fasmnewbie wrote:

MariaSM

Seems like it may take you forever to understand revolution and ASMGuru here. I modified this code from my library and I intentionally leave one part buggy for you to figure out. It deals only with lower case characters. Learn from it and read carefully of what Guru and revolution are saying. Take your time.

Code:

org 100h
        xor si,si
;------------------------
get:    call get$
        cmp al,0dh
            je ok
        mov [chr+si],al
        sub al,61h
        movzx di,al
        add [chr2+di],1
        inc si
        cmp si,20
            je ok
        jmp get
ok:     call nline
        xor di,di
;-------------------------
done:   cmp [chr2+di],'a'
            ja change
        inc di
good:   cmp di,24
            je exit
        jmp done
change: xor ax,ax
        mov al,[chr2+di]
        sub al,61h
        mov dl,[chr2+di]
        call prtc
        mov dl,'='
        call prtc
        call prtnum
        call nline
        inc di
        cmp [chr2+di],'a'
            ja change
        jmp good
;--------------------------
get$:   mov ah,1
        int 21h
        ret
nline:  mov dl,0dh
        call prtc
        mov dl,0ah
        call prtc
        ret
prtc:   push ax
        mov ah,2
        int 21h
        pop ax
        ret
;-------------------------
prtnum: pusha
        xor si,si
        xor dx,dx
        mov cx,10
.start: div cx
        push dx
        xor dx,dx
        inc si
        test ax,ax
            jz .prt
        jmp .start
.prt:   dec si
        pop ax
        add al,30h
        mov dl,al
        call prtc
        test si,si
            jz .done
        jmp .prt
.done:  popa
        ret
;-------------------------
exit:   mov ah,1
        int 21h
        mov ah,4ch
        int 21h

chr db 20 dup(?),0
chr2 db 24 dup(61h)    

Output:

Code:

welcome to fasm
b=1
b=1
c=2
b=1
b=1
c=2
c=2
b=1
b=1
b=1    

I don't understant the output. What is "b=1 b=1..."?

welcome to fasm
a=1
c=1
e=2
...    

MariaSM wrote:

fasmnewbie wrote:

MariaSM

Seems like it may take you forever to understand revolution and ASMGuru here. I modified this code from my library and I intentionally leave one part buggy for you to figure out. It deals only with lower case characters. Learn from it and read carefully of what Guru and revolution are saying. Take your time.

Code:

org 100h
        xor si,si
;------------------------
get:    call get$
        cmp al,0dh
            je ok
        mov [chr+si],al
        sub al,61h
        movzx di,al
        add [chr2+di],1
        inc si
        cmp si,20
            je ok
        jmp get
ok:     call nline
        xor di,di
;-------------------------
done:   cmp [chr2+di],'a'
            ja change
        inc di
good:   cmp di,24
            je exit
        jmp done
change: xor ax,ax
        mov al,[chr2+di]
        sub al,61h
        mov dl,[chr2+di]
        call prtc
        mov dl,'='
        call prtc
        call prtnum
        call nline
        inc di
        cmp [chr2+di],'a'
            ja change
        jmp good
;--------------------------
get$:   mov ah,1
        int 21h
        ret
nline:  mov dl,0dh
        call prtc
        mov dl,0ah
        call prtc
        ret
prtc:   push ax
        mov ah,2
        int 21h
        pop ax
        ret
;-------------------------
prtnum: pusha
        xor si,si
        xor dx,dx
        mov cx,10
.start: div cx
        push dx
        xor dx,dx
        inc si
        test ax,ax
            jz .prt
        jmp .start
.prt:   dec si
        pop ax
        add al,30h
        mov dl,al
        call prtc
        test si,si
            jz .done
        jmp .prt
.done:  popa
        ret
;-------------------------
exit:   mov ah,1
        int 21h
        mov ah,4ch
        int 21h

chr db 20 dup(?),0
chr2 db 24 dup(61h)    

Output:

Code:

welcome to fasm
b=1
b=1
c=2
b=1
b=1
c=2
c=2
b=1
b=1
b=1    

I don't understant the output. What is "b=1 b=1..."?

That's the bug I told you about. It should display the character and the number of times it appears in the string you entered.

Code:

welcome to fasm
a=1
c=1
e=2
...    

The string entering code misses the max # of characters field:
http://spike.scu.edu.au/~barry/interrupts.html#ah0a
First you need two bytes BEFORE the buffer and they must be set according to the link above.
So, you need 20 characters in text and 1 byte for Carriage Return plus 2 bytes for the header.
That makes it something like this:

Code:

bufMaxLen DB 20   ; Max # of characters
bufEntered DB 0     ; This is how many characters user entered
bufText DB 0 DUP(21)  ; 20 characters + CR byte

... to enter the string ...
lea dx, bufMaxLen
mov ah, 0Ah
int 21h

... to count characters ...
lea si, bufText
.. and go on ...
    

