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
答案 0 :(得分:3)
1)对于冒泡排序,您需要两个嵌套循环。外部循环重置内部循环的起始参数,直到没有任何东西可以交换。
2)您对字符进行排序。这是8位值(字节)。您无法将它们直接加载到16位寄存器(mov ax, [bx+si]
)。
3)[bx+si]
& [bx+si+1]
:这是错误的,我无法解释错误:-)。
我没有更正你的代码,而是从头开始写了一个例子"":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&#34;动画&#34;插图: