TASM字符串到十六進制數

Question

我對基於TASM的匯編代碼有疑問。 我想將“字符串”指針中的字符串轉換為十進制數，然后將其打印為十六進制數。 但是，該代碼僅正確地從右側打印前2個字符。 這可能是什么問題？

concd   SEGMENT
        ASSUME cs: concd
        ORG 100h

main:   
    mov ax, 0
    mov bx, offset string
    mov cx, 0
    mov dx, 0       
    mov si, 0       

jump:   
    mov cx, [bx + si]
    cmp cx, 0h          ;checking if we reached the end of the string
    jz exit

    mov dx, ax
    mov cx, 9

mult:   
    add ax, dx          ;loop for multiplying by 10 
    loop mult

    mov cx, [bx + si]
    sub cx, 30h
    add ax, cx      

    inc si      
    jmp jump

exit:   
    call hex
    mov ah, 04Ch
    mov al, 0
    int 21h

hex PROC
    mov dx, ax
    mov cl, 12
jump2: 
    mov ax, dx
    shr ax, cl
    and ax, 15
    cmp ax, 9
    jng t
    add ax, 7h
t:  add ax, 30h
    call putc
    sub cl, 4
    jnc jump2

    RET
hex ENDP

putc PROC

    mov ah, 0Eh
    int 10h

    RET
putc ENDP

string  DB  "1234", 0h

concd   ENDS
    END     main

Answer 1

我做了一些小改動，現在可以了。 我在“階段1”中評論了它實際上確實得到了字符abd做過的更改，在“階段2”中評論了更正了轉換的更改。

現在是OP找出問題所在的時候了。

concd   SEGMENT
        ASSUME cs: concd
        ORG 100h

main:   
    mov ax, cs ;; added - phase 1
    mov ds, ax ;; added - phase 1
    mov ax, 0
    mov bx, offset string
    mov cx, 0
    mov dx, 0       
    mov si, 0       

jump:   
    mov ch, 0  ;; Added - phase 2
    mov cl, [bx + si] ;; added - phase 2
    ;mov cx, [bx + si] ;; removed - phase 2
    cmp cx, 0h          ;checking if we reached the end of the string
    jz exit

    mov dx, ax
    mov cx, 9

mult: ; ax contains the number
    ;; removed the next 2 lines - phase 2
    ;add ax, dx          ;loop for multiplying by 10 
    ;loop mult
    ;; added the next 2 lines - phase 2
    mov cl, 10
    mul cl          ; multiply by 10

    mov cx, [bx + si]
    sub cx, 30h
    add ax, cx      

    inc si      
    jmp jump

exit:   
    call hex
    mov ah, 04Ch
    mov al, 0
    int 21h

hex PROC
    mov dx, ax
    mov cl, 12
jump2: 
    mov ax, dx
    shr ax, cl
    and ax, 15
    cmp ax, 9
    jng t
    add ax, 7h
t:  add ax, 30h
    call putc
    sub cl, 4
    jnc jump2

    RET
hex ENDP

putc PROC

    mov ah, 0Eh
    int 10h

    RET
putc ENDP

string  DB  "1234", 0h

concd   ENDS
    END     main