我需要做这段代码,它显示一个单词是否是另一个单词的子串,两者都是从键盘读取的,首先是字符串,然后是我需要检查它是否是子串的单词。
问题是我输入它们,输出是:
无效。找到单词。未找到单词。
我试着检查第二个输入是否比第一个大,所以基本上,它会比较它们并显示消息:“invalid”。无论我写什么,输出都是一样的:“无效。找到单词。未找到单词。”
下面是我的完整代码:
.model small
.stack 200h
.data
prompt1 db "Input String: $"
prompt2 db 10,10, 13, "Input Word: $"
prompt3 db 10,10, 13, "Output: $"
found db "Word Found. $"
notfound db "Word Not Found. $"
invalid db 10,10, 13, "Invalid. $"
InputString db 21,?,21 dup("$")
InputWord db 21,?,21 dup("$")
actlen db ?
.code
start:
mov ax, @data
mov ds, ax
mov es, ax
;Getting input string
mov ah,09h
lea dx, prompt1
int 21h
lea si, InputString
mov ah, 0Ah
mov dx, si
int 21h
;Getting input word
mov ah,09h
lea dx, prompt2
int 21h
lea di, InputWord
mov ah, 0Ah
mov dx, di
int 21h
;To check if the length of substring is shorter than the main string
mov cl, [si+1]
mov ch, 0
add si, 2
add di, 2
mov bl, [di+1]
mov bh, 0
cmp bx, cx
ja invalid_length
je valid
jb matching
valid:
cld
repe cmpsb
je found_display
jne notfound_display
mov bp, cx ;CX is length string (long)
sub bp, bx ;BX is length word (short)
inc bp
cld
lea si, [InputString + 2]
lea di, [InputWord + 2]
matching:
mov al, [si] ;Next character from the string
cmp al, [di] ;Always the first character from the word
je check
continue:
inc si ;DI remains at start of the word
dec bp
jnz matching ;More tries to do
jmp notfound_display
check:
push si
push di
mov cx, bx ;BX is length of word
repe cmpsb
pop di
pop si
jne continue
jmp found_display
again:
mov si, ax
dec dx
lea di, InputWord
jmp matching
invalid_length:
mov ah, 09h
lea dx, invalid
int 21h
found_display:
mov dx, offset found
mov ah, 09h
int 21h
notfound_display:
mov dx, offset notfound
mov ah, 09h
int 21h
end start
1条答案
按热度按时间qyuhtwio1#
为了澄清杰斯特已经说过的话:
标签本身不影响
ip
寄存器的操作。这是因为标签在运行时不存在;它们仅仅是程序员的一种便利。所以你的计算机实际上是这样处理上面的代码的:如果你只想运行这三个命令中的一个,你需要在每个命令的末尾重定向
ip
。有几种方法可以做到这一点,这里有一种: