在程序集中将十进制转换为二进制

Question

我的PicoBlaze模拟器在中的应用中最长的示例程序是这个十进制到二进制转换器：

;This is an example program that uses
;UART, the interface that PicoBlaze uses
;for connecting to terminals (a DOS-like
;user interface, with a keyboard and a
;screen capable of displaying text).
;It loads base-10 integer numbers from
;the terminal, converts them into binary,
;and then prints the binary
;representations back onto the terminal.
;Example input would be: 

;1
;2
;4
;8
;15
;127
;255
;

;And the expected output is: 

;1_(10)=1_(2)
;2_(10)=10_(2)
;4_(10)=100_(2)
;8_(10)=1000_(2)
;15_(10)=1111_(2)
;127_(10)=1111111_(2)
;255_(10)=11111111_(2)
;

;Note that you need to click the
;"Enable UART" button in order to use it.
;Also, the trailing empty line in the
;input is necessary for the result to be
;printed.

;Now follows some boilerplate code
;we use in our Computer Architecture
;classes...
CONSTANT LED_PORT,00
CONSTANT HEX1_PORT,01
CONSTANT HEX2_PORT,02
CONSTANT UART_TX_PORT,03
CONSTANT UART_RESET_PORT,04
CONSTANT SW_PORT,00
CONSTANT BTN_PORT,01
CONSTANT UART_STATUS_PORT,02
CONSTANT UART_RX_PORT,03
; Tx data_present
CONSTANT U_TX_D, 00000001'b
; Tx FIFO half_full
CONSTANT U_TX_H, 00000010'b
; TxFIFO full
CONSTANT U_TX_F, 00000100'b
; Rxdata_present
CONSTANT U_RX_D, 00001000'b
; RxFIFO half_full
CONSTANT U_RX_H, 00010000'b
; RxFIFO full
CONSTANT U_RX_F, 00100000'b

ADDRESS 000
START:
;At the beginning, the number is 0.
load s0,0
;And we are storing its string
;representation at the beginning
;of RAM.
namereg s3,pointer
load pointer,0
;Now follows a loop to load
;the digits of the number.
loading_the_number:
  ;Load a character from the UART
  ;terminal.
  call UART_RX
  ;Check whether the character is a digit.
    compare s9,"0"
    ;If it is not a digit, jump to the
    ;part of the program for printing
    ;the number you have got.
    jump c,print_the_number
    load s1,"9"
    compare s1,s9
    jump c,print_the_number
  ;If it is a digit, store it into RAM.
  store s9,(pointer)
  add pointer,1
  ;Multiply the number you have got by 10.
  load sf,s0
  call multiply_by_10
  load s0,se
  ;Then, convert the digit from ASCII
  ;into binary.
  sub s9,"0"
  ;And then add it to the number you
  ;have got.
  add s0,s9
  call c,abort ;In case of overflow.
  jump loading_the_number ;Repeat until a
                          ;non-digit is
                          ;loaded.
print_the_number:
;If there are no digits to be printed,
;do not print anything.
sub pointer,0
jump z,START
print_the_decimal:
load s4,pointer
load pointer,0
printing_the_decimal_loop:
  compare pointer,s4
  jump nc, end_of_printing_the_decimal
  fetch s9,(pointer)
  ;Do some basic sanity check: Is the
  ;character you are printing indeed
  ;a decimal digit?
    compare s9,"0"
    call c,abort
    load s1,"9"
    compare s1,s9
    call c,abort
  ;If it is indeed a decimal digit,
  ;print it.
  call UART_TX
  add pointer,1
  jump printing_the_decimal_loop
end_of_printing_the_decimal:
;After you have repeated the decimal
;number, print the string "_(10)=".
load s9,"_"
call UART_TX
load s9,"("
call UART_TX
load s9,"1"
call UART_TX
load s9,"0"
call UART_TX
load s9,")"
call UART_TX
load s9,"="
call UART_TX
;If the number to be printed is
;equal to zero, print 0.
sub s0,0
jump nz,print_the_binary
load s9,"0"
call UART_TX
jump end_of_printing_loop
print_the_binary:
;Make the pointer point to the
;beginning of RAM.
load pointer,0
;Now goes a loop which stores the binary
;representation of the number we have
;got into RAM, but reversed.
beginning_of_converting_to_binary:
  sub s0,0
  jump z,end_of_converting_to_binary
  load s9,"0"
  sr0 s0
  jump nc,store_digit_to_memory
  add s9,1
  store_digit_to_memory:
  store s9,(pointer)
  add pointer,1
  jump beginning_of_converting_to_binary
end_of_converting_to_binary:
;Do some basic sanity check, such as that
;the pointer does not point to zero.
compare pointer,0
call z,abort ;Something went wrong
             ;so end the program.
;Check whether there are more than 8 bits.
compare pointer,9
call nc,abort
;Now goes a loop which will print
;the binary number in RAM, with digits
;in the correct order. The pointer now
;points at a memory location right after
;the binary number (not at the last digit,
;but after it).
beginning_of_printing_loop:
  sub pointer,1
  jump c,end_of_printing_loop
  fetch s9,(pointer)
  ;Do some basic sanity check:
  ;Is the character the pointer points to
  ;indeed a binary digit?
    compare s9,"0"
    jump z,memory_is_fine
    compare s9,"1"
    jump z,memory_is_fine
    call abort ;Something went wrong,
               ;so end the program.
  memory_is_fine:
  ;If everything is fine, print that
  ;digit.
  call UART_TX
  ;Repeat until you have printed all
  ;digits of the binary number
  ;stored in RAM.
  jump beginning_of_printing_loop
end_of_printing_loop:
;After you have printed that binary
;number, print the string "_(2)" and
;a new-line.
load s9,"_"
call UART_TX
load s9,"("
call UART_TX
load s9,"2"
call UART_TX
load s9,")"
call UART_TX
load s9,a ;newline character, 0xa=='\n'.
call UART_TX
;The program runs in an infinite loop...
JUMP START

