ARM汇编代写代做代考 CS1021 Tutorial 5

CS1021 Tutorial 5
Logic and Shift Instructions
Q1 Calculate, in hexadecimal, the results of the following 8 bit expressions
(i) 0x96 & 0xF0 0x90
(ii) 0x96 | 0x0F 0x9F
(iii) 0xAA ^ 0xF0 0x5A
(iv) ~0xA5 0x5A
(v) 0x96 >> 2 0x25
and 32-bit expressions
(vi) 0x0123 << 2 0x048c (vii) 0x12345678 >> 24 0x12
(viii) 0x12345678 >> 16 0x1234
(ix) (0x12345678 >> 16) & 0xFF 0x34
(x) (0x12345678 & ~0xFF00) | 0x4400 0x12344478
LDR R1, =0xFFFFFF0F AND R0, R0, R1
or
BIC R0, R0, #0xF0
(ii) clear the first and last bytes of R0
LDR R1, =0xFF0000FF BIC R0, R0, R1
(iii) invert the most significant bit of R0
EOR R0, R0, #0x80000000
(iv) setbits2to4ofR0
ORR R0, R0, #0x1C
ARM Assembly Language instructions to perform the following operations (assume the (i) clear bits 4 to 7 of R0
Q2 Write
LSB of a register is bit 0).
; mask ; and
; bit clear
; mask
; bit clear
; invert MS bit
; or
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CS1021 Tutorial 5  2018 jones@scss.tcd.ie

(v) swap the most and least significant bytes of R0
AND R1, R0, #0xFF
AND R2, R0, #0xFF000000 BIC R0, R0, #0xFF
BIC R0, R0, #0xFF000000 ORR R0, R0, R1, LSL #24 ORR R0, R0, R2, LSR #24
; extract LS byte
; extract MS byte
; clear LS byte
; clear MS byte
; insert extracted and shifted LS byte into MS byte ; insert extracted and shifted MS byte into LS byte
(vi) replace bits 8 to 15 in R0 with the value 0x44
BIC R0, R0, #0xFF00 ORR R0, R0, #0x4400
(vii) R0 = R1*10
MOV R0, R1, LSL #3 ADD R0, R0, R1, LSL #1
(viii) R0 = R1*100
MOV R0, R1, LSL #6 ADD R0, R0, R1, LSL #5 ADD R0, R0, R1, LSL #2
(ix) R0 = R1/256
MOV R0, R1, LSR #8
; clear bits
; insert 0x44 in correct position
;R0=R1*8 ;R0=R1*8+R1*2
;R0=R1*64 ;R0=R1*64+R1*32 ;R0=R1*64+R1*32+R1*4
;R0=R0/256
(x) R0 = R1 % 256 (mod operator – remainder on division)
AND R0, R1, #0xFF ; R0 = R1 % 256
Q3 Write an ARM assembly language program to calculate, in R0, the (sum % 256) of the 4 bytes in R1. For example, if R1 = 0x12345678, R0 = (0x12 + 0x34 + 0x56 + 0x78) % 256 = 0x14
AND R0, R1, #0xFF
ADD R0, R0, R1, LSR #8 ADD R0, R0, R1, LSR #16 ADD R0, R0, R1, LSR #24 AND R0, R0, #0xFF
; R0 = LS byte
; add next byte (ignore carry/over flow from LS byte) ; add next byte (ignore carry/over flow from LS byte) ; add next byte (ignore carry/over flow from LS byte) ; R0 = R0 % 256 (mod operator)
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CS1021 Tutorial 5  2018 jones@scss.tcd.ie

CS1021 Tutorial 5  2018 jones@scss.tcd.ie Q4 Write and ARM assembly language program to calculate, in R0, the number of one bits in R1.
For example, if R1 = 0x12345678, R0 = 13.
LDR R1, =0x12345678 MOV R0, #0
; R1 = 0x12345678 (13 bits set) ; R0 = 0
; if R1 == 0?
; finished
; shift LS bit into CARRY flag ; add CARRY to R0
; next bit
L CMP
BEQ L0
MOVS R1, R1, LSR #1 ADC R0, R0, #0
B L
L0B L0
R1, #0
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