LcqY:
cmpl $5,%eax
jl .LcqW
cmpl $5,%eax
jne .LcqS
movl $_F_f3_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.Lcr0:
cmpl $2,%eax
jne .LcqS
movl $_F_f6_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.Lcr2:
testl %eax,%eax
jne .LcqS
movl $_F_f8_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.Lcr4:
cmpl $1,%eax
jl .Lcr2
cmpl $1,%eax
jne .LcqS
movl $_F_f7_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.Lcr6:
cmpl $2,%eax
jl .Lcr4
cmpl $3,%eax
jl .Lcr0
cmpl $3,%eax
jne .LcqS
movl $_F_f5_closure,%esi
addl $4,%ebp
andl $-4,%esi
jmp *(%esi)
.section .data
.align 4
.align 1
_F_f_srt:
.long _F_zdwf_closure
.data
.align 4
.align 1
.globl _F_f_closure
_F_f_closure:
.long _F_f_info
.long 0
.text
.align 4,0x90
.long _F_f_srt-(_srh_info)+0
.long 0
.long 65568
_srh_info:
.Lcrv:
movl 3(%esi),%eax
movl %eax,0(%ebp)
jmp _F_zdwf_info
.text
.align 4,0x90
.long _F_f_srt-(_F_f_info)+0
.long 65541
.long 0
.long 65551
.globl _F_f_info
_F_f_info:
.Lcrz:
movl 0(%ebp),%esi
movl $_srh_info,0(%ebp)
testl $3,%esi
jne _srh_info
jmp *(%esi)
If we change the original code to
f :: Int -> String
f 1 = "Zero"
f 2 = "One"
f 3 = "Two"
f 4 = "Three"
f 5 = "Four"
f 6 = "Five"
f 7 = "Six"
f 8 = "Seven"
f _ = "Undefined"
The branches are on <5, <7, <8, with <5 going to <3, <4, etc., so it's probably doing this based on sorting the arguments. We can test that by scrambling the numbers, and even adding spacing between them:
f :: Int -> String
f 20 = "Zero"
f 80 = "One"
f 70 = "Two"
f 30 = "Three"
f 40 = "Four"
f 50 = "Five"
f 10 = "Six"
f 60 = "Seven"
f _ = "Undefined"
Sure enough, the branches are still on <50, <70, <80, with <50 going to <30, <40, etc.
Apparently function pattern matching happens in O(1) (constant time), while Map's lookup (of course referring to Data.Map) is guaranteed to be O(logn).
Considering the above assumptions, I'd go with pattern matching:
message 200 = "OK"
message 404 = "Not found"
The case ... of and the multiple equations are exactly equivalent. They compile to the same core. For a large number of cases you should probably do this:
import qualified Data.Map as Map
message =
let
theMap = Map.fromList [ (200, "OK"), (404, "Not found"), ... ]
in
\x -> Map.lookup x theMap
This constructs the map only once. If you don't like the Maybe String return type you can apply fromMaybe to the result.
For a small number of cases (especially if they are integers) the case statement is probably faster if the compiler can translate it to a jump table.
In an ideal world, ghc would pick the right version automatically.
Related
So I have written a program based on NASM that receives user input (two numbers to be exact), and then it realizes the addition and difference between the two numbers, and then it prints it back. The program works fine in NASM but I am having trouble with GAS. The subroutine that counts the length of a string using the repne scasb instruction is giving me a headache because of a segmentation fault.
I have checked the code for segmentation faults, and I have located the fault in the repne scasb line.
What I am basically doing is translating a NASM code I made into its respective GAS code. However, as I said before, it's giving me a segmentation fault. After I get the first number from the user, to be more specific.
