摩托罗拉68000存储组件

Question

编写此程序的目的是接受带有四个令牌的空间分隔用户输入，如果这些令牌都是数字，则将它们相加在一起，并将结果打印到终端。现在，它适用于诸如“1111 1”和"123 123 123“这样的数字，但是当我试图添加4个7位长的数字(因为这是极端情况)时，它的加起来只有23068。这听起来像是我的一个标签或什么的大小问题，但我不确定。

以下是代码：

*
        ORG     $0
        DC.L    $3000           * Stack pointer value after a reset
        DC.L    start           * Program counter value after a reset
        ORG     $3000           * Start at location 3000 Hex
*
*----------------------------------------------------------------------
*
#minclude /home/cs/faculty/cs237/bsvc/macros/iomacs.s
#minclude /home/cs/faculty/cs237/bsvc/macros/evtmacs.s
*
*----------------------------------------------------------------------
*
* Register use
*
*----------------------------------------------------------------------
*
start:  initIO                  * Initialize (required for I/O)
    setEVT          * Error handling routines
*   initF           * For floating point macros only    

                * Your code goes HERE
*Output info:               
    lineout header      *Display header info
    lineout prompt      *Display prompt
    linein  buffer      *Read input to buffer

    lea buffer,A1       *
    adda.l  D0,A1       *Add null terminator
    clr.b   (A1)        *

    lea buffer,A1       *Reload the beginning of buffer address

    move.l  #1,D1       *D1 is input counter and starts at 1

    clr.l   D2          *
    clr.l   D3          *Prepping registers for calculations
    move.l  #0,result   *

    move.l  A1,A2       *Duplicating address to use for strlen

top:    
    tst.b   (A1)        *Check for end of string
    BEQ rest            *If end, go to rest

    cmpi.b  #47,(A1)    *Check current byte against low end of ascii numbers
    BGT toprange        *This means byte *might* be an ascii number

    cmpi.b  #32,(A1)    *Byte is below range for numbers. Is it a space?
    BNE notno           *If this triggers, it's not a space and not a number. Exit.

    cmpi.b  #32,1(A1)   *Is the character after this a space? If yes, loop to top.
    BNE addit           *If not, it's either another valid byte or null terminator. 

    adda.l  #1,A1       *Increment token counter and loop to top.
    BRA     top 

toprange:   
    cmpi.b  #57,(A1)    *Is byte value higher than ascii numbers range?
    BGT notno           *If yes, it's not an ascii number. Exit. 

    cmpi.b  #32,1(A1)   *Is the byte after this a space?
    BEQ endoftoken      *If yes, that means this is the end of the token.

    tst.b   1(A1)       *If not, is this byte a null terminator?
    BEQ endoftoken      *If yes, this is the last token. Add it.

    adda.l  #1,A1       *Else increment the address pointer and loop.
    BRA top

endoftoken: 
    adda.l  #1,A1       *Increment pointer
    move.l  A1,D2       *
    sub.l   A2,D2       *Find length of token

    cvta2   (A2),D2     *Convert current token segment to number
    add.l   D0,result   *Add converted number to result address.
    BRA     top         *Loop to top.

addit:          
    tst.b   1(A1)       *Test for null
    BEQ endoftoken      *If null, go endof token to add it to running total

    addi.l  #1,D1       *If next byte isn't null, there might be more tokens. Incr & cont.
    adda.l  #1,A1   

    move.l  A1,A2       *Shift token starting point pointer forward
    BRA top 

rest:
    cmpi.l  #4,D1       *Check to make sure we have 4 tokens
    BNE incnums         *If not, exit on error

    move.l  result,D0   *Convert numbers back to text representations
    ext.l   D0
    cvt2a   result,#8
    stripp  result,#8
    lea result,A0
    adda.l  D0,A0
    clr.b   (A0)

    lea sum,A1          *Point to first bit of text for strcat
    lea output,A2       *Point to destination during copying
strcat1:
    tst.b   (A1)        *Null?
    BEQ strcat2         *Go to next segment of code
    move.b  (A1)+,(A2)+ *If not null, copy from A1 to A2. Post increment
    BRA strcat1

strcat2:
    move.b  #32,(A2)+   *Append space. Post increment
    lea result,A1       *Point to calculated result

strcat3:    
    tst.b   (A1)        *Is this byte null?
    BEQ printr          *If yes, go to print response.  
    move.b  (A1)+,(A2)+ *If not, copy byte to output string. 
    BRA strcat3