multiply_by_10:
  load se,sf
  add se,se
  call c,abort
  add se,se
  call c,abort
  add se,sf
  call c,abort
  add se,se
  call c,abort
return

abort:
  load s9,"E"
  call UART_TX
  load s9,"R"
  call UART_TX
  load s9,"R"
  call UART_TX
  load s9,"O"
  call UART_TX
  load s9,"R"
  call UART_TX
  load s9,"!"
  call UART_TX
  load s9,a ;newline
  call UART_TX
  infinite_loop:
  jump infinite_loop
return

;Now follows some boilerplate code
;we use in our Computer Architecture
;classes...
UART_RX:
  INPUT sA, UART_STATUS_PORT
  TEST sA, U_RX_D
  JUMP Z, UART_RX
  INPUT s9, UART_RX_PORT
RETURN

UART_TX:
  INPUT sA, UART_STATUS_PORT
  TEST sA, U_TX_F
  JUMP NZ, UART_TX
  OUTPUT s9, UART_TX_PORT
RETURN

那你觉得呢？你对如何使它变得更好有一些建议吗？

Answer 1

可读性差

您应该插入更多的空行，以便干净地分离逻辑代码块。

你对缩进的使用并不完全一致。如果你做得对，它将大大提高可读性。

而且，考虑到您可以使用一些别名(namereg s3,pointer)，我认为您可以更经常地使用它来提高可读性。

我觉得你的“基本精神健康检查”更像是“精神错乱”检查。你不应该那么怀疑。额外的代码也增加了出错的风险。

当出现问题时，您将进入程序的中止部分，不再返回。看到大多数时候您使用call来完成这个任务，而不是使用更加清晰的jump，这是令人困惑的。

优化

如果使用ASCII字符集，则冒号字符(":")紧跟字符"9“。您可以简化有效数字的验证(不再需要临时寄存器s1)，例如：

compare s9,"0"
jump c,print_the_number
compare s9,":"
jump nc,print_the_number

尽量减少指令的数量，当然也控制传输指令，是很重要的。我已经把越有可能跳回不太可能的跳跃移回流产。

jump nc,loading_the_number ;Repeat until a non-digit is loaded.
jump abort

；如果要打印的数字是；等于零，则打印0。子s0,0跳nz，print_the_binary load s9，"0“调用UART_TX跳转end_of_printing_loop

如果号码是零的话，你不需要特殊情况。它将刮去无数的指令。

你的大部分循环都是有条件的，无条件的跳转，返回到顶部。使用重复-直到循环将更快，并要求只有一个条件跳转指令。

新的守则：

START:
  load s0,0
  namereg s3,pointer
  load pointer,0

loading_the_number:
  call UART_RX
  compare s9,"0"
  jump c,print_the_number
  compare s9,":"
  jump nc,print_the_number
  store s9,(pointer)
  add pointer,1
  load sf,s0
  call multiply_by_10
  load s0,se
  sub s9,"0"
  add s0,s9
  jump nc,loading_the_number ;Repeat until a non-digit is loaded.
  jump abort

print_the_number:
  sub pointer,0 ;If there are no digits to be printed, do not print anything.
  jump z,START

  load s4,pointer
  load pointer,0
printing_the_decimal_loop:
  fetch s9,(pointer)
  call UART_TX
  add pointer,1
  compare pointer,s4
  jump c,printing_the_decimal_loop

  ... some fixed text gets outputted here


  load pointer,0                    ; Works even if s0 == 0
beginning_of_converting_to_binary:  ; This stores in reverse order!
  load s9,"0"
  sr0 s0
  jump nc,store_digit_to_memory
  load s9,"1"
store_digit_to_memory:
  store s9,(pointer)
  add pointer,1
  sub s0,0
  jump nz,beginning_of_converting_to_binary

  ; pointer is certainly not zero
  sub pointer,1
beginning_of_printing_loop:
  fetch s9,(pointer)
  call UART_TX
  sub pointer,1
  jump nc,beginning_of_printing_loop

  ... some fixed text gets outputted here

  JUMP START