Re: All is right !

Hello.

My aim : i want remove 2 bytes if they are equal.
So, it will reduce the size of the file.
If this way is easy, OK

If if the file have two bytes similars... and replace each byte by ~ , or an
other particular value : OK

Just the ascii to remove or replace have value from 22H to 2Fh

But, i discovered that i have also to remove bytes if i can find them 3
times or more.
So for 3 or more, it ca be letters.

The final aim is to have a file, and just keep the "Text-Ascii"

So by example : %% becomes ~ or ~ ~ or nothing !
%%% : do the same 1, 2, or 3 ~ or nothing

But now i identify if i have eee or sss or kkk remove into ~
Sure that in a french word, il will never find 3 sames letters.
I accept the risk of a " www.my_web.com "

Below, you can find the try. ( it is not usefull )
I built others versions... with out correct work.

mov ah,9
mov dx,offset tester ; display original characters
int 21h


mov cx,3Eh
mov si, offset tester

myloop:
lodsb
nop
nop
cmp al,22h
jb next ; scan ascii from 22h to 2Fh
cmp al,2Fh
ja next
cmp cx,1 ; take care about CX and the loop
jbe next
mov ah,al
lodsb
dec cx ; ajust SI and CX
cmp al,ah
jnz next
mov al,126
mov [si],al
call mycall

dec si
inc cx ; ajust SI and CX
xchg ah,al ; il will have the value "7E7Eh" in AX
next:
mov [si],al
call mycall
inc si

loop myloop
nop
nop

mov dx,offset tester
mov ah,9
int 21h
mov dx, offset here
int 21h

int 20h

mycall:
db: 0A3 ; will do mov [here],ax
out_put_buffer db: ,92h,01 ; will do A3 92 01

inc byte out_put_buffer
ret

tester db: "!0!!0###! 0##0%%0(((*()0**0))0++-+0--0>>0==0<<>=0<??¨? 0'''0---"
db: ,10,13,36
here db: " "
db: ,10,13,36

-------------- other version -------------

myloop:
lodsb
nop
cmp al,22h
jb next ; scan ascii from 22h to 2Fh
cmp al,2Fh
ja next
cmp ah,al
jnz next
mov al,126
mov [si],al
call mycall
dec si
inc cx

next:
mov [si],al
mov ah,al
call mycall
inc si
loop myloop
nop

mov dx,offset tester
mov ah,9
int 21h
mov dx, offset here
int 21h

int 20h

mycall:
db: 0A2h ; will do mov [here],al
out_put_buffer db: ,87h,01 ; will do A2 87 01 mov [here+?],al

inc byte out_put_buffer
ret

tester db: "!0!!0###! 0##0%%0(((*()0**0))0++-+0--0>>0==0<<>=0<??¨? 0'''0---"
db: ,10,13,36
here db: " "
db: ,10,13,36

------- same result : do some thing, but.... not right ---

"Wolfgang Kern" <nowhere@xxxxxxxx> a écrit dans le message de news:
fuheoc$g2j$1@xxxxxxxxxxxxxxxxxxxxxxxx

almas wroter:

Sure, at the last line "you are a comment"
But i did not found any instruction called Wolfgang :-)
:)
Yes, i coult try to igore errors, so i can shutdown some bytes.
May be, i will do it, actualy i prefer that the COM file can detect
errors
a run correctly
.... / ....
Ok.

Now, i'm working on an other goal.
The aim is if i have two ascii
between 22h and 2Fh ( i can identify it )
and if the next ascii have same value, i remove it into "~"

Example : (( become ~ ~
but ( * stay (*
### become ~ ~ ~

I try .... and have troubles
If i put ##(*) i obtain ~##(**
or i have in others version ~ ~ ~*)

I fact the next byte keep the value of the byte before.

Not seeing your attempts I can only guess ...
and I'm not sure to understand what it shall do.
A) You like to replace two equal characters with 7eh,20h ("~ ") ?
B) or is your desire to reduce the string size by replacing doubles
with single "~"-bytes ?

Version A) should be simple enough ...
remember the location of the first and write 207eh (endianess!)
to this location if a match occured (advance the pointer by two).
but for B) you also need to move string-parts and adjust its size.

The next step : destroy all the ~

How to destroy a byte ? :)
replace it with 20h? or remove it ?

I spend a large part of saturday and obtain "funny" ; stranges results

Learning by doing can be funny, but it is for sure the right way
for those who like to learn about all details at the lowest level.
Keep on your way! It may sound like a slow way, but what you learn
with this method will remain solid.

__
wolfgang





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