How do i do a bitshift right in binary? - binary

Question

Hopefully this is a simple question but I cannot for the life of me figure out how to do a bitshift in binary. This is being done in the LC3 environemnt. I just need to know how to arithmetical divide by two and shift to the right. I know going left is simple by just adding the binary value to itself, but I have tried the opposite for bitshift right(subtracting from itself, NOTing and then subtracting etc etc.) Would be much appreciated.
Or if you have a better way to move x00A0 to x000A that would also be fantastic. Thanks!

Answer 1

This is an older post, but I ran into the same issue so I figured I would post what I've found.
When you have to do a bit-shift to the right you're normally halving the the binary number (divide by 2) but that can be a challenge in the LC-3. This is the code I wrote to preform a bit-shift to the right.
; Bit shift to the right
.ORIG x3000
MAIN
LD R3, VALUE
AND R5, R5, #0 ; Reseting our bit counter
B_RIGHT_LOOP
ADD R3, R3, #-2 ; Subtract 2 from the value stored in R3
BRn BR_END ; Exit the loop as soon as the number in R3 has gone negative
ADD R5, R5, #1 ; Add 1 to the bit counter
BR B_RIGHT_LOOP ; Start the loop over again
BR_END
ST R5, ANSWER ; Store the shifted value into the ANSWER variable
HALT ; Stop the program
; Variables
VALUE .FILL x3BBC ; Value is the number we want to do a bit-shift to the right
ANSWER .FILL x0000
.END
Keep in mind that with this code the left most bit B[0] is lost. Also this code doesn't work if the number we are trying to shift to the right is negative. So if bit [15] is set this code won't work.
Example:
VALUE .FILL x8000 ; binary value = 1000 0000 0000 0000
; and values higher than x8000
; won't work because their 15th
; bit is set
This should at least get you going on the right track.

Answer 2

.ORIG x3000
BR main
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; UL7AAjr
; shift right register R0
; used rigisters R1, R2, R3, R4, R5
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shift_right
AND R4, R4, #0 ; R4 - counter = 15 times
ADD R4, R4, #15
AND R1, R1, #0 ; R1 - temp result
LEA R2, _sr_masks ; R2 - masks pointer
_sr_loop
LDR R3, R2, #0 ; load mask into R3
AND R5, R0, R3 ; check bit in R0
BRZ _sr_zero ; go sr_zero if bit is zero
LDR R3, R2, #1 ; R3 next mask index
ADD R1, R1, R3 ; add mask to temp result
_sr_zero
ADD R2, R2, #1 ; next mask address
ADD R4, R4, #-1 ; all bits done?
BRNP _sr_loop
AND R0, R0, #0 ; R0 = R1
ADD R0, R0, R1
RET
_sr_masks
.FILL x8000
.FILL x4000
.FILL x2000
.FILL x1000
.FILL x0800
.FILL x0400
.FILL x0200
.FILL x0100
.FILL x0080
.FILL x0040
.FILL x0020
.FILL x0010
.FILL x0008
.FILL x0004
.FILL x0002
.FILL x0001
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main
LD R0, data
JSR shift_right
HALT
data .FILL xFFFF
.END

Answer 3

; right shift R0 1-bit with sign-extention
; Algorithm: build bit form msb one by one
.ORIG x3000
AND R1, R1, #0 ; r1 = 0
ADD R2, R1, #14 ; r2 = 14
ADD R0, R0, #0 ; r0 = r0
BRzp LOOP
ADD R1, R1, #-1 ; r1 = xffff
LOOP ADD R1, R1, R1 ; r1 << 1
ADD R0, R0, R0 ; r0 << 1
BRzp MSB0
ADD R1, R1, #1 ; r1++
MSB0 ADD R2, R2, #-1 ; cnt--
BRp LOOP
ADD R0, R1, #0 ; r0 = r1
HALT
.END

Answer 4

; right shift R0 1-bit with sign-extention
; Algorithm: left-rotate 14 times with proper sign
.ORIG x3000
LD R1, CNT
ADD R2, R0, #0
LOOP ADD R0, R0, R0 ; r0 << 1
BRzp NEXTBIT
ADD R0, R0, #1
NEXTBIT ADD R1, R1, #-1
BRp LOOP
LD R3, MASK
AND R0, R0, R3
ADD R2, R2, #0
BRzp DONE
NOT R3, R3
ADD R0, R0, R3
DONE HALT
MASK .FILL x3FFF
CNT .FILL 14
.END

Answer 5

; right shift R0 1-bit with sign-extention
; Algorithm: look-uo table and auto-stop
.ORIG x3000
AND R1, R1, #0 ; r1 = 0
LEA R2, TABLE ; r2 = table[]
AND R0, R0, #-2
LOOP BRzp MSB0
LDR R3, R2, #0 ; r3 = table[r2]
ADD R1, R1, R3 ; r1 += r3
MSB0 ADD R2, R2, #1 ; r2++
ADD R0, R0, R0 ; r0 << 1
BRnp LOOP
ADD R0, R1, #0 ; r0 = r1
HALT
TABLE
.FILL xC000
.FILL x2000
.FILL x1000
.FILL x0800
.FILL x0400
.FILL x0200
.FILL x0100
.FILL x0080
.FILL x0040
.FILL x0020
.FILL x0010
.FILL x0008
.FILL x0004
.FILL x0002
.FILL x0001
.END