printr: 
    move.b  #46,(A2)+   *Append period to output string. 
    clr.b   (A2)        *Null terminate the string. 
    lineout output      *Print built string to terminal.
    BRA end 

incnums:
    lineout incno       *If here, there were not the correct number of tokens.
    BRA end


notno:
    cmpi.b  #1,D1       *This checks the token counter to determine which token was not a #
    BNE ch2
    lineout bn1
    BRA end

ch2:    
    cmpi.b  #2,D1
    BNE ch3
    lineout bn2
    BRA     end

ch3:    
    cmpi.b  #3,D1
    BNE ch4
    lineout bn3
    BRA end

ch4:    
    lineout bn4


end:



*Output result




        break                   * Terminate execution
*
*----------------------------------------------------------------------
*       Storage declarations

prompt: dc.b    'Enter the four space separated numbers:',0
sum:    dc.b    'The sum is',0
incno:  dc.b    'There are not four inputs.',0
buffer: ds.b    80
result: ds.l    3
output: ds.l    3
bn1:    dc.b    'The #1 input is not a number',0
bn2:    dc.b    'The #2 input is not a number',0
bn3:    dc.b    'The #3 input is not a number',0
bn4:    dc.b    'The #4 input is not a number',0
        end

编辑1

它看起来与当我将ascii表示转换成一个实际数字时有关。我把.我加到结果标签上。标签足够大，可以存储字符，但我没有移动足够大的块到它。

当我输入"9999999 9999999 9999999“并设置一个断点观看它时，内存将正确显示26259FC的十六进制值，所以当我使用提供的宏将它转换回时会出现问题。

我不指望有人能找到解决办法，但也许有人有办法。

Edit2:这段代码是在罗兰的指导下修改的(非常感谢)。我想我得到了他所涵盖的一切，修改后的代码已经作为一个答案提交了。

Answer 1

错误(您已经发现了)是ext.l D0。考虑到这个指令符号-扩展了D0中的低单词，结果被取消并不令人惊讶。

我很少有机会钻研一些好的68K代码，所以这里我有一些评论可以改进你的程序。

D1 adda.l #1，A1

如果使用addq指令，您可以编写从#1到#8添加的这些小的添加：

addq.l  #1,D1
addq.l  #1,A1

要在某个输入不是数字时显示错误消息，如果将D1 (1-4)中的数字转换为字符("1"-"4")并将其写入单个错误消息，则可以编写得简单得多：

lea     bn,A1
addi.b  #48,D1   *From number to character
move.b  D1,5(A1) *Replaces the dot in the error message
lineout bn

...

bn:     dc.b    'The #. input is not a number',0

产出: ds.l 3

此输出缓冲区不足以满足您所做的工作！

您只有12个字节，但是首先要复制10个字符的长和消息，添加一个空格，添加几个字符的长结果，添加一个句点，然后添加一个零。很明显是缓冲区溢出。

现在，您可以使这个缓冲区更长，或者智能地停止复制周围的所有内容，只需将结果缓冲区放置在sum消息附近(附加一个空格，并且没有终止零)。然后一次显示合并的和结果文本。一个简单得多的解决方案。

move.l  result,D0   *Convert numbers back to text representations
cvt2a   result,#8
stripp  result,#8
lea     result,A2
adda.l  D0,A2
move.b  #46,(A2)+   *Append period to output string. 
clr.b   (A2)        *Null terminate the string. 
lineout sum         *Print built string to terminal.
BRA end 

...
sum:    dc.b    'The sum is '
result: ds.l    3

cmpi.b #32,1(A1) *后面的字符是空格吗？如果是的话，转到顶部。如果不是，它要么是另一个有效字节，要么是空终止符。adda.l #1，A1 *增量令牌计数器并循环到顶部。胸罩顶部：

在这里，您可以在代码中创建一个快捷方式，从而加快程序的速度。

不需要BRA一直到顶端，在那里你将做3个不必要的测试。

SkipWhitespace:
    cmpi.b  #32,1(A1)
    BNE     addit
    addq.l  #1,A1
    BRA     SkipWhitespace
toprange:   

move.l A1,A2 \*Duplicating address to use for strlen top: ... move.l A1,A2 \*Shift token starting point pointer forward BRA top rest:

始终尝试而不是编写冗余指令。

topEx:
    move.l  A1,A2       *Duplicating address to use for strlen
top:    

    ...

