【发布时间】：2017-11-17 06:15:37
【问题描述】：
编写此程序以使用四个标记获取空格分隔的用户输入，如果它们都是数字，则将标记加在一起，并将结果打印到终端。现在它适用于诸如“1 1 1 1”和“123 123 123 123”之类的数字，但是当我尝试添加 4 个七位数长的数字（7 位，因为这是极端情况）时，它只会添加最多 23068。这听起来像是我的一个标签或其他东西的尺寸问题，但我不确定。
代码如下：
*
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
看起来它与我将 ascii 表示形式转换为实际数字有关。我将.l 添加到结果标签。标签足够大，可以存储字符，但我没有向它移动足够大的块。
当我输入“9999999 9999999 9999999 9999999”并设置断点观看时，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其转换回来时会出现问题。
我不希望任何人对此有解决方案，但也许有人有。
编辑2： 此代码已在 Sep Rowland 的指导下进行了修订（非常感谢）。我想我得到了他所涵盖的所有内容，并且已将修改后的代码作为答案提交。
【问题讨论】：