这实际上是PicoBlaze Simulator中的“十进制到二进制”示例：
;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
compare s9, "9" + 1
;Suggestion by a StackExchange user.
jump nc, 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
compare s9, "9" + 1
call nc, 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
我已经在真实硬件上尝试过（通过组装和下载十六进制文件），但它在那里不起作用。它在那里什么都不做，它不响应我在 UART 中键入的任何内容，也不打印任何内容。这里发生了什么？