任何人都可以运行此程序，或者试着帮助我理解为什么我的计数器没有更新？
我应该从提示中读取文本，找到文本的长度并用冒号输出，然后打印出输入的实际文本。
如果我第一次输入“test”时长度为4，但是当它循环返回以启动时它要求我再次输入，它会输出正确的文本，但除非文本更长，否则计数器不会改变。
所以，如果我输入“I”，它将输出4的长度，因为测试更长并且是4.但是如果我输入7个字母的“Control”，它会将计数器更新为7。
输出：
Enter: Hey
3:Hey
Enter: Test
4:Test
Enter: Control
7:Control
Enter: Hey
7:Hey <---- Length should be 3!
谢谢！
.orig x3000 ; Starting point of the program.
BR start ; Branch to the start routine.
newln: .stringz "\n"
msg1: .stringz "Enter: "
; Prints out the instructions to the user and reads some user input.
start:
lea r0, newln ; Load address of newline into R0.
puts ; Print newline.
lea r0, msg1 ; Load address of message3 into R0.
puts
lea r2, MESSAGE ; Load starting point address of MESSAGE.
and r1, r1, #0 ; Initialize R1 to zero.
input:
getc ; Read in a single character to R0.
out
add r5, r0, #-10 ; Subtract 10 because enter key is 10.
BRz printint ; If zero, branch to checkChar routine.
; Else continue the loop.
str r0, r2, #0 ; Store char in MESSAGE.
add r2, r2, #1 ; Increment index of MESSAGE.
add r1, r1, #1 ; Increment input counter.
BR input ; Unconditional branch to input.
checkChar:
lea r5, inv81 ; Load address of inv68 into R6.
ldr r5, r5, #0 ; Load contents of inv68 into R6 (R6 now holds -68).
add r0, r3, r5 ; Add -68 to the value in R3, to check if it's 'q'.
BRz quit ; If zero, branch to decrypt.
;
;print integer starts here
;
printint:
ld r3,psign
jsr STRLEN
ADD r7, r0, #0 ; get the integer to print
brzp nonneg
ld r3,nsign
not r7,r7
add r7,r7,1
nonneg:
lea r6,buffer ; get the address of o/p area
add r6,r6,#7 ; compute address of end of o/p
ld r5,char0 ; get '0' to add to int digits
loop1:
and r0,r0,#0 ; init quotient for each divide
loop2:
add r7,r7,#-10 ; add -10
brn remdr ; until negative
add r0,r0,#1 ; incr to compute quotient
br loop2 ; repeat
remdr:
add r7,r7,#10 ; add 10 to get remainder
add r7,r7,r5 ; convert to ascii
str r7,r6,0 ; place ascii in o/p
add r7,r0,#0 ; move quot for next divide
brz end ; if done then print
add r6,r6,#-1 ; move to prev o/p position
br loop1 ; repeat
end:
add r6,r6,#-1 ; move to prev o/p position
str r3,r6,0 ; place sign
add r0,r6,#0 ; move address of 1st char
puts ; into r0 and print
output:
ld r5, colon
and r3,r3, 0;
add r0, r3, r5;
out
lea r2, MESSAGE ; Load (starting) address of MESSAGE.
outputLoop:
ldr r0, r2, #0 ; Load contents of address at MESSAGE index into R0.
out ; Print character.
add r2, r2, #1 ; Increment MESSAGE index.
add r1, r1, #-1 ; Decrease counter.
BRp outputLoop ; If positive, loop.
br start
quit:
halt ; Halt execution.
STRLEN:
LEA R2, MESSAGE ;R1 is pointer to characters
AND R0, R0, #0 ;R0 is counter, initially 0
LD R5, char0
LOOP: ADD R2, R2, #1 ;POINT TO NEXT CHARACTER
LDR R4, R2, #0 ;R4 gets character input
BRz FINISH
ADD R0, R0, #1
BR LOOP
FINISH:
ADD R0, R0, #1
ret
MESSAGE: .blkw 99 ; MESSAGE of size 20.
inv48: .fill #-48 ; Constant for converting numbers from ASCII to decimal.
inv81: .fill #-81 ; Constant for the inverse of 'Q'.
buffer: .blkw 8 ; o/p area
null: .fill 0 ; null to end o/p area
char0: .fill x30
colon .fill x3A
nsign .fill x2d
psign .fill x20
.end
答案 0 :(得分：2)
在您的示例结束时，从消息开始的内存中的内容是： Heytrol0000000
在我看来，问题是在STRLEN中我们通过计算来计算字符串的长度，直到我们到达第一个字符为0.“Heytrol”中有7个字符。
但是，当我们存储消息时，我们会计算我们读入的字符数（保留在r1中）。当我们稍后打印字符串时，我们使用r1中的值，因此我们不会打印任何“额外”字符。
要解决这个问题，我要输出r1中的值，该值是在读取字符串作为其长度时计算出来的（完全除去STRLEN代码）或确保当我们在输入循环中读取输入时，我们在进行打印之前在字符串中写入零：
input:
getc ; Read in a single character to R0.
out
add r5, r0, #-10 ; Subtract 10 because enter key is 10.
BRz finishString ; If zero, branch to finishString routine.
; Else continue the loop.
str r0, r2, #0 ; Store char in MESSAGE.
add r2, r2, #1 ; Increment index of MESSAGE.
add r1, r1, #1 ; Increment input counter.
BR input ; Unconditional branch to input.
finishString:
and r0, r0, #0 ; set r0 to zero so we can store it
str r0, r2, #0 ; write zero (from r0) into the end of the string
BR printint ; Now, branch to checkChar routine.