通过对两个字符串的字符求和/减法创建新数组

正如@Jester所指出的,双循环可以工作,这只是我如何做的一个例子。然而,这个例子是破坏性的,因为s1的内容正在被修改。如果您想使用这个模型,但要修改以完全符合您的示例,那么修改应该很容易概念化。

    ASSUME cs:text_,ds:data_

        data_ SEGMENT
    s1 db '1','2','3','4'
    s2 db '5','6','7','8'
        data_ ENDS

        text_ SEGMENT
    start:
            mov     ax, data_
            mov     ds, ax
            mov     es, ax

            mov     si, offset s1
            mov     di, offset s2
            mov     cx, di
            sub     cx, si              ; Now you have the number of entries in CX
            xor     bx, bx
            jcxz    endi

        Next:   
            lodsb
            mov     ah, [di]
            bt      bx, 0               ; Is bit on, then we want to subtract
            jnc     @F - 2
            sub     al, ah
            jmp     @F
            add     al, ah

     @@:    stosb
            xor     bl, 1               ; Toggle add or subtract indicator
            loop    Next

        endi:
            mov     ax, 4c00h
            int     21h

        text_ ENDS
    end start   

注意:我使用NASM,所以希望这个示例可以使用MASM

代码有一些问题。

1. 不能添加或删除 角色 . 字符不是存储为数字,而是存储为ASCII码。您可以将ASCII码想象为监视器显示特定字符的命令。处理器需要另一种格式将其作为整数处理。您必须对其进行转换(如果要显示它,还必须进行转换)。字符的ASCII部分位于高位半字节。要获取整数,请将其删除( AND 0Fh 或 SUB 48 ). 要获取字符,请将ASCII部分添加到整数中( OR 30h 或 ADD 48 ). 阶梯是你家庭作业的主要部分,因为处理负号和一位数以上的数字有点困难。

2. SI、DI和BX必须使用2循环解决方案增加2。 LODSB 和 STOSB 仅增加1。因此,您必须为该寄存器添加额外的增量。您忘记增加BX。

3. 这里没关系,但是。。。 JL 和 JLE 表演 签署 比较。CX不太可能变成负值。所以把它改成未签名的camparisons JB 和 JBE .

下面的代码将结果存储为整数 d :

ASSUME cs:text_,ds:data_

data_ SEGMENT

s1 db '1','2','3','4'
l equ $-s1
s2 db '5','6','7','8'
d db l dup (?)

data_ ENDS

text_ SEGMENT
start:
mov ax, data_
mov ds, ax
mov es, ax
mov si, offset s1
mov bx, offset s2
mov di, offset d
cld
mov cx, l
jcxz endi

xor cx,cx                       ; cx-register is the counter, set to 0
loop1:
    lodsb                       ; current character of s1 is stored in al
    mov ah, byte ptr es:[bx]    ; the character of s2, which is on the same position as the one stored in al, is added to al
    and ax, 0F0Fh               ; Remove the ASCII parts in AH and AL
    add al, ah
    stosb                       ; data from al is moved to di
    add bx, 2                   ; Increment
    add si, 1                   ; Increment
    add di, 1                   ; Increment
    add cx, 2                   ; cx is incremented
    cmp cx,l                    ; Compare cx to the limit
;jl loop1                       ; Loop while less
jb loop1                        ; Loop while below

mov si, offset s1 + 1           ; Reinitialize SI, DI and BX at odd indices
mov bx, offset s2 + 1
mov di, offset d + 1
mov cx, 1                       ; cx-register is the counter, set to 1
loop2:
    lodsb                       ; current character of s1 is stored in al
    mov ah, byte ptr es:[bx]    ; the character of s2, which is on the same position as the one stored in al, is added to al
    and ax, 0F0Fh               ; Remove the ASCII parts in AH and AL
    sub al, ah
    stosb                       ; data from al is moved to di
    add bx, 2                   ; Increment
    add si, 1                   ; Increment
    add di, 1                   ; Increment
    add cx, 2                   ; Increment
    cmp cx,l                    ; Compare cx to the limit
;jle loop2                      ; Loop while less or equal
jbe loop2                       ; Loop while below or equal

endi:
   mov ax, 4c00h
   int 21h
text_ ENDS
end start

在单循环解决方案中进行比较非常容易。想想二进制!奇数在最右边的位有一个1,偶数有一个0。你可以 AND 带1的数字,得到0(未设置)或1(设置)。 TEST 是一条特殊的x86指令,它执行 以及 但只设置标志,不更改寄存器。未设置零鞭毛(请参见 JNZ )如果结果不为null->奇数。

ASSUME cs:text_,ds:data_

data_ SEGMENT

s1 db '1','2','3','4'
l equ $-s1
s2 db '5','6','7','8'
d db l dup (?)

data_ ENDS

text_ SEGMENT
start:
mov ax, data_
mov ds, ax
mov es, ax
mov si, offset s1
mov bx, offset s2
mov di, offset d
cld
mov cx, l
jcxz endi

xor cx,cx                       ; cx-register is the counter, set to 0
loop1:
    lodsb                       ; current character of s1 is stored in al
    mov ah, byte ptr es:[bx]    ; the character of s2, which is on the same position as the one stored in al, is added to al
    and ax, 0F0Fh               ; Remove the ASCII parts in AH and AL
    test cx, 1                  ; Odd?
    jnz odd                     ; Yes -> jump to the subtraction
    even:
    add al, ah
    jmp l2                      ; Skip the subtraction
    odd:
    sub al, ah
    l2:
    stosb                       ; data from al is moved to di
    add bx, 1                   ; Increment
    add cx, 1                   ; cx is incremented
    cmp cx,l                    ; Compare cx to the limit
jbe loop1                       ; Loop while below

endi:
   mov ax, 4c00h
   int 21h
text_ ENDS
end start

我考虑过只选一个,根据比较 创建和或差。

这就是只使用一个循环(根据索引值求和或求差)的方法:

只需用以下代码替换两个循环:-

  mov cl, 2              ; store 2 in cl which is used to check later if the index is even or odd
  loop1:
        lodsb            ; current character of s1 is stored in al
        mov dl, al       ; save al in dl   
        mov ax, si       ; store the value of si in ax
        dec ax           ; since lodsb increments si we have to decrement it to get the current index  
        div cl           ; divide al with cl
        mov al, dl       ; get the stored value from dl to al 
        cmp ah, 0        ; check if remainder is zero (div instruction stores remainder in ah and quotient in al if divisor is 8bit)
        jne oddNum       ; if no then we are at odd position otherwise we are at even position 
        add al, byte ptr es:[bx] 
        jmp continue
        oddNum:
        sub al, byte ptr es:[bx]      ;the character of s2, which is on the same position as the one stored in al, is added to al
        continue:
        stosb                ; data from al is moved to di    
        inc bx               ; increment bx so that it points to next element in s2 
        cmp si,l             ; Compare cx to the limit
    jl loop1                 ; Loop while less 

在上面的代码中,您将根据si寄存器的值进行求和或求差(可以是偶数或奇数)