我正在撰寫一個程式，將 0-4096 之間的非負整數轉換為 3 位十六進制數。問題是這個程式甚至不接受我的輸入。我正在使用帶有自定義宏的 SASM32。相反，它會溢位重復輸出訊息并使用 0 作為我們的輸入。此外，由于回傳提示的默認值是“Y”，它會一直持續到崩潰。
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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
uj5u.com熱心網友回復：
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
關于這些操作的方向，這里有一些非常錯誤的地方！
在move user_input, input_store我所期待source -> destination，但
move temp, user_input它相當destination <- source。不能同時...
假設destination <- source這是發生的事情：
- 由于input_store 為0，因此該行
move user_input, input_store會破壞輸入并使其為零。 - 商和余數的計算將產生零。
- 從商加載user_input后，它將為零（再次）。
- 最后
compare user_input, 0bez return1總是跳轉到return1。
compare reprompt_char, 'Y' ;if N was entered bez reprompt_flag reprompt_flag: br start
像上面這樣的代碼構成了一個無限回圈。如果答案是“是”，你跳轉到該進入指令開始，如果答案是不是“Y”（這是不一樣已進入“N”），你通過在下降的是去指令啟動. 無論哪種方式，您都重復該程式。
請嘗試以下操作。這樣，回答“Y”的用戶將重新運行程式，而任何其他回答將在退出時失敗。
reprompt:
put_str reprompt_msg
get_ch reprompt_char
compare reprompt_char, 'Y'
bez start
exit
end start
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