Question

我试过在互联网上阅读这个，但这是我的问题。我被给了一串双字。

我必须按顺序从这些双字中排序低字（最不重要）的字符串。高句话保持不变。例如：strin DD 12345678h 1256ABCDh，12AB4344h 结果将是1234ABCDh，12565678h，12AB4344h。

现在我尽力编写了一些代码，但是我的插入程序无法正常工作。如果你能看一看并告诉我我做错了什么，我会很高兴。我尝试在td模式下运行它，但我无法弄清楚。

assume cs:code, ds:data
data segment
    s dd 12345678h, 1256ABCDh, 12AB4344h
    ls equ ($-s)/4  ;this is supposed to be the length of my source string
    d dd ls dup (?) ;this is my destination string
    aux dw ?
    aux2 dw ?
data ends

code segment
insert proc
    push di ;here I use the stack to get more free registers
    push cx
    cmp di, offset d ;if di=offset d it means that I didn't store any number yet
    je addPrim 
    std ;we plan on working form right to left on the string for the next part
    mov cx, di
    sub cx, offset d ;here I find out with how many words I have to compare the word from AX
    dec di
    dec di ;since I work with doublewords, for some reason I thought I should decrease di
    dec di ;3 times but here my procedure gets fuzzy and doesn't work properly anymore
    repeta1: ;this repeat is supposed to compare the word from AX with the rest of the least 
        scasw ;significant words from es:di
        jge DIplus2 ;if my number from AX is bigger or equal than what's in es:di, I increment
 ;di twice and store it
        mov bx, word ptr es:[di+1] ;this part is supposed to interchange words but it's not
 ;working how I planned so I don't know how to change it
        mov word ptr es:[di+2], bx
        loop repeta1

    jmp DIplus1

    DIplus2:
        inc di

    DIplus1:
        inc di

    addPrim: ;this label  just adds the first word in the destination string
        stosw

    pop cx
    pop di
    inc di
    inc di
    cld
    ret 
    insert endp

start:
    mov ax, data
    mov ds, ax
    mov es, ax
    mov si, offset s
    mov di, offset d
    mov cx, ls ; store in cx the length of the strings
    jcxz exit

repeta:
    lodsw ;because of little endian, my first word will be my least significant word in the 
;in the doubleword so right after it is moved in ax, i apply the procedure insert
    call insert
    lodsw ;here it moves in ax my most significan word in the dd, so i auto store it 
    stosw ;in my destination string 
    loop repeta

exit:
    mov ax, 4c00h
    int 21h

code ends
end start

Answer 1

ls equ ($-s)/4  ;this is supposed to be the length of my source string

这实际上计算了元素的数量。

mov cx, di
sub cx, offset d ;here I find out with how many words ...

在插入过程的第二次调用时，这将设置CX = 4，这对于仅有3个值的列表来说太大了。我建议你把CX除以4。

dec di
dec di ;since I work with doublewords...
dec di ;3 times but here my procedure gets fuzzy

这当然是错的。 SCASW表示你要么递减4，要么不递减！

mov bx, word ptr es:[di+1] ;this part is supposed to interchange words...
mov word ptr es:[di+2], bx

这不起作用，因为偏移仅相隔1个字节！

jmp DIplus1

这会产生一个DI的增量，因而是一个错误，因为你想在那个地方存储一个单词。

x86 Asm插入排序

1 个答案: