Question

我正在使用EMU8086为学校的项目进行装配游戏。在这个游戏中，我需要允许用户输入一个字符串才能进步。当他输入字符串时，他可能会输入错误的内容并使用 backspace 来纠正它。问题是退格将光标移动到前一个字符上，但先前输入的字符仍然存在。为什么退格不清除前一个字符？如何修复程序，以便删除屏幕上的前一个字符？

我的代码是：

data segment

ends

stack segment
dw   128  dup(0)
ends    
StringHelper db 20 dup(?)
Line db 13,10,'$'
FullInput db 'You cant type more than 20 letters!!! please try again!!',13,10,'$'
t db '$'
code segment
PROC PrintMessage
  ;BX MUST have OFFSET OF MESSAGE
  ; if you want to go down a line do (lea bx,line)      
  mov dx,bx      
  mov ah,09h
  int 21h     
  ret 
 endp printMessage
 proc InputString          
;askes the user to input chars untill he press (enter) then puts it in StringHelper
b:
lea bx, StringHelper
mov cx,5  
xor dx,dx
a:                 ;restarts string helper
mov [bx],00
inc bx
loop a
lea bx, StringHelper
up:
cmp dx,20        ;cheacks if you wrote more than 20 chars
jz TryAgain
deleted:
xor ax,ax
mov ah,01h
int 21h
xor ah,ah
mov cx,08h        ;checks if the user inputed the backspace key
cmp al,cl
jz BackSpace
mov cx,0dh
cmp al, cl                ;checks if the user enters enter
jz InputIsOver
inc dx
mov [bx],al
inc bx
jmp up
TryAgain:
lea bx, line
call PrintMessage 
lea bx, FullInput
call PrintMessage
jmp b:
BackSpace:
cmp dx,0            ;checks if te user didnt just BackSpaced nothing
jz deleted
lea bx,stringhelper  ;gets the start of the array
add bx,dx            ;adds dx which is the indexer to how many chars  you     already wrote
mov [bx],00h         ;puts 0(nothing) at that place
dec bx
dec dx
jmp deleted          ;returen to get an extra input
InPutIsOver:

ret
endp
start:
  mov  ax, @data
 mov  ds, ax
mov al,13h
int 10h 
call InputString
 ; add your code here

mov ax, 4c00h
int 21h  

ends

end start

Answer 1

对于典型的屏幕和终端仿真器，打印退格字符只是将光标向左移动一个位置。要清除字符，请尝试打印退格键+空格+退格键。

Answer 2

Keith是正确的，即使在DOS中，如果你打印退格字符，它也是非破坏性的。这意味着光标将返回但下面的字符仍然存在。这是正常行为。

我没有时间仔细查看代码，但在最初打印第一个退格后，您可以使用Int 21h/Ah=2打印另一个 space 字符然后是另一个退格。

在退格代码中你有：

BackSpace:
cmp dx,0            ;checks if te user didnt just BackSpaced nothing
jz deleted

要解决此问题，我认为您可以将代码更改为：

BackSpace:
cmp dx,0            ;checks if te user didnt just BackSpaced nothing
jz deleted

mov ah, 02h         ; DOS Display character call 
mov dl, 20h         ; A space to clear old character 
int 21h             ; Display it  
mov dl, 08h         ; Another backspace character to move cursor back again
int 21h             ; Display it

删除屏幕上的字符

2 个答案: