0
任何人都可以運行此程序，或嘗試幫助我理解爲什麼我的計數器不更新嗎？LC-3如何重置櫃檯？
我應該從提示中讀取文本，並在打印出輸入的實際文本之前查找文本的長度並用冒號輸出。
如果我第一次輸入「test」的長度是4，但是當它循環回來開始時它會要求我再次輸入，它會輸出正確的文本，但是計數器不會改變，除非文本更長。因此，如果我輸入「I」，它將輸出長度爲4，因爲測試更長，並且是4.但是如果我輸入7個字母的「Control」，它會將計數器更新爲7.
OUTPUT：
Enter: Hey
3:Hey
Enter: Test
4:Test
Enter: Control
7:Control
Enter: Hey
7:Hey <---- Length should be 3!
謝謝！
.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