.section .data
msg: .ascii "Insert a number: "
msgLen = .-msg
msg2: .ascii "Insert another number: "
msg2Len = .-msg2
errorMsg: .ascii "Error: invalid character!\n"
errorMsgLen = .-errorMsg
displaySuma: .ascii "The result of the addition is: "
displaySumaLen = .-displaySuma
displayResta: .ascii "The result of the difference is: "
displayRestaLen = .-displayResta
enterChar: .ascii "\n"
terminator: .byte 0
.section .bss
.lcomm num1, 8
.lcomm num2, 8
.lcomm buffer, 10
.lcomm buffer2, 10
.section .text
.global _start
_start:
call _clear #clear registers. Probably an useless routine
call _msg1 #Display msg1
call _num1 #Read num1
movl num1, %edi
call _lenString #ECX now has num1 length
lea (num1), %esi
call _stringToInt #EAX now has num1 in integer
movl %eax, %r15d #R15D now has the integer
call _msg2 #Display msg2
call _num2 #Read num2
xor %edi, %edi #Clear EDI
movl num2, %edi #Moving num2 to EDI register to call _lenstring
xor %ecx, %ecx #Clear ECX
call _lenString #ECX has num2 length
xor %esi, %esi #clear ESI
lea (num2), %esi
call _stringToInt #EAX now has integer value of num2
mov %eax,%r14d ###R14D has num2 now
#Addition
#r8d = num1 + num2
mov %r15d, %r8d
add %r14d, %r8d #R8D has num1 + num2
#Difference
#If num1 > num2 =======> r9d = num1 - num2
#If num1 < num2 =======> r9d = num2 - num1
cmp %r14d, %r15d
jg .greater
mov %r14d, %r9d
sub %r15d, %r9d #R9D has num1 - num2
jmp .next
.greater:
mov %r15d, %r9d
sub %r14d, %r9d #R9D has num2 - num1
jmp .next
.next:
mov %r8d, %eax #Sum is now at EAX to convert it to ascii characters
lea (buffer), %esi
call _intToString
#EAX ascii of the sum
mov %eax, %r10d #Using R10D to store the new string
mov %r9d, %eax #Difference result is now at EAX
lea (buffer2), %esi
call _intToString
#EAX has the pointer to the difference result.
mov %eax, %r11d #Storing string in R10D
xor %edi, %edi
xor %r15d, %r15d
xor %r14d, %r14d
mov %r10d, %edi
call _lenString #ECX length of sum string
mov %ecx, %r15d #R15D now has that value
call _suma #This prints the sum result
xor %edi, %edi #Clear EDI
mov %r11d, %edi
call _lenString #ECX has length of dif. string
mov %ecx, %r14d #R14D has that value
call _resta #Print dif. result
movl $1, %eax #End of the program
movl $0, %ebx
int $0x80
_stringToInt:
xor %ebx, %ebx
.next_digit:
movzxb (%esi), %eax
cmp $0x30, %eax #These 4 lines check for invalid characters
jb _errorMsg
cmp $0x39, %eax
ja _errorMsg
inc %esi
sub $0x30, %eax ###Sub 48 (converts to integer)
imul $10, %ebx
add %eax, %ebx #ebx = ebx*10 + eax
loop .next_digit #loop [ECX] times
mov %ebx, %eax
ret
_intToString:
add $10, %esi
mov (terminator),%esi
mov $10, %ebx
.next_digit1:
xor %edx, %edx
div %ebx
add $0x30, %edx ##
dec %esi
mov %dl, (%esi)
test %eax, %eax
jnz .next_digit1
mov %esi, %eax
ret
#######################################################################################################
_msg1:
movl $4, %eax #msg1 routine
movl $1, %ebx
movl $msg, %ecx
movl $msgLen, %edx
int $0x80
ret
_num1:
movl $3, %eax #Reads first number
movl $0, %ebx
movl $num1, %ecx
movl $8, %edx
int $0x80
ret
_msg2:
movl $4, %eax #msg2 display
movl $1, %ebx
movl $msg2, %ecx
movl $msg2Len, %edx
int $0x80
ret
_num2:
movl $3, %eax #Reads the next number
movl $0, %ebx
movl $num2, %ecx
movl $8, %edx
int $0x80
ret
_salir:
movl $1, %eax #Exit
movl $0, %ebx
int $0x80
_errorMsg:
movl $4, %eax #Error msg
movl $1, %ebx
movl $errorMsg, %ecx
movl $errorMsgLen, %edx
int $0x80
jmp _salir
_lenString:
xor %ecx, %ecx
not %ecx
xor %al, %al
mov $0xA, %al
cld
repne scasb #Segmentation fault is caused by this line
not %ecx
dec %ecx
ret
_suma:
movl $4, %eax
movl $1, %ebx
movl $displaySuma, %ecx
movl $displayRestaLen, %edx
int $0x80
movl $4, %eax
movl $1, %ebx
mov %r10d, %ecx
mov %r15d, %edx
int $0x80
movl $4, %eax
movl $1, %ebx
movl $enterChar, %ecx
movl $1, %edx
int $0x80
ret
_resta:
movl $4, %eax
movl $1, %ebx
movl $displayResta, %ecx
movl $displayRestaLen, %edx
int $0x80
movl $4, %eax
movl $1, %ebx
mov %r11d, %ecx
mov %r14d, %edx
int $0x80
movl $4, %eax
movl $1, %ebx
movl $enterChar, %ecx
movl $1, %edx
int $0x80
ret
_clear:
xor %eax, %eax
xor %ebx, %ebx
xor %ecx, %ecx
xor %edx, %edx
xor %esi, %esi
xor %edi, %edi
xor %r8d, %r8d
xor %r9d, %r9d
xor %r10d, %r10d
xor %r11d, %r11d
xor %r14d, %r14d
xor %r15d, %r15d
ret
I am using this makefile to create the .o and .exe files (Given by my professor):
#*************************************************
# Executable name : hola
# Version : 2.0
# Created date : February 12, 2019
# Authors :
# Eduardo A. Canessa M
# Description : simple makefile for GAS
# Important Notice: To be used for GAS only
#*************************************************
#change the name "ejemplo" for the name of your source file
name=addSubInteger
#program to use as the assembler (you could use NASM or YASM for this makefile)
ASM=as
#flags for the assember
#ASM_F= #*** place here flags if needed ***
#program to use as linker
LINKER=ld
#link executable
$(name): $(name).o
$(LINKER) -o $(name) $(name).o
#assemble source code
$(name).o: $(name).s
$(ASM) $(ASM_F) -o $(name).o $(name).s
There is a segmentation fault error after the program reads the first user input.
This is the NASM code of my program (Hope you don't mind the spanish comments in it, but it's essentially the same program as the one written on GAS).
I know I have made some next level spaghetti code, but this is the solution I came to.
movl num1, %edi
call _lenString #ECX now has num1 length
lea (num1), %esi
call _stringToInt #EAX now has num1 in integer
The first instruction does not load the address in %edi. You can use lea like you did for the call to _stringToInt that follows next. Or if you care about a shorter encoding then write mov $num1, %edi.
lea (num1), %edi
call _lenString #ECX now has num1 length
The same problem exists for the second number:
movl num2, %edi SAME PROBLEM
xor %ecx, %ecx
call _lenString
The _intToString subroutine has 2 problems!
You destroy the address in %esi by writing a random value in it.
You (try to) write in memory beyond the buffer that was reserved via .lcomm buffer, 10. This will destroy the first byte in buffer2.
Since converting a 32-bit integer can produce (at most) 10 characters, you will need to enlarge your buffer to 11 bytes so you can safely store the byte-sized terminator.
.lcomm buffer, 11
.lcomm buffer2, 11
Then use this code:
_intToString:
mov $10, %ebx
add %ebx, %esi #Instead of 'ADD $10, %ESI' now that EBX==10
mov (terminator), %dl
mov %dl, (%esi)
.next_digit1:
xor %edx, %edx
div %ebx
add $0x30, %edx ##
dec %esi
mov %dl, (%esi)
test %eax, %eax
jnz .next_digit1
mov %esi, %eax
ret
The original NASM source uses
STRING_TERMINATOR equ 10
An equ does not consume memory at run-time. Your terminator: .byte 0 does use run-time memory! A good translation for the equ would be
.set terminator,0
Now you can write
_intToString:
mov $10, %ebx
add %ebx, %esi #Instead of 'ADD $10, %ESI' now that EBX==10
movb $terminator, (%esi)
data
tekst: .ascii "heLLo WoRlD 93a9s\0"
tekst_len = . - tekst
.text
.global _start
_start:
mov $tekst_len, %edx
petla:
cmp $tekst_len, %edx
je koniec
cmpb $'a', tekst(,%edx,1)
jg zamien #?????????????????????????????????????
inc %edx
jmp petla
zamien:
movb $'X', tekst(,%edx,1)
inc %edx
jmp petla
koniec:
movl $4, %eax
movl $1, %ebx
movl $tekst, %ecx
movl $tekst_len, %edx
int $0x80
I have such a problem: the instruction 'jg zamien' doesn't work. No matter what condition I choose 'jl', 'je' it never executes. Can someone tell me why?
This is the reason:
mov $tekst_len, %edx
petla:
cmp $tekst_len, %edx <-- edx will equal $tekst_len here on the first iteration
je koniec
You're always jumping to koniec on the first iteration of the loop.
The mov should probably be mov $0,%edx.
It needs to allow a two-digit number for input that will be used to indicate how many times the name is printed. I can't figure out how to separate the second digit though and have it checked to make sure it is between 0x30 and 0x39. I also keep getting this weird box after the name that has 0017 inside it.
.data
input_msg_len: .long 26
input_msg: .ascii "Enter a two-digit number: "
name: .ascii "Michael Chabon\n"
name_len: .long 16
max: .long 0
count: .long 0
tmp: .long 0
input_str: .ascii "??"
.text
.global _start
_start:
mov $4, %eax
mov $1, %ebx
mov $input_msg, %ecx
mov input_msg_len, %edx
int $0x80
mov $3, %eax
mov $0, %ebx
mov $input_str, %ecx
mov $2, %edx
int $0x80
mov $input_str, %eax
add count, %eax
mov $input_str, %eax
mov (%eax), %bl
cmp $0x30, %bl
jl _start
cmp $0x39, %bl
jg _start
mov count, %eax
inc %eax
mov %eax, count
sub $0x30, %bl
mov %bl, max
mov $10, %bl
imul %bl
mov %bl, max
#Not sure how to check second char in input_str.
#Want to check it then subtract $0x30 and move to tmp before adding tmp to max.
mov $0, %edi
again:
cmp max, %edi
je end
mov $4, %eax
mov $1, %ebx
mov $name, %ecx
mov name_len, %edx
int $0x80
inc %edi
jmp again
end:
mov $1, %eax
int $0x80
Thanks in advance!
There are some bugs in your code.
Below, 2 first lines of that block are redundant, as mov $input_str, %eax overwrites eax anyway.
mov $input_str, %eax
add count, %eax
mov $input_str, %eax
Then here, it makes no sense to load count into eax here:
mov count, %eax
inc %eax
mov %eax, count
You can do this in a lot shorter and clearer way with:
incl count
Then, next bug is that you recently loaded count into eax, and then multiply the lowest 8 bits of count loaded into al with 10, in this piece of code:
mov (%eax), %bl // bl = first character
cmp $0x30, %bl
jl _start
cmp $0x39, %bl
jg _start
mov count, %eax // eax = count
inc %eax // eax++
mov %eax, count // count = eax
sub $0x30, %bl // 0 <= bl <= 9
mov %bl, max // max = bl <- you lose this value in the next mov %bl, max
mov $10, %bl // bl = 10
imul %bl // ax = 10 * and(count, 0xff) // ax = al*bl (signed multiply)
mov %bl, max // max = 10 <- here you overwrite the value of max with 10
So, according to my intuition you don't want to do ax = 10 * and(count, 0xff), but 10 * (first number). imul %bl does o signed multiply between al and bl, and stores the result in ax. So the code above could be changed to something like this:
mov (%eax), %bl // bl = first character
cmp $0x30, %bl
jl _start
cmp $0x39, %bl
jg _start
incl count
pushl %eax // push eax to stack
sub $0x30, %bl // 0 <= bl <= 9
mov $10, %al // al = 10
imul %bl // ax = 10 * bl (signed multiply)
mov %al, max // 0 <= max <= 90
Then, you can check the second character similarly to the first character:
pop %eax // pop eax from stack
incl %eax
mov (%eax), %bl // bl = second character
cmp $0x30, %bl
jl _start
cmp $0x39, %bl
jg _start
sub $0x30, %bl // 0 <= bl <= 9
add %bl, max // 0 <= max <= 99
I strongly recommend you to learn to use some debugger. gdb has several frontends, of which I think ddd works best according to my experience. gdbtui is also convenient.
I've been recently working towards learning a bit of assembly and I'm currently stumped on an exercise which requires me to find the maximum number of a list of long values.
