我有一个8086汇编语言程序，用于确定主字符串中是否存在给定的子字符串。 当子字符串是单个字符时，它工作正常。否则不行。 帮我找到错误。
这是我的代码段
print macro arg
lea dx,arg
mov ah,09h
int 21h
endm
data segment
CarriageReturn equ 0dh ; Next Line
LineFeed equ 0ah ; Printer Next Line
Message1 db CarriageReturn,LineFeed,"Enter the string:$"
Message2 db CarriageReturn,LineFeed,"Enter the sub string:$"
Message3 db CarriageReturn,LineFeed,"sub string found $"
Message4 db CarriageReturn,LineFeed,"sub string not found $"
String db 100 dup(?)
SubString db 100 dup(?)
outs dw 100 dup(?)
begin db 0000h
n db 0000h
StringLength db 0000h
SubStringLength db 0000h
dif db 0000h
n1 db 0000h
of1 dw 0000h
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data ; Point ds ( Data Segement )
mov ds,ax ; To data segment
lea si,String ; si to Main String
lea di,SubString ; di to Sub String
mov cl,00h
print Message1 ; Call Macro to Display String "Enter the String"
LoopReadMainString:
mov ah,01h ; Dos Function To Read Character From Standard Input
int 21h ; Dos Interrupt 21h
; OUTPUT : AL = character from the standard input device
cmp al,0dh ; Cmp if carrage return or Enter Key is Pressed
mov StringLength,cl ; Store the Length of String in Memory
je l1 ; If Enter Key is pressed jump to l1
mov [si],al ; Store Character in Memory Address of Main String
inc si ; Increment to Next Memory Location for next character in Main String
inc cl ; Increment String Length Counter
jmp LoopReadMainString
l1:
print Message2 ; Call Macro to Display String "Enter the sub string"
mov cl,00h
jmp LoopReadSubString
LoopReadSubString:
mov ah,01h ; Dos Function To Read Character From Standard Input
int 21h ; Dos Interrupt 21h
cmp al,0dh ; Cmp if carrage return or Enter Key is Pressed
mov SubStringLength,cl ; Store the Length of Sub String in Memory
je l3 ; If Enter Key is pressed Jump to l3
mov [di],al ; Store Character in Memory Address of Sub Main String
inc di ; Increment to Next Memory Location for next character in Main String
inc cl ; Increment String Length Counter
jmp LoopReadSubString
l3:
mov al,StringLength
cmp al,SubStringLength ; Cmp if StringLength = SubStringLength
jz l4 ; Jump to l4 if zero flag is set, ZF = 1 if equal
jnc l4 ; Jump if carry flag not set to l4
jc exit1 ; if carry flag is set go to exit
l4:
lea si,String ; Load effective address of string to si, ideally point si to memory location of String
lea di,SubString ; Point di to memory location of SubString
mov bl,[di] ; copy character in memory location of di ( SubString ) to bl
mov begin,bl ; copy character to begin variable
mov cl,StringLength ; cl or cx is generally used as counter, copy String Length to cl
jmp l5 ; Jmp to l5.
l5:
cmp cl,00h ; cmp to see if the value of cl counter ie length of string value is 0
je exit1 ; jump to exit.
mov al,[si] ; copy character in main string to al
cmp al,begin ; Compare character in main string with character in substirng ( both are pointing to memory location )
mov bl,SubStringLength ; bl will contain the string length of substring
jz l8 ; if they are same jump to l8
inc si ; inc Main String Pointer to point to next character
dec cl ; counter cx is decrease to compare to zero the first line in l5
jmp l5 ; loop . jump to l5
l8:
dec cl
mov ax,si
mov of1,ax
jmp l6
l6:
cmp bl,00h
je exit2
mov al,[si]
cmp al,[di]
inc si
inc di
dec bl
jnz l7
jmp l6
l7:
mov si,of1
mov al,[si]
inc si
lea di,SubString
jmp l5
exit1:
print Message4 ; Print "Substring Found"
mov ah,4ch ; Dos Sub Function To Exit Program
int 21h ; Dos Interrupt 21h
exit2:
print Message3 ; Print "Substring Not Found"
mov ah,4ch ; Dos SubFunction To Exit Program
int 21h ; Dos Interrupt 21h
code ends
end start
答案 0 :(得分：1)
Rajil kv试试这段代码...我希望它正在运作......实际上我只是编辑了你自己的代码......所以我不保证它是100％正常工作......所以请检查并且提供反馈
print macro arg
lea dx,arg
mov ah,09h
int 21h
endm
data segment
cr equ 0dh
lf equ 0ah
m1 db cr,lf,"Enter the string:$"
m2 db cr,lf,"Enter the sub string:$"
m3 db cr,lf,"sub string found $"
m4 db cr,lf,"sub string not found $"
str db 100 dup(?)
