在8086彙編中對字符串進行排序

Question

我想編寫一個8086彙編程序，它將用戶的5個字符串作爲輸入，然後對這些字符串進行排序並打印排序後的結果作爲輸出。我實際上做了所有事情，但我在分揀部分遇到了很大的問題。我知道如何使用例如冒泡排序來排序從一個特定地址開始的數組中的項目，但在這裏我有5個不在同一個數組中的不同字符串。每個字符串都有自己的地址和自己的字符。我嘗試比較每個字符串的最後一個字符，然後如果一個字符大於另一個字符，我將整個字符串交換，然後繼續爲所有字符串的全部字符做第一個字符。

例如，如果我們輸入字符串是：

eab  
abe  
cbd  
cda  
adb 

我會第一時間排序每個字符串的最後一個字符，我想出了這一點：

cda  
eab  
adb  
cbd  
abe 

然後，我將通過比較它們中間字符：

eab  
cbd  
abe  
cda  
adb 

最後用第一個字符，所有東西都排序：

abe 
adb 
cbd 
cda  
eab 

但它實際上是我的想法，我不知道誰來實現我的工作。

 
; multi-segment executable file template. 

data segment 
    data1 db 64,?,64 dup(?) 
    data2 db 64,?,64 dup(?) 
    data3 db 64,?,64 dup(?) 
    data4 db 64,?,64 dup(?) 
    data5 db 64,?,64 dup(?) 

    change db 66 dup(?) 

    msg db 0ah,0dh,"You enter a wrong option",0ah,0dh,"try again",0ah,0dh,"$" 
    prompt db 0ah,0dh,"Choose an option:",0ah,0dh,"$" 
    prompt1 db ".a: Sort in ascending order",0ah,0dh,"$" 
    prompt2 db ".d: Sort in descending order",0ah,0dh,"$" 
    prompt3 db ".q: Quit",0ah,0ah,0dh,"$" 
    enter db 0ah,0ah,0dh,"Enter 5 strings:",0ah,0dh,"$" 
    pkey db 0ah,0dh,"press any key...$" 
ends 

stack segment 
    dw 128 dup(0) 
ends 

code segment 
main proc far 
; set segment registers: 
    mov ax, data 
    mov ds, ax 
    mov es, ax 

again: 
; printing the prompts for the user 
    lea dx, prompt 
    mov ah, 09h 
    int 21h 

    lea dx, prompt1 
    mov ah, 09h 
    int 21h 

    lea dx, prompt2 
    mov ah, 09h 
    int 21h 

    lea dx, prompt3 
    mov ah, 09h 
    int 21h 

; getting a character from the user as an input 
    mov ah, 01h 
    int 21h 

; determining which option the user selects  
    cmp al, 'a' 
    je ascending 
    cmp al, 'd' 
    je descending 
    cmp al, 'q' 
    je quit 

; this is for the time that the user enters a wrong char  
    lea dx, msg 
    mov ah, 09h 
    int 21h 
    jmp again  ; again calling the application to start 

ascending: 
    call input 
    call AscendSort 
    jmp again  ; again calling the application to start 

descending: 
    call input 
    call DescendSort 
    jmp again  ; again calling the application to start 

quit:    
    lea dx, pkey 
    mov ah, 9 
    int 21h  ; output string at ds:dx 

    ; wait for any key....  
    mov ah, 1 
    int 21h 

    mov ax, 4c00h ; exit to operating system. 
    int 21h 
main endp 
;................................................. 
; this subroutine gets input from user 
input proc 

    lea dx, enter 
    mov ah, 09h 
    int 21h 
    call newline 

    mov ah, 0ah 
    lea dx, data1 
    int 21h  
    call newline 

    mov ah, 0ah 
    lea dx, data2 
    int 21h 
    call newline 

    mov ah, 0ah 
    lea dx, data3 
    int 21h 
    call newline 

    mov ah, 0ah 
    lea dx, data4 
    int 21h 
    call newline 

    mov ah, 0ah 
    lea dx, data2 
    int 21h 
    call newline 

    ret 
input endp 
;................................................ 
; sorting the strings in the ascending order 
AscendSort proc   

    mov si, 65 
    lea dx, change 
    mov al, data1[si] 
    cmp al, data2[si] 
    ja l1  
    ????? 

    ret 
AscendSort endp 
;................................................ 
; sorting the strings in the descending order 
DescendSort proc 



    ret 
DescendSort endp 
;................................................ 
; newline 
newline proc 

    mov ah, 02h 
    mov dl, 0ah 
    int 21h 

    mov dl, 0dh 
    int 21h 

    ret   
newline endp  
ends 

end main ; set entry point and stop the assembler. 

任何其他用於分揀這些整個字符串的算法也將被理解。

Answer 1

我其實自己找出答案，我使用字符串命令比較字符串2乘2，看看他們是否更大，更小或相等。喜歡的東西在特定的宏接受兩個字符串來檢查他們，並做必要的操作就像換弦，使它們分類代碼如下：

check macro a, b 
    local next, finish 
    cld 
    mov cx, 64 ; the size of our buffer that saves the string 
    mov si, a 
    mov di, b 

    repe cmpsb ; comparing two strings with each other 
    ja next 
    jmp finish 

next: 
    ; swaping our strings if needed 
    mov cx, 64 
    mov si, a 
    lea di, change 
    rep movsb 

    mov cx, 64 
    mov si, b 
    mov di, a 
    rep movsb 

    mov cx, 64 
    lea si, change 
    mov di, b 
    rep movsb 

finish: 

endm