The code is as follows:
.section .data
data_items: .long 200, 201, 101, 10, 0
min_val: .long 0x8000000000000000 # MIN_VALUE in long
.section .text
.global _start
_start:
movl $0, %edi # init counter to 0
movl min_val, %ebx
start_loop:
cmpl $0, %eax
je loop_exit # go to end if 0 encountered
incl %edi
movl data_items(,%edi,4), %eax
cmpl %ebx, %eax
jle start_loop # if new value < max value in ebx, read next element
movl %eax, %ebx
jmp start_loop
loop_exit:
movl $1, %eax
int $0x80
Two problems with this code:
When trying to assemble the code, I get the message: Warning: value 0x8000000000000000 truncated to 0x0
If I rewrite my code in an alternative logic (one which doesn't require min_value variable), any value greater than 255 in the list of data_items is truncated or returned as value % 256 even though the range of .long should be much larger?
Can anyone help me understand what I'm doing wrong?
EDIT: After changes, the code looks like below. Note how maximum in this case turns out to be 145 instead of 401.
.section .data
data_items: .long 401, 201, 101, 10, 0
max_val: .long 0x80000000
.section .text
.global _start
_start:
movl $0, %edi # init counter to 0
movl max_val, %ebx
start_loop:
movl data_items(,%edi,4), %eax
cmpl $0, %eax
je loop_exit # go to end if 0 encountered
incl %edi
cmpl %ebx, %eax
jle start_loop # if new value < max value in ebx, read next element
movl %eax, %ebx
jmp start_loop
loop_exit:
movl $1, %eax
int $0x80
First of all, 0x8000000000000000 doesn't even fit in a long, it's a long long. A long -1 is 0xffffffff.
As for the other point, I can't comment on code you haven't posted.
I made simple arguments adder with assembly but it isn't working
it always return 0
.section .data
.section .text
.global _start
_start:
call adder
movl %eax,%ebx
movl $1,%eax
int $0x80
adder:
pushl %ebp
movl %esp,%ebp
movl $0,%eax #eax = return
movl $1,%ebx #ebx = index
movl 8(%ebp),%ecx #number of args
loop:
cmpl %ebx,%ecx
jg exit
addl 8(%ebp,%ebx,4),%eax
incl %ebx
jmp loop
exit:
movl %ebp,%esp
popl %ebp
ret
There are a few problems with this code, but you're on the right track.
1. Loop condition
You are using this to quit the loop when ebx >= ecx:
cmpl %ebx,%ecx
jg exit
The syntax is rather confusing, but this actually means "exit if ecx is greater than ebx". Changing it to jng exit fixes this problem.
2. Arguments on stack
You are referring to arguments with 8(%ebp,%ebx,4), but the arguments actually start at 12(%ebp). You are right in that you should start with index 1, because the argument with index 0 is merely the name of the program.
3. Arguments are always strings
The arguments on the stack are only pointers to strings. movl 12(%ebp),%eax will not put a number from the command line in eax. It will only put a memory address in eax, which points to a series of characters that make up the first argument.
To get the number represented by the string "123" you need to parse it with a function such as atoi. atoi will then return 123 in eax.
Here's what the code looks like once these things are fixed. I put a comment next to each changed line.
.section .data
.section .text
.global _start
_start:
call adder
movl %eax,%ebx
movl $1,%eax
int $0x80
adder:
pushl %ebp
movl %esp,%ebp
movl $0,%eax #eax = return
movl $1,%ebx #ebx = index
movl 8(%ebp),%ecx #number of args
loop:
cmpl %ebx,%ecx
jng exit # quit if ecx is not greater than ebx
pushl %eax # save registers on stack
pushl %ecx # because they will be destroyed soon
pushl 12(%ebp,%ebx,4) # push next argument pointer on stack
call atoi # invoke atoi, this destroys registers eax,ecx,edx
add $4,%esp # restore stack pointer
mov %eax,%edx # atoi returns the value in eax, save that to edx
popl %ecx # restore ecx from stack
popl %eax # restore eax from stack
addl %edx,%eax # add parsed number to accumulator
incl %ebx
jmp loop
exit:
movl %ebp,%esp
popl %ebp
ret
And the program seems to work now:
$ gcc -nostartfiles test.S -m32 && ./a.out 1 2 3 4 5 ; echo $?
15
The program returns the result in its exit value, which means it can not count higher than 255 :)
It would probably be better to let the program print the result to stdout, using printf.