To output the value in CL you can just use a code like that:
(knowing that the value can be in range 0 to 20)

Code:

if (value in CL <= 9)
{
        CL + '0'
        print CL
}
else if (CL = 20)
{
        print '20'
}
else  ; CL between 10 and 19
{
        print '1'
        CL - 10
        CL + '0'
        print CL
}
    

But of course, it is better to learn how to convert any number to a string.
If you search the forum -- there are few examples of this.

sorry but I have some problems with my code.if you can help me with this, i would remain grateful
i don't want to solve other problem with a different code

MariaSM wrote:

sorry but I have some problems with my code.if you can help me with this, i would remain grateful
i don't want to solve other problem with a different code

It is not about the code. Check the algorithm of solving the problem particularly in converting integer to string. As I see it, that's your biggest problem and that's what revolution meant by 'DIV' instruction. Why don't you convert that to TASM because I am no TASM guy. Put the thing to convert in AX and call the similar putnum procedure in TASM. That solves 50% of your problem.

yes, I've understood what the purpose was..but I don't really have much time to make this program works.and I really appreciate if someone can say me, how i can do to display ah content

mov ah,9 ;want to display '9'?
mov dl,ah
add dl,30h
mov ah,2
int 21h    

cmp al, 0
je print_zero
;
; Divide AH by 10
;
shr ax, 8
mov dx, 0
mov cx, 10
div cx
;
; DL is a remainder of division, AL is a result.
; Ex.: If you have 17 and divide by 10 - remainder is 7, and result is 1
; 1st you print the result and then remainder:
;
push dx
add al, '0'
mov dl, al
mov ah, 2
int 21h
;
; print remainder
;
pop dx
add dl, '0'
int 21h
jmp all_done

print_zero:
;
; print the text "0$", like you do with other strings
;



all_done:
;
; ...
;
    

If AH contains values from 0 to 20, then try this code:

Code:

cmp al, 0
je print_zero
;
; Divide AH by 10
;
shr ax, 8
mov dx, 0
mov cx, 10
div cx
;
; DL is a remainder of division, AL is a result.
; Ex.: If you have 17 and divide by 10 - remainder is 7, and result is 1
; 1st you print the result and then remainder:
;
push dx
add al, '0'
mov dl, al
mov ah, 2
int 21h
;
; print remainder
;
pop dx
add dl, '0'
int 21h
jmp all_done

print_zero:
;
; print the text "0$", like you do with other strings
;



all_done:
;
; ...
;
    

cmp al, 0    

cmp ah, 0    

I made a mistake there (my apologies).
But you still not understanding the code, and you need to.
If you would have read the code - you would have found my mistake.
You act like we have to write the code for your homework.
The mistake is in the first line:

Code:

cmp al, 0    

it must be:

Code:

cmp ah, 0    

LEA SI, STRING
    

MOV SI, OFFSET STRING     ; like you do in other places
    

+---------+---------+-----------------------+
| MAXIMUM | ENTERED | STRING ENTERED ...    |
+---------+---------+-----------------------+
|
DX
    

To be consistent you may use MOV instead of LEA:

Code:

LEA SI, STRING
    

is same as:

Code:

MOV SI, OFFSET STRING     ; like you do in other places
    

But that is not a bug.
You are using MASM, and I use FASM, so I need some time to look into it.

Issue #2:
The incorrect use of function 0AH - DX must point to a following data:

Code:

+---------+---------+-----------------------+
| MAXIMUM | ENTERED | STRING ENTERED ...    |
+---------+---------+-----------------------+
|
DX
    

In your code DX is always pointing to a STRING - not correct.
I think I posted the correct code in some post above.

When you ask user to enter 20 characters - MAXIMUM byte must be 20 (or even 21 to counf the 0Dh)
Of course, when you ask for a single character - MAXIMUM must be 1 (or 2)

String: ABCDEFGH-HELLO
Character: E
Count: 2
Positions: 5, 11
    

String: ABCDEFGH-HELLO
Character: E
- found at position: 5
- found at position: 11
Count: 2
    

PRINT_AH:
... do some stuff ...
RET
    

MOV AH, 17
CALL PRINT_AH     ; It will print "17" (value in AH) to the console
    

like that

Code:

String: ABCDEFGH-HELLO
Character: E
Count: 2
Positions: 5, 11