我正在开发一个程序，它接受一个数字的输入，并在基数为10,2,4和8中输出。从输入中获取数字的逻辑很简单，以及这样的转换给定基数的数字。我所坚持的部分需要将转换后的数字存储到缓冲区中，最后使用TRAP x22
这是我的想法：
将DIGITS的第一个值加载到另一个寄存器R4
然后执行分部
进行正确的算法检查（如果商为零，则不再进行除法，退出循环）
将R5放入R0
我尝试过我在下面的代码中所描述的内容，但是我对输出感到喋喋不休（并且奇怪的是整个DIGITS字符串）我在这里检查的是什么？显然存储操作不起作用，但我不明白为什么不存在。请让我知道我可以在哪里开始修复此错误。
;Subroutine PRINT*************************************************************
;Displays an unsigned integer in any base up to 16, e.g. binary, octal, decimal
;Parameters - R0: the integer - R1: the base
PRINT
ST R0, PR0
ST R1, PR1
ST R2, PR2
ST R3, PR3
ST R4, PR4
ST R5, PR5
ST R6, PR6
ST R7, PR7
LEA R5, FINNUM ;Prepare pointer to storage
LEA R6, BUFFER
STR R6, R5, #0
ADD R5, R5, #-1
LD R4, DIGITS ;Prepare pointer to the digits
SDIV
JSR DIVIDE
ADD R6, R4, R3 ;Add the remainder with digits to get number
STR R6, R5, #0 ;Store number into buffer
ADD R5, R5, #-1 ;Decrement
ADD R0, R2, #0 ;Pass quotient into R0
BRP SDIV ;If number is at least 1, then divide again
ADD R0, R5, #0 ;If we got here, then we should have the converted number
TRAP x22 ;Last but not least we display
LD R7, PR7
RET
;Data
DIGITS .STRINGZ "0123456789ABCDEF" ;Digits
FINNUM .BLKW 18 ;Output Buffer
BUFFER .FILL x0000 ;Null
PR0 .BLKW 1
PR1 .BLKW 1
PR2 .BLKW 1
PR3 .BLKW 1
PR4 .BLKW 1
PR5 .BLKW 1
PR6 .BLKW 1
PR7 .BLKW 1
如果您需要，请输入以下代码：
.ORIG x3000
LEA R0, MPROMPT
TRAP x22
LOOP
JSR GETDEC ;Input an unsigned (decimal) integer
ADD R0, R0, #0 ;Exit if the input integer is 0
BRZ EXIT
JSR NEWLN ;Move cursor to the start of a new line
AND R1, R1, #0 ;R1 = 10 (output base 10 DECIMAL)
ADD R1, R1, #10
JSR PRINT ;Print the integer in decimal
JSR NEWLN ;Move cursor to the start of a new line
AND R1, R1, #0 ;R1 = 2 (output base 2 BINARY)
ADD R1, R1, #2
JSR PRINT ;Print the integer in binary
JSR NEWLN ;Move cursor to the start of a new line
AND R1, R1, #0 ;R1 = 8 (output base 8 OCTAL)
ADD R1, R1, #8
JSR PRINT ;Print the integer in octal
JSR NEWLN ;Move cursor to the start of a new line
AND R1, R1, #0 ;R1 = 16 (output base 16 HEXADECIMAL)
ADD R1, R1, #15
ADD R1, R1, #1
JSR PRINT ;Print the integer in hexadecimal
JSR NEWLN ;Move cursor to the start of a new line
BRNZP LOOP
EXIT ;Loop exit point
TRAP x25 ;HALT program execution
MPROMPT .STRINGZ "Enter a sequence of unsigned integer values\nEnter 0 To Quit\n"
;Subroutine NEWLN*************************************************************
;Advances the console cursor to the start of a new line
NEWLN
ST R7, NEW7 ;Save working registers
ST R0, NEW0
LD R0, NL ;Output a newline character
TRAP x21
LD R0, NEW0 ;Restore working registers
LD R7, NEW7
RET ;Return
;Data
NL .FILL x000A ;Newline character
NEW0 .BLKW 1 ;Save area - R0
NEW7 .BLKW 1 ;Save area - R7
;Subroutine GETDEC************************************************************
;Inputs an unsigned integer typed at the console in decimal format
;The input value is returned in R0
GETDEC
;Save Values of ALL Registers that don't need to change
ST R1, GET1
ST R2, GET2
ST R3, GET3
ST R4, GET4
ST R5, GET5
ST R6, GET6
ST R7, GET7 ;Save R7 to be able to return to main
AND R3, R3, #0 ;Prepare a value
LEA R0, GPROMPT
TRAP x22
GETDKEY
TRAP x20 ;To input a keystroke
ADD R2, R0, #-10 ;Subtract new keystroke by 10 (new line, or enter)
BRZ GFINIS ;If R2 is 0 we just pressed enter, so we break out
TRAP x21 ;Echo keystroke
ADD R2, R0, #0 ;Get the number back
AND R2, R2, #15 ;We mask to only get the first four bits
ADD R3, R3, R3 ;1st addition: x + x = 2x
ADD R4, R3, #0 ;Store the 2x
ADD R3, R3, R3 ;2nd addition: 2x + 2x = 4x
ADD R3, R3, R3 ;3rd addition: 4x + 4x = 8x
