为什么这个汇编程序不接受输入？

Question

我正在编写一个程序，将 0-4096 范围内的非负整数转换为 3 位十六进制数。问题是这个程序甚至不接受我的输入。我正在使用带有自定义宏的 SASM32。相反，它会溢出并重复输出消息并使用 0 作为我们的输入。此外，由于 return 提示符的默认值是 'Y'，它会一直持续到崩溃。

.486
include C:\Program Files (x86)\SASM\include\sasmacros.inc


.data 
;Messages 
inputmsg db 'Input decimal integer to convert:',0
outputmsg db 'in hexadecimal is:',0
reprompt_msg db 'Convert another integer? (Y/N):',0 
reprompt_char db 'N'

;Variables 
input_store dd 0
user_input dd 0

temp dd 0
remainder_store dd 0 


quotient dd 0 
remainder dd 0 
temp_remainder dd 0 

hexnumber dd 0 DUP(100)
output dd 1, 2, 3

;main loop control is positions and user_input 
;record each hex position as a character or digit(either one will be 0 or digit)
positions dd 0 
pos1 db '0'
pos2 db ' '
pos3 db ' '


.code

;take number and divide it by 16 and at the remainder of it to the output
;divide -> check remainder (if n > 9) -> add hex number loop -> divide loop
; store the quotient for the next iteration of dividing
;  divide until our remainder or the result of our mod is == 0 

;for reprompt just compare if N - N == 0 is true

;if input is less than 15 then just return it 

                   

start:
    get_i user_input
    move user_input, input_store ; store our input before it is lost 
    ;might delete
    ;might delete
    br divide

divide:

    ;divide by 16
    move temp, user_input ; setting temp to the current input 
    idivi temp, 16
    move quotient, temp ;quotient = temp 
    move temp, user_input ;temp = input
    irem temp, 16 ; takes remainder of temp and saves it as temp
    move remainder, temp ;remainder = temp
    ;set our input = quotient 
    move user_input, quotient ; storing our input for next loop as quotient 
    br store_result
    
store_result:

    compare user_input, 0 ;if our user_input is 0 (which becomes the quotient of the previous calculation), then finish
    bez return1           ; problem is that we would also hit this case if our input is 0
    ;check for 0
    compare remainder, 10
    bgez convert_to_upper_hex
    
    ;otherwise just store our remainder and loop again
    ;48 - remainder = character representation of remainder ASCII
    iadd positions, 1 ; increment positions 
    iadd remainder, 48 ; create ascii value 
    br figure_what_to_store


    ;converts digits 10-15 
convert_to_upper_hex:
    iadd positions, 1 ;have to increment still in this loop
    iadd remainder, 41 ;make it A-F using ASCII starting at 'A'
    br figure_what_to_store

figure_what_to_store:
    compare 1, positions
    bez store_pos_one

    compare 2, positions
    bez store_pos_two

    compare 3, positions
    bez store_pos_three


store_pos_one:
    move pos1, remainder ;move remainder into pos1
    br divide 

store_pos_two:
    move pos2, remainder 
    br divide 

store_pos_three:
    move pos3, remainder 
    br return1
    ;end loop here 

return1: 
     put_i input_store
     put_str outputmsg
     ;update displaying integer
     compare positions, 1
     bez return_single
     compare positions, 2 
     bez return_double
     
     ;if we aren't displaying 1 or 2 digits, then we are displaying 3
     put_ch pos3
     put_ch pos2
     put_ch pos1
     ;check if more positions, if so bnz return
     br reprompt

return_single:
    put_ch pos1
    br reprompt 
    
return_double:
    put_ch pos2
    put_ch pos1
    br reprompt    
     
reprompt:
     put_str reprompt_msg 
     get_ch reprompt_char
     compare reprompt_char, 'Y' ;if N was entered 
     bez reprompt_flag
     
reprompt_flag:
    ;also clear output 
    br start
    
    exit
end start

Answer 1

move user_input, input_store ; store our input before it is lost 
;divide by 16
move temp, user_input ; setting temp to the current input

这些操作的方向有很大的问题！
在 move user_input, input_store，我预计 source -> destination，但在
move temp, user_input 而是 destination <- source。不能同时...

假设destination <- source是这样的：

因为input_store保持0，行move user_input, input_store销毁输入并使其为零。
商和余数的计算结果均为零。
从商加载 user_input 后，它将（再次）为零。
最后compare user_input, 0bez return1会一直跳转到return1.

  compare reprompt_char, 'Y' ;if N was entered 
  bez reprompt_flag

reprompt_flag:
  br start

如上代码构成死循环。如果答案是 'Y'，你 分支到 进入 start 的指令，如果答案不是 'Y' ]（这与输入 'N' 不同）您落入进入 start 的指令。无论哪种方式，您都重复该程序。

尝试以下操作。这样，回答 'Y' 的用户将重新运行程序，而任何其他回答将 在退出时失败 。

reprompt:
     put_str reprompt_msg 
     get_ch reprompt_char
     compare reprompt_char, 'Y'
     bez start
     
    exit
end start

为什么这个汇编程序不接受输入？

Why Isn't this Assembly Program Accepting Input?

assembly

x86-64