程序集-气泡排序以对字符串进行排序 [英] Assembly - bubble sort for sorting string
问题描述
我正在使用tasm编写汇编程序.我的任务是编写一个程序，该程序将使用冒泡排序按字母顺序对输入的字符串进行排序.前任.如果输入"hello"，则应输入"ehllo".我已经写过乞求输入字符串并对其进行排序(我认为它可以正常工作，直到最后应该打印出结果，但是最后它只写了一次.data并完成了工作)PS对不起，不好英文
I am writing a program in assembly using tasm. My task is to write a program that will use bubble sort to sort entered string alphabetically. Ex. if you enter "hello" it should write "ehllo". I have writened the begging to enter string and to sort it (I think it works okey until the end where it should print out the result, but at the end it just writes my .data once and the finisheds its work) P.S sorry for bad english
.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
推荐答案
1)对于冒泡排序，您需要两个嵌套循环.外循环会重置内循环的启动参数，直到没有剩余可交换的内容为止.
1) For bubble sort you need two nested loops. The outer loop resets the start parameters for the inner loop until there is nothing left to swap.
2)您对字符进行排序.那是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.
2) You sort characters. That are 8-bit values (bytes). You can't load them directly into a 16-bit register (mov ax, [bx+si]).
3)[bx+si]& [bx+si+1]:这太错了，我无法解释该错误:-).
3) [bx+si] & [bx+si+1]: this is so wrong that I cannot explain the error :-) .
我没有纠正您的代码，而是从头开始"编写了一个示例:按照
Instead of correcting your code I wrote an example "from scratch": following the illustration in http://en.wikipedia.org/wiki/Bubble_sort:
.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:
And here is another "animated" illustration:
https://www.youtube.com/watch?v=lyZQPjUT5B4
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