汇编 - 用于排序字符串的冒泡排序

Question

我正在使用tasm编写程序集.我的任务是编写一个程序,使用冒泡排序按字母顺序对输入的字符串进行排序.防爆.如果你输入"你好",它应该写"ehllo".我已经写了乞求输入字符串并对它进行排序(我认为它工作直到它应该打印出结果的结尾,但最后它只是写了我的.data一次并完成它的工作)PS对不起英语

.model small
.stack 100h

.data
request     db 'This program is using bubblesort to get alphabetical order of your enterd string', 0Dh, 0Ah, 'Enter your string:', 0Dh, 0Ah, '$'
result      db 0Dh, 0Ah, 'Result:', 0Dh, 0Ah, '$'
buffer      db 100, ?, 100 dup (0)

.code

start:
MOV ax, @data                   
MOV ds, ax                      


MOV ah, 09h
MOV dx, offset request
int 21h


MOV dx, offset buffer           
MOV ah, 0Ah                     
INT 21h                         


MOV si, offset buffer           
INC si                          
MOV bh, [si]                    
INC si                          

sort:
mov cx, [si] 
mov bx, [si]     

nextelement:
mov ax, [bx+si]     
cmp ax, [bx+si+1]   
jge noswap          
xchg ax, [bx+si+1]
mov ax, [bx+si]

noswap:
inc si              
cmp cx, si          
jl nextelement      
loop nextelement 



MOV ah, 09h
MOV dx, offset result
int 21h


char:
LODSB                           
MOV ah, 2                       
MOV dl, al                      
INT 21h                        

DEC bh                          
JZ ending                       
JMP char                        


ending:
MOV ax, 4c00h               
INT 21h                         

end start

Answer 1

1）对于冒泡排序，您需要两个嵌套循环。外循环重置内循环的启动参数，直到没有任何内容可以交换。

2）你对字符进行排序。这是 8 位值（字节）。您不能将它们直接加载到 16 位寄存器 ( mov ax, [bx+si]) 中。

3) [bx+si]& [bx+si+1]：这是错误的，我无法解释这个错误:-)。

我没有更正您的代码，而是“从头开始”编写了一个示例：按照http://en.wikipedia.org/wiki/Bubble_sort中的插图：

.MODEL small
.STACK 1000h                        ; Don't skimp with stack!

.DATA
    Struct0A EQU $                  ; Buffer for INT 21h/0Ah (max,got,buf)
        max db 100                  ; Maximum characters buffer can hold (incl. CR (0Dh))
        got db 0                    ; Number of characters actually read, (excl. CR (0Dh))
        buf db 100 dup (0)          ; Actual characters read, including the final carriage return (0Dh)
    Linefeed db 13, 10, '$'
    GetString   db 'Enter string: $'

.CODE
start:
    mov ax, @DATA                           ; Initialize DS
    mov ds, ax

    ; Input String
    mov ah, 09h
    mov dx, OFFSET GetString
    int 21h
    mov dx, OFFSET Struct0A
    mov ah, 0Ah
    INT 21h

    mov si, OFFSET buf                      ; Base for [si + bx] 
    xor bx, bx                              ; Prepare BX for following byte load
    mov bl, got                             ; Load length of string = 0Dh at the end
    mov BYTE PTR [si + bx], '$'             ; Delimiter for int 21h / 09h

    outer:
    dec bx                                  ; The last character is already at the right place
    jz done                                 ; No characters left = done
    mov cx, bx                              ; CX: loop variable
    mov si, OFFSET buf
    xor dl, dl                              ; DL (hasSwapped) = false

    inner:
    mov ax, [si]                            ; Load **two** characters
    cmp al, ah                              ; AL: 1. char, AH: 2. char
    jbe S1                                  ; AL <= AH - no change
    mov dl, 1                               ; hasSwapped = true
    xchg al, ah                             ; Swap characters
    mov [si], ax                            ; Store swapped characters
    S1:
    inc si                                  ; Next pair of characters
    loop inner

    test dl, dl                             ; hasSwapped == true?
    jnz outer                               ; yes: once more
    done:

    ; Print result
    mov dx, OFFSET Linefeed
    mov ah, 09h
    int 21h
    mov dx, OFFSET buf
    mov ah, 09h
    int 21h

    mov ax, 4C00h
    int 21h

END start

这是另一个“动画”插图：

https://www.youtube.com/watch?v=lyZQPjUT5B4