我需要在整个文件中用我的名字替换用户从键盘输入的文件中的特定字母，然后将其另存为新文件。
问题所在的代码部分。总是跳到错误，但是有了路径，一切都好了：
write_to_new_file:
mov ah, 3Ch
lea dx, new_file
mov al, 0
int 21h
jc file_error
mov bx, ax
这是代码的整个部分，我正在替换字母+尝试通过更改创建一个新文件：
org 100h
jmp start
; File paths
file1 db "c:\emu8086\vdrive\C\file1.txt", 0 ; and files 2 and 3 almost same paths
new_file db "c:\emu8086\vdrive\C\created_folder\file_corrected.txt", 0
handle dw ?
...
; General messages
...
; Replace functionality messages
...
start:
mov ax, cs
;mov dx, ax
mov es, ax
...
;********** REPLACEMENT *********************
; Create new file
lea dx, new_file
mov ah, 3Ch ; Create file function
mov cx, 0 ; Normal attributes
int 21h
mov bx, ax ; Save handle of new file in BX
choose_file_for_replacement:
mov dx, offset replace_message ; Load address of footer
mov ah, 09h ; DOS service to display string
int 21h ; Call DOS interrupt
; Wait for user input
mov ah, 01h ; DOS service to read character
int 21h ; Call DOS interrupt
; Compare input directly with ASCII values for '1', '2', '3'
cmp al, '1' ; Compare with '1'
je replace_file1
cmp al, '2' ; Compare with '2'
je replace_file2
cmp al, '3' ; Compare with '3'
je replace_file3
replace_file1:
; Initialize data segment
mov ax, @data
mov ds, ax
; Open the file
mov ah, 3Dh ; Open file function
mov al, 0 ; Open for reading
lea dx, file1 ; Load address of file name
int 21h
jc file_error ; If error, jump to error handling
mov bx, ax ; Store file handle in BX
jmp read_into_buffer
replace_file2:
...
replace_file3:
...
read_into_buffer:
mov ah, 3Fh ; Read file function
lea dx, buf ; Load address of buffer
mov cx, 255 ; Number of bytes to read
int 21h
jc file_error ; If error, jump to error handling
mov si, offset buf ; SI points to the buffer
; Prompt user to enter the character to replace
mov ah, 02h ; DOS service to write character
mov dx, offset prompt
mov ah, 9
int 21h
; Read the character from the user
mov ah, 01h
int 21h
mov bl, al ; Store user input in BL
mov ah, 02h ; DOS service to write character
; Process the buffer
process_message:
mov al, [si]
cmp al, 0
je write_to_new_file
cmp al, bl
jne skip_replacement
; Replace character with name
mov di, offset replacement
output_replacement:
mov al, [di]
cmp al, 0
je continue_processing
mov dl, al
mov ah, 2
int 21h
inc di
jmp output_replacement
skip_replacement:
mov dl, al
mov ah, 2
int 21h
continue_processing:
inc si
jmp process_message
write_to_new_file:
mov ah, 3Ch
lea dx, new_file
mov al, 0
int 21h
jc file_error
mov bx, ax
; Calculate the length of the modified buffer
mov cx, si ; SI points to the end of the modified part
; Write the modified content to the new file
mov ah, 40h
lea dx, buf
int 21h
jc file_error
; Close the new file
mov ah, 3Eh
int 21h
; Display success message
lea dx, success_message
mov ah, 9
int 21h
jmp end_program
file_error:
; Display error message
lea dx, message_error
mov ah, 9
int 21h
jmp end_program
...
我尝试在互联网上查找有关此内容的一些信息，但找不到与保存现有文件的更改副本类似的内容
ps。如果需要代码的其他部分，请告诉我
org 100hjmp start如果“直接将输入与‘1’、‘2’、‘3’的 ASCII 值进行比较”失败，您仍然需要中止程序或重做用户输入。
可以合理地假设您正在使用小型源文件（例如限制为 255 字节）。那么最好在将文件检索到buf后立即关闭该文件。
您想用您的名字替换输入字符的每一次出现。接下来是如何做到这一点：
Top:
mov al, [si] ; Read from 'buf' \ LODSB
inc si /
mov [di], al ; Write to 'anotherbuf' \ STOSB
inc di /
cmp al, dl ; DL is the replace candidate
jne .b ; Nothing special, so continue the loop
dec di ; Take back the last char written and ...
mov bx, offset replacement ; ... substitute by ASCIIZ string
mov al, [bx] ; There would better be at least 1 true char
.a: inc bx
mov [di], al ; Write to 'anotherbuf' \ STOSB
inc di /
mov al, [bx]
cmp al, 0 ; For all chars except zero
jne .a
.b: dec cx ; For all chars in 'buf' \ LOOP TOP
jnz Top /
现在是创建新文件并将“anotherbuf”的内容写入其中的时候了。
; Calculate the length of the modified buffer mov cx, si ; SI points to the end of the modified part
获取新内容的大小不是通过复制“修改部分末尾的地址”，而是通过计算结束地址和起始地址之间的差：
mov cx, di
sub cx, anotherbuf
或更短，通过 LEA：
lea cx, [di - anotherbuf]