Reading past Terminated String

[ic2]

I have one final question about this.  I don't mean to be bothersome or wear the Forum out, but i nearly got all of this down to a science, except i can't count :(

I'm trying to insert the same number in two locations per string.

I got the first two numbers inserted properly, but i can't get the 3rd and 4th number in place.  add esi, 4 and 5 may be  what is giving me the problem.  I can't figure out what to use.  I know it's not add esi, 3 again, i don't think. I tried a lot of ways, but i don't know what im doing.

Could someone show me the syntax to insert the 3rd and 4th number with an brief explanation of why it work.  I do appreciate it.


What i don't understand is exactly what (add esi, num value) does in this case.  It not adding like 1+1=2 i don't think or is it.

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        mov al, [esi]                   ;  1st Num Inserted = 9
        mov     [edi + 1], al       ;  9
        mov     [edi + 11], al     ;  9


        mov al, [esi + 1]            ;  2nd Num Inserted = 0
        add esi, 3                        ; 
        mov     [edi + 2], al       ;  90
        mov     [edi + 12], al     ;  90


        mov al, [esi + 2]           ;  3rd Num Inserted = 1
        add esi, 4                       ; 
        mov     [edi + 3], al      ;  901
        mov     [edi + 13], al    ;  901


        mov al, [esi + 3            ; 4th Num Inserted = 5
        add esi, 5                       ; 
        mov     [edi + 4], al      ;  9015
        mov     [edi + 14], al    ;  9015[\code]

[dsouza123]

What are the contents of the strings ?

If the string that is the source of numbers to insert is something like

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  source db "9015",0


and the destination string is something like

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  dest   db "ABCDEFGHIJKLMNOPQRSTUVWXYZ",0


and you want the result to be

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  dest   db "A9015FGHIJK9015PQRSTUVWXYZ",0


and esi and edi point to the strings

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  mov esi, offset source
  mov edi, offset dest


then the code would be

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        mov al, [esi]                   ;  1st Num Inserted = 9
        mov     [edi + 1], al       ;  9
        mov     [edi + 11], al     ;  9


        mov al, [esi + 1]            ;  2nd Num Inserted = 0
        mov     [edi + 2], al       ;  90
        mov     [edi + 12], al     ;  90


        mov al, [esi + 2]           ;  3rd Num Inserted = 1
        mov     [edi + 3], al      ;  901
        mov     [edi + 13], al    ;  901


        mov al, [esi + 3]           ; 4th Num Inserted = 5
        mov     [edi + 4], al      ;  9015
        mov     [edi + 14], al    ;  9015


esi and esi act as indexes into the strings,
whether by themselves [esi] [edi] or with displacements [esi + 1] [edi + 12]
they are indicating the location (which byte) in the string is being worked on.

[dsouza123]

The way the code you posted would have worked with an extended source string

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.data
  source db "9015abcdefghijklm",0
  dest   db "ABCDEFGHIJKLMNOPQRSTUVWXYZ",0


the result in dest

A90bgFGHIJK90bgPQRSTUVWXYZ

[ic2]

Well, i must be getting closer.  I remove the 4th insertion code and changed this line to (add esi, 1) in the 3rd insertion code .... It runs, but now i get 8 extra strings per number.

Another strange thing is  that I can swear that this was the first thing i tried while trying to stumble upon the solutions many, many times ...

i tried difference numbers like 123456 in the add esi, num space and got nothing all night but now all of a sudden i get this...

I modified your last example.  It all based on your original last last example but with difference strings to allow the right amount of space for insertion.

