在8086 Assembly中对字符串进行排序
我想编写一个8086汇编程序，它将用户的5个字符串作为输入，然后对这些字符串进行排序并将排序后的结果作为输出进行打印。 我实际上做了所有事情，但我在分拣部分遇到了很大的问题。 我知道如何使用例如冒泡排序来排序从一个特定地址开始的数组中的项目，但在这里我有5个不在同一个数组中的不同字符串。 每个字符串都有自己的地址和自己的字符。 我尝试比较每个字符串的最后一个字符，然后如果一个字符大于另一个字符，我将整个字符串交换，然后继续为所有字符串的全部字符做第一个字符。
例如，如果我们的输入字符串是：
eab
abe
cbd
cda
adb
我将首先排序每个字符串的最后一个字符，然后提出这个问题：
cda
eab
adb
cbd
abe
然后我会用中间字符比较它们：
eab
cbd
abe
cda
adb
最后用第一个字符排序：
abe
adb
cbd
cda
eab
但实际上在我脑海里，我不知道为我的工作实施这个目标。
; 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.
将理解用于分拣这些整个字符串的任何其他算法。
我实际上自己找出答案，我使用字符串命令来比较字符串2和2，看看它们是更大，更小还是相等。 类似下面的特定宏中的代码，它使用两个字符串来检查它们并执行所需的操作，例如交换字符串以使它们排序：
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