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revolution 12 Jun 2011, 18:30
Do you have a question? Or some code you wish to contribute?
Or did you just want to find an excuse to post "weee"?  |
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12 Jun 2011, 18:30 |
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Teehee 12 Jun 2011, 18:43
my question is implicit, i want a solution <.>
my current code just compare the first 4 bytes in edx, lol. Code: bnf_instruction: push esi push edi mov eax,instr_list ; string array xor ecx,ecx .for: mov esi,keyword ; source lea edi,[eax+ecx*4] mov edx,[esi] cmp [edi],edx jne @f mov eax,1 jmp .done @@: inc ecx cmp ecx,instr_list.size jne .for xor eax,eax .done: pop edi pop esi ret PS: i dont like to put code bc I like to see the solutions without influence with my codes. Also bc my codes are weird(?) for me. |
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12 Jun 2011, 18:43 |
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revolution 12 Jun 2011, 18:52
If you only need a simple solution then just do a byte-by-byte compare until you get either zero or a mismatch.
There is no need to complicate with dword compares. If your string is only 1 or 2 characters then you could get incorrect results. Also strings that match the first 4 characters might not match later characters. PS: it is always best to post your code. At least that shows you have tried to do it and are not just being lazy and asking others to do work for you. |
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12 Jun 2011, 18:52 |
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Teehee 14 Jun 2011, 23:52
there we go:
Code: bnf_instruction: push esi push edi mov ebx,instr_list mov edx,keyword mov ecx,instr_list.size .for: mov esi,edx mov edi,[ebx+ecx*4] @@: mov al,[esi] cmp [edi],al jne .neq inc esi inc edi test al,al jnz @b jmp .done .neq: dec ecx jns .for .done: pop edi pop esi ret I really need this to be fast. Can we optimize it? _________________ Sorry if bad english. |
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14 Jun 2011, 23:52 |
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idle 15 Jun 2011, 02:02
Code: .for: mov edi,[ebx+ecx*4] ;warning: +4 for element #1 ; you could keep number of items at offset 0 ; warning disappears then ... test al,al ;pedantic shift jz .done inc esi inc edi jmp @b |
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15 Jun 2011, 02:02 |
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revolution 15 Jun 2011, 02:57
Teehee wrote: I really need this to be fast. Can we optimize it? Instead this will give exponential speed up: Sort the elements as they are added. Use a binary search to find elements. |
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15 Jun 2011, 02:57 |
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Coty 15 Jun 2011, 03:12
You could use REPE/REPZ, CMPS/CMPSB/CMPSW/CMPSD and JZ/JNZ instructions witch would be slightly faster, and more faster on Hyper threading CPUs (Since Intel says to avoid loops on hyper threading because its slower)...
Here is a small example using DOS, but you could use it even in 64bit mode... (Well the actual compare code anyway) Code: org 0x0100 use16 cld ; set direction flag, so we read from left to right. mov di, string_one mov si, string_two mov cx, 6 ; Maximum string size. (imma do 6 for "weeeee") ;----------------------; ; Here we compare... ; ;----------------------; repe cmpsb ; repete while equal; compare string, byte by byte ; NOTE: we can also use "cmpsw" to compare word by word ; and "cmpsd" to compare dword by dword. jnz not_equal ; If there not equal jump to the not equal label... ;----------------------; ; Display our cool msg ; ;----------------------; mov ah, 0x09 mov dx, imahappyhappyhorse int 0x21 jmp _end ;----------------------; ; Display our fail msg ; ;----------------------; not_equal: mov ah, 0x09 mov dx, NOOOOOOOOOOOOOOOOO int 0x21 _end: xor ax, ax int 0x16 int 0x20 string_one: db "weeeee" string_two: db "weee3e" imahappyhappyhorse: db "Strings are equal!$" NOOOOOOOOOOOOOOOOO: db "Strings not equal!$" |
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15 Jun 2011, 03:12 |
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Teehee 15 Jun 2011, 12:01
Quote: Sort the elements as they are added. ouch, its time to implement a binary search 
@Coty but i need to know the string lenght in order to use 'rep cmpsb'. will I need to spend time to count it lenghts or i just need to set a max value (like 15, once my biggest world has that size) and it will repeat just while the bytes are equals regardless of the loop lenght? _________________ Sorry if bad english. |
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15 Jun 2011, 12:01 |
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Coty 15 Jun 2011, 12:27
What I normally do is set (E)CX to something like 20, then create buffers for my strings 20 bytes in length.
Code: string_one: db "weeeee", 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 string_two: db "weee3e", 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 string_three: db "longwordorIthinkso", 0, 0 Note that if you don't add the buffers repe cmpsb will continue to read on past your string until data is un-equal! Or CX expires, this may give undesired results. |
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15 Jun 2011, 12:27 |
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edfed 15 Jun 2011, 13:05
coty, this method isn't good at all.
it is very hard to write strings like that. why not use align 32, then, words with up to 31 letters (align 32-db 0) but it is by far, better to do with the list in the teehees first post. by this way, it will let the list become better, containing a lot of information by expending fields, and so on. to compare the strings, it is very simple. just give two pointers to your comparator. for example, (e)di for the sting to compare, and (e)si for the dictionary word. to define the si register, just load the pointer to current word in list le list will contain N elements. if every elements are dword pointers, means that the size in byte will give us the numer of elements*4 this count will be a value, somewhere in the source (a label, a dword...) it is better to record this size as a dword, for example, the first dword of the list is the size of the list. the edi string (our string to seek) is just a string like db 'string',0 the esi string comes from a list, then, from si if we are vicious. ebx then will be used as index in the dictionary. ebx can be picked from the size of the list, and decremented for every comparaison. Code: mov edi,stringtotest mov esi,dictionary mov ebx,[esi] shr ebx,2 dec ebx jl .end @@: mov esi,[esi+ebx*4+4] ;point to the element(ebx*4) of the list(+4) call compare jc .ok dec ebx jnl @b .end: clc .ok: ret ;.... compare: ;edi=string to compare ;esi=string from dictionary ;...type your code here .ok: stc ret the code of the compare function cannot be more than... 10 lines, and will pass the result with carry flag. if carry, then, string found. |
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15 Jun 2011, 13:05 |
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typedef 15 Jun 2011, 18:10
i thought you can just loop through the comparative string and check byte by byte against the base string, and if you run into a null character you are done.
if base string is greater than the latter string, then there's no match. vice versa. |
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15 Jun 2011, 18:10 |
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