    BRA     topEx
rest:

Answer 2

初始问题：的ext.l指令。这是由九月罗兰在提供的其他答案之一解释。

其他版本:在的指导下，我在几个方面优化了代码。

+1用于内存查看器和断点。

* Problem statement: Read input and determine if 4 numbers are provided. Add numbers.
* Input: ### ### ### ###
* Output: "The sum is ###"
* Error conditions tested: Correct number of data provided. Number vs Char
*  Also handles leading white spaces/multiple spaces between tokens
* Included files: None
* Method and/or pseudocode: 
* References: 
*----------------------------------------------------------------------
*
        ORG     $0
        DC.L    $3000           * Stack pointer value after a reset
        DC.L    start           * Program counter value after a reset
        ORG     $3000           * Start at location 3000 Hex
*
*----------------------------------------------------------------------
*
#minclude /home/cs/faculty/cs237/bsvc/macros/iomacs.s
#minclude /home/cs/faculty/cs237/bsvc/macros/evtmacs.s
*
*----------------------------------------------------------------------
*
* Register use
*
*----------------------------------------------------------------------
*
start:  initIO                  * Initialize (required for I/O)
    setEVT          * Error handling routines
*   initF           * For floating point macros only    


    lineout header      *Display header info
    lineout prompt      *Display prompt
    linein  buffer      *Read input to buffer
    lea buffer,A1       *
    adda.l  D0,A1       *Add null terminator
    clr.b   (A1)        *   
    lea buffer,A1       *Reload the beginning of buffer address
    clr.l   D1          *D1 is input counter and starts at 0
    clr.l   D2          *D2 used as workspace
    move.l  #0,result   *Clearing garbage out of memory address
    move.l  A1,A2       *A2 used for strlen

top:    
    tst.b   (A1)        *Check for end of string
    BEQ rest            *If end, go to rest

    cmpi.b  #47,(A1)    *Check current byte against low end of ascii numbers
    BGT checktoprange   *This means byte *might* be an ascii number

    cmpi.b  #32,(A1)
    BNE notnumber

whitespace:             *This will eat whitespace anywhere in buffer
    addq.l  #1,A1       *If we are here, we know current location is space
    cmpi.b  #32,(A1)    *So increment pointer and check for additional spaces
    BEQ whitespace
    move.l  A1,A2       *If we are here, we encountered a token
    BRA top             *Shift our pointer for token start location


checktoprange:  
    cmpi.b  #57,(A1)    *Is byte value higher than ascii numbers range?
    BGT notnumber       *If yes, it's not an ascii number. Exit. 

    cmpi.b  #32,1(A1)   *Is the byte after this a space?
    BEQ endoftoken      *If yes, that means this is the end of the token.

    tst.b   1(A1)       *If not, is this byte a null terminator?
    BEQ endoftoken      *If yes, this is the last token. Add it.

    addq.l  #1,A1       *Else increment the address pointer and loop.
    BRA top

endoftoken: 
    addq.l  #1,A1       *Increment pointer
    move.l  A1,D2       *
    sub.l   A2,D2       *Find length of token

    cvta2   (A2),D2     *Convert current token segment to number

    add.l   D0,result   *Add converted number to result address.
    addq.l  #1,D1       *Increment token counter
    BRA     top         

rest:
    cmpi.l  #4,D1       *Check to make sure we have 4 tokens
    BNE incnums         *If not, exit on error

    move.l  result,D0   *Convert numbers back to text representations
    cvt2a   result,#8
    stripp  result,#8
    lea result,A0
    adda.l  D0,A0
    move.b  #46,(A0)+
    clr.b   (A0)
    lineout sum
    BRA end


incnums:
    lineout incno       *If here, there were not the correct number of tokens.
    BRA end


notnumber:
    lea bn,A1
    addi.b  #49,D1       *From number to character
    move.b  D1,5(A1)     *Replaces the dot in the error message
    lineout bn

end:    

        break                   * Terminate execution
*
*----------------------------------------------------------------------
*       Storage declarations

header: dc.b    'This is a header',0
prompt: dc.b    'Enter the four space separated numbers:',0
incno:  dc.b    'There are not four inputs.',0
buffer: ds.b    80
bn:     dc.b    'The #. input is not a number',0
sum:    dc.b    'The sum is '
result: ds.l    1


        end