rev db 100 dup(?)
count1 db 0000h
count2 db 0000h
data ends
code segment
assume cs:code,ds:data
start: mov ax,data
mov ds,ax
mov si,offset str
mov di,offset rev
mov cl,00h
print m1
l0:mov ah,01h
int 21h
cmp al,0dh
je l1
mov [si],al
inc si
inc cl
mov count1,cl
jmp l0
l1:print m2
mov cl,00h
l2:mov ah,01h
int 21h
cmp al,0dh
je l3
mov [di],al
inc di
inc cl
mov count2,cl
jmp l2
l3:mov si,offset str
mov di,offset rev
lpag:mov al,[si]
mov bl,[di]
cmp al,bl
jne lpne
inc di
dec count2
jnz lpne
print m3
jmp lpend1
lpne: inc si
dec count1
mov dl,count1
cmp dl,00h
jnz lpag
print m4
lpend1:mov ah,4ch
int 21h
code ends
end start
答案 1 :(得分：0)
这是我的代码片段，用于查找（{3}}
中发布的块缓冲区中是否存在（子）字符串 MOV EDI,OFFSET BlockBuffer
MOV ECX,SIZEOF BlockBuffer
MOV ESI,OFFSET String
MOV EDX,SIZEOF String - 1
CLD
LODSB ; 1st Char of String
Find1stChar:
REPNE SCASB ; find 1st Char in BlockBuffer
JNE NotFound ; all BlockBuffer searched
PUSH ECX,ESI,EDI
CMP ECX,EDX
JB NearEnd
MOV ECX,EDX
REPE CMPSB ; compare the rest of String
NearEnd:
POP EDI,ESI,ECX
JNE Find1stChar ; if not found, continue with 1st Char
Found: DEC EDI ; to point at string in BlockBuffer
答案 2 :(得分：0)
您可以尝试我的代码：
DATA SEGMENT
STR1 DB 'MADAM'
LEN1 DW ($-STR1); storing the length of STR1
STR2 DB 'MADAA'
LEN2 DW ($-STR2); stroing the length of STR2
DATA ENDS
CODE SEGMENT
LEA SI, STR1
LEA DI, STR2
MOV DX, LEN1
MOV CX, LEN2
CMP CX, DX; comparing main & substring length
JA EXIT; if substring size is bigger than there is no chance to be found it in main string
JE SAMELENGTH; if main & sub string both have same length the we can compare them directly
JB FIND; general case (substring length < mainstring length): we can apply our main process
SAMELENGTH:
CLD
REPE CMPSB
JNE RED
JMP GREEN
FIND:
MOV AL, [SI]; storing the ascii value of current character of mainstring
MOV AH, [DI]; storing the ascii value of current character of substring
CMP AL,AH; comparing both character
JE CHECK;
JNE NOTEQL
NOTEQL:
INC SI; if both character don't match then we would point to the next char of main string
DEC DX; DX keeps track of how many character of mainstring is left to process
CMP DX, 0000H; checking if there are any character left in the main string for further comparison
JE RED; if no character is left in main string then obviously the substring doesn't exists in main string
JMP FIND
CHECK:
MOV CX, LEN2; CX is used internally for REPE CMPSB. So storing length of the substring in CX would limit the number of characters for comparison to exact length of substring.
