We help IT Professionals succeed at work.

problem with linear search method

azcalv408
azcalv408 asked
on
Medium Priority
781 Views
Last Modified: 2008-03-17

hello, I'm trying to convert C to Assembly with the following method and I keep getting a 1 as return value, can anyone tell what's wrong ? Thanks ! :

int lsearch(const int x[], int cursize, int target)
{
int i;
for(i = 0; i < cursize; i++)
{
if(x[i]==target)
return i;
}
return -1;
}

Assembly codes:
_asm
{
         mov ebx,x
         mov ecx,cursize
         jecxz done
top:
         mov edx,[ebx]
         cmp edx,target
         jne notFound
         mov eax,ecx
         add ebx,4
         jmp next
notFound:
         mov eax,-1
         add ebx,4
next:
        loop top
done:
}

Comment
Watch Question

hi,
 where are you putting the output. I dont see in the asm code.
Also you have taken int as 4 bytes. Is it 32 bit code.
Int in 2 bytes in 16 bit code.

regards manish

Author

Commented:
int is 4 bytes (32 bit code)
the output is in main : it's just outputting index of found element or -1 if not found. It is also in C just for testing. I'm doing inline assembly by the way.
Commented:
Unlock this solution and get a sample of our free trial.
(No credit card required)
UNLOCK SOLUTION

Commented:
something like this may be faster and get you what you want

asm
{
         push   ebx                 ;always save
         mov ecx,cursize
         mov ebx,x                 ;it is never a good idea to use ebx in a functon
                                         ;this register is used by the compiler and if it is changed
                                         ;and you compile a release mode your porgram will
                                         ;CRASH. You should update your x[] pointer in another
                                         ;function
         or   ecx,ecx               ;much faster than jecxz and give
         jz notFound               ;zero search length should also be an error
         mov  edx,[target]
         or   eax,0                  ;make eax = 0
top:
         cmp   edx,[ebx+eax*4]         ;this is an int so 4 byte steps
         je     done
         inc    eax
         cmp  eax,ecx
         jne  top
notFound:
         mov eax,-1
done:
         pop  ebx                             ;restore ebx
}
>>  or   eax,0                  ;make eax = 0

I think you mean:  and eax,0 :P
Commented:
Unlock this solution and get a sample of our free trial.
(No credit card required)
UNLOCK SOLUTION
Unlock the solution to this question.
Thanks for using Experts Exchange.

Please provide your email to receive a sample view!

*This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply.

OR

Please enter a first name

Please enter a last name

8+ characters (letters, numbers, and a symbol)

By clicking, you agree to the Terms of Use and Privacy Policy.