Reputation: 159
The following code is meant to find the sum of all numbers less than 1000 that are multiples of 3 or 5. (Project Euler first problem).
However, it is giving a segmentation fault when I execute it. I have a feeling it may have something to do with after returning from my check function, I pop ebx off the stack before adding 8 to esp to restore the stack frame. Is this true and if so what can I do to fix it?
section .data
section .text
msg: db "%d", 0
global main
extern printf
main:
mov ebx, 0
mov ecx, 0
;MAINLOOP
loopstart:
inc ebx
cmp ebx, 999 ;check if we are done
je done ;if we are, exit
push 3
push ebx ;otherwise, check if our number is a multiple of 3
call check
pop ebx
add esp, 8 ;fix stack pointer
cmp eax, 1 ;if it is a multiple, add it to our sum
je true
push 5
push ebx
call check ;otherwise check if multiple of 5
pop ebx
add esp, 8
cmp eax, 1
je true ;if it is, add it to our sum
jmp loopstart
true:
add ecx, ebx
jmp loopstart
done:
push ecx
push msg
call printf
add esp, 8
ret
check:
push ebp
mov ebp, esp
mov eax, [ebp+8]
mov ebx, [ebp+12]
mov edx, 0
div ebx
cmp edx, 0
je multiple
mov eax, 0
pop ebp
mov esp, ebp
ret
multiple:
mov eax, 1
pop ebp
mov esp, ebp
ret
Upvotes: 0
Views: 599
Reputation: 159
I have fixed the problem The error was in my check function, I was using
pop ebp
mov esp, ebp
which should have been
mov esp, ebp
pop ebp
I believe there were also some other errors with restoring stack frame etc. Here is the complete working code:
section .data
section .text
msg: db "%d", 0
global main
extern printf
main:
mov ebx, 0
mov ecx, 0
loopstart:
inc ebx
cmp ebx, 1000
je done
push ecx
push ebx
push 3
call check
add esp, 4
pop ebx
pop ecx
cmp eax, 1
je true
push ecx
push ebx
push 5
call check
add esp, 4
pop ebx
pop ecx
cmp eax, 1
je true
jmp loopstart
true:
add ecx, ebx
jmp loopstart
done:
push ecx
push msg
call printf
add esp, 8
ret
check:
push ebp
mov ebp, esp
mov eax, [ebp+12]
mov ebx, [ebp+8]
mov edx, 0
div ebx
cmp edx, 0
je multiple
mov eax, 0
mov esp, ebp
pop ebp
ret
multiple:
mov eax, 1
mov esp, ebp
pop ebp
ret
Upvotes: 1