; For example to compare between "madam" & "ada" we need to compare *ada* portion of main string with substring ada, no more, no less
MOV SP, SI; storing the index of current character of main string so if the following REPE CMPSB find mismatch then the process can be started over from the next character of main string (SEE line 1 of TEMPRED) by going to TEMPRED > FIND
ADD SP, 0001H
CLD
REPE CMPSB
JNE TEMPRED
JMP GREEN
TEMPRED:; substring not found starting from the current character of main string, but it is possible to find match if we start from next character in main string
MOV SI,SP; going to the next character of main string (after REPE CMPSB of CHECK segment)
DEC DX
LEA DI, STR2; reloading substring index in DI (after REPE CMPSB of CHECK segment)
JMP FIND; if a character matches but the following substring mismatches in main string then we start over the same process from the next character of main string by going to FIND segment
GREEN:
MOV BX, 0001H; substring found
JMP EXIT
RED:
MOV BX, 0000H; substring not found
JMP EXIT
EXIT:
CODE ENDS
END
RET
答案 3 :(得分：0)
org 100h
; add your code here
.data
string db 'umare q$'
stringS dw 18
subString db 'umar'
subSize dw 4
count db 0
;######################################## .code
lea si , string
mov ax , ds
mov es , ax
lea di , subString
;#########################################
mov cx , stringS
cld
again:
lea di , subString
mov cx , subSize
repe cmpsb
je Equal
jmp ee
Equal:
inc count
ee:
cmp [ds+si] , '$'
jne again
;#########################################
.exit
ret
答案 4 :(得分：-2)
print macro arg
lea dx,arg
mov ah,09h
int 21h
endm
data segment
CarriageReturn equ 0dh ; Next Line
LineFeed equ 0ah ; Printer Next Line
Message1 db CarriageReturn,LineFeed,"Enter the string:$"
Message2 db CarriageReturn,LineFeed,"Enter the sub string:$"
Message3 db CarriageReturn,LineFeed,"sub string found $"
Message4 db CarriageReturn,LineFeed,"sub string not found $"
String db 100 dup(?)
SubString db 100 dup(?)
outs dw 100 dup(?)
begin db 0000h
n db 0000h
StringLength db 0000h
SubStringLength db 0000h
dif db 0000h
n1 db 0000h
of1 dw 0000h
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data ; Point ds ( Data Segement )
mov ds,ax ; To data segment
lea si,String ; si to Main String
lea di,SubString ; di to Sub String
mov cl,00h
print Message1 ; Call Macro to Display String "Enter the String"
LoopReadMainString:
mov ah,01h ; Dos Function To Read Character From Standard Input
int 21h ; Dos Interrupt 21h
; OUTPUT : AL = character from the standard i/0 devi
cmp al,0dh ; Cmp if carrage return or Enter Key is Pressed
mov StringLength,cl ; Store the Length of String in Memory
je Next ; If Enter Key is pressed jump to l1
mov [si],al ; Store Character in Memory Address of Main String
inc si ; Increment to Next Memory Location for next character in Main String
inc cl ; Increment String Length Counter
jmp LoopReadMainString
Next:
print Message2 ; Call Macro to Display String "Enter the sub string"
mov cl,00h
jmp LoopReadSubString
LoopReadSubString:
mov ah,01h ; Dos Function To Read Character From Standard Input
int 21h ; Dos Interrupt 21h
cmp al,0dh ; Cmp if carrage return or Enter Key is Pressed
mov SubStringLength,cl ; Store the Length of Sub String in Memory
je Nextto ; If Enter Key is pressed Jump to l3
mov [di],al ; Store Character in Memory Address of Sub Main String
inc di ; Increment to Next Memory Location for next character in Main String
inc cl ; Increment String Length Counter
jmp LoopReadSubString
Nextto:
lea si,String
lea di,SubString
CheckFirstLetter:mov al,[si]
mov bl,[di]
cmp al,bl
jne CheckNextLetter
inc di
dec SubStringLength
jnz CheckNextLetter
print Message3
jmp exit
CheckNextLetter:inc si
dec StringLength
mov dl,StringLength
cmp dl,00h
jnz checkit
print Message4
checkit:mov al,[si]
mov bl,[di]
cmp al,bl
je CheckFirstLetter
print Message4
exit:
mov ah,4ch ; Dos Sub Function To Exit Program
int 21h ; Dos Interrupt 21h
code ends
end start