Example:
A FGHIJK PQRSTUVWXYZ          1-9
A  FGHIJK  PQRSTUVWXYZ          10-99
A   FGHIJK   PQRSTUVWXYZ          100-999
A    FGHIJK    PQRSTUVWXYZ          1000-9999

If all is well this would be the last string of the list
A9999FGHIJK9999PQRSTUVWXYZ


1-9 works great

10-99 works great

100- and beyond don't work             

         



A9015FGHIJK9015PQRSTUVWXYZ          ; this is what i get
A9015FGHIJK9015PQRSTUVWXYZ
A9015FGHIJK9015PQRSTUVWXYZ
A9015FGHIJK9015PQRSTUVWXYZ
A9015FGHIJK9015PQRSTUVWXYZ
A9015FGHIJK9015PQRSTUVWXYZ
A9015FGHIJK9015PQRSTUVWXYZ
A9015FGHIJK9015PQRSTUVWXYZ
A9015FGHIJK9015PQRSTUVWXYZ
etc



A9015FGHIJK9015PQRSTUVWXYZ          ; this is what i should get
A9016FGHIJK9016PQRSTUVWXYZ
A9017FGHIJK9017PQRSTUVWXYZ
A9018FGHIJK9018PQRSTUVWXYZ
A9019FGHIJK9019PQRSTUVWXYZ
A9020FGHIJK9020PQRSTUVWXYZ
A9021FGHIJK9021PQRSTUVWXYZ
A9022FGHIJK9022PQRSTUVWXYZ
A9023FGHIJK9023PQRSTUVWXYZ
etc

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This is the one i changed that got me this far


        mov al, [esi + 2]           ;  3rd Num Inserted = 1
        add esi, 1                 ; 
        mov     [edi + 3], al       ;  901
        mov     [edi + 13], al      ;  901

The example you JUST posted will make all the numbers the same
9015... but you did not know all of what i was  after.  This should finally explain it.  I came a long way with your example and still going .
If i would have explained it this way in the beginning you would have known.  I just did not know where to start.  So i posted small numbers than realized each time i should have asked about  this no, i should have asked about said that... every time i kind of messed it up...

[ic2]

Here what i been working with since.  I had it up to triple it size with a lot of confusing try out... I just cleaned  it up so you can see what i been trying to do without getting everyone else confused also.

Also, when it come to my naming procedures that others help me out with i like this one the best because it fits and since i can't pronounce your name i just say diagnosis 1 2 or 3


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[dsouza123]

The main issue was the code to calculate dscount_x,
shr dscount, 1 only applied when the length of each individual string was 2 bytes.

The other issue is the pointer to the row only gets updated (once)
after all the characters are placed in the rStringx

Extra variables to clear things up.

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.data
indiv_1       dd  x_1    - x_0
indiv_2       dd  x_11   - x_10
indiv_3       dd  x_101  - x_100
indiv_4       dd  x_1001 - x_1000


Here is a working dsouza_3 with a more general routine to calculate dsCount_3.

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dsouza_3:

mov edx, 0                         ;  byte count list of
mov eax, dsCount_3                 ;  strings / bytes per string
mov ecx, indiv_3
div ecx
mov dsCount_3, eax

C0:     mov ecx, 0

C1:     mov al, [rString3 + ecx]   ;  copy rString to destination
        mov   [edi + ecx], al      ;  starting at edi

        inc ecx
        cmp ecx, SIZE_OF3          ;  0..20, 21 bytes
jb C1
        mov al, [esi]              ;  row 1
        mov     [edi +  1], al     ;  replace 1st char at
        mov     [edi + 10], al     ;  replace 2st char at col 2

        mov al, [esi + 1]          ;  row 2
        mov     [edi +  2], al     ;  replace x char by list
        mov     [edi + 11], al     ;  replace x char by list

        mov al, [esi + 2]          ;  row 3
        mov     [edi +  3], al     ;  replace x char by list
        mov     [edi + 12], al     ;  replace x char by list

        add esi, indiv_3           ;  is increased by 1 row

add edi, SIZE_OF3
dec edi                            ;  edi increased by
                                   ;  strln - 1 (== 20)
dec dsCount_3
jnz C0

    ret

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[ic2]

dsouza123, I feel this to all who went of there way to help out.  I'm forever in your debt.

I mean this and plan to do something about it someday soon...Stay Tune.

Thanks you