ADD R3, R3, R4 ;4th addition: 8x + 2x = 10x
ADD R3, R3, R2 ;10x + new number
BRNZP GETDKEY ;Go back to get next number
GFINIS
ADD R0, R3, #0 ;Put number into R0
;Restore the values of registers that don't need change
LD R1, GET7
LD R2, GET7
LD R3, GET7
LD R4, GET7
LD R5, GET7
LD R6, GET7
LD R7, GET7 ;Get back the value of R7
RET
;Data
GPROMPT .STRINGZ "Enter an unsigned integer> " ;Input Prompt
GET1 .BLKW 1
GET2 .BLKW 1
GET3 .BLKW 1
GET4 .BLKW 1
GET5 .BLKW 1
GET6 .BLKW 1
GET7 .BLKW 1
;Subroutine PRINT*************************************************************
;Displays an unsigned integer in any base up to 16, e.g. binary, octal, decimal
;Parameters - R0: the integer - R1: the base
PRINT
ST R0, PR0
ST R1, PR1
ST R2, PR2
ST R3, PR3
ST R4, PR4
ST R5, PR5
ST R6, PR6
ST R7, PR7
LEA R5, FINNUM ;Prepare pointer to storage
LEA R6, BUFFER
STR R6, R5, #0
ADD R5, R5, #-1
LD R4, DIGITS ;Prepare pointer to the digits
SDIV
JSR DIVIDE
ADD R6, R4, R3 ;Add the remainder with digits to get number
STR R6, R5, #0 ;Store number into buffer
ADD R5, R5, #-1 ;Decrement
ADD R0, R2, #0 ;Pass quotient into R0
BRP SDIV ;If number is at least 1, then divide again
ADD R0, R5, #0 ;If we got here, then we should have the converted number
TRAP x22 ;Last but not least we display
LD R7, PR7
RET
;Data
DIGITS .STRINGZ "0123456789ABCDEF" ;Digits
FINNUM .BLKW 18 ;Output Buffer
BUFFER .FILL x0000 ;Null
PR0 .BLKW 1
PR1 .BLKW 1
PR2 .BLKW 1
PR3 .BLKW 1
PR4 .BLKW 1
PR5 .BLKW 1
PR6 .BLKW 1
PR7 .BLKW 1
;Subroutine DIVIDE************************************************************
;Perform unsigned integer division to obtain Quotient and Remainder
;Parameters (IN) : R0 - Numerator, R1 - Divisor
;Parameters (OUT) : R2 - Quotient, R3 - Remainder
DIVIDE
ST R5, DIV7
ST R6, DIV7
ST R7, DIV7
AND R2, R2, #0
ADD R3, R0, #0
ADD R5, R1, #0
NOT R5, R5
ADD R5, R5, #1 ;2's complement to R5 to be able to substract
DSUBT
ADD R6, R3, R5 ;R6 = R3 - R5
BRN ENDDIV
ADD R2, R2, #1 ;If here, R0 was bigger than R1, we can add 1
ADD R3, R3, R5 ;Subtract again to avoid losing the remainder
BRNZP DSUBT
ENDDIV
LD R5, DIV7
LD R6, DIV7
LD R7, DIV7
RET ;Return
;Data
DIV7 .BLKW 1
.END
这是我可以用Java完成的事情，但是我对这种语言太新了，坦率地说，我仍然试图理解在函数执行之前存储寄存器的问题并在最后加载它们。我之所以提到这一点，是因为它导致了输入（即返回主执行）和分区的问题。请让我知道，并提前感谢您的帮助
编辑：将功能更改为从BUFFER地址开始，添加18以获取最后一个可用存储空间，然后将存储过程修改为先减少然后存储。此外，所有寄存器都在开头存储，然后在最后重新加载，以防万一：
;Subroutine PRINT*************************************************************
;Displays an unsigned integer in any base up to 16, e.g. binary, octal, decimal
;Parameters - R0: the integer - R1: the base
PRINT
ST R0, PR0
ST R1, PR1
ST R2, PR2
ST R3, PR3
ST R4, PR4
ST R5, PR5
ST R6, PR6
ST R7, PR7
LEA R6, FINNUM
ADD R6, R6, #10 ;Add to get to the end of the buffer
ADD R6, R6, #8 ;Add to get to the end of the buffer
AND R5, R5, #0 ;To make sure R5 is cleared
LD R4, DIGITS ;Prepare pointer to the digits
SDIV
JSR DIVIDE
ADD R5, R4, R3 ;Add the remainder with digits to get number
ADD R6, R6, #-1 ;Decrement to get to the next one
STR R5, R6, #0 ;Store number into buffer
ADD R0, R2, #0 ;Pass quotient into R0
BRP SDIV ;If number is at least 1, then divide again
ADD R0, R6, #0 ;If we got here, then we should have the converted number
TRAP x22 ;Last but not least we display
LD R0, PR0
LD R1, PR1
LD R2, PR2
LD R3, PR3
LD R4, PR4
LD R5, PR5
LD R6, PR6
LD R7, PR7
RET
;Data
DIGITS .STRINGZ "0123456789ABCDEF" ;Digits
FINNUM .BLKW 18 ;Output Buffer
BUFFER .FILL x0000 ;Null
PR0 .BLKW 1
PR1 .BLKW 1
PR2 .BLKW 1
PR3 .BLKW 1
PR4 .BLKW 1
PR5 .BLKW 1
PR6 .BLKW 1
PR7 .BLKW 1
LC3在ADD R0, R2, #0之后立即停在STR。该程序是否存储在适当的位置？