CS1021 Tutorial 7
Advanced LDR and STR Instructions
Q1 If R1 = 0x1000 and R4 = 8, what memory location (in hexadecimal) is loaded into R0 and what is the value of R1 (in hexadecimal) after each of the following instructions has been executed.
(i) LDR
(ii) LDR
(iii) LDR
(iv) LDR
(v) LDR
(vi) LDR
(vii) LDR
(viii) LDR
(ix) LDR
R0, [R1, #8]
R0, [R1], #-8
R0, [R1, #12]!
R0, [R1, R4]
R0, [R1], R4
R0, [R1, R4]!
R0, [R1, R4, LSL #3] R0, [R1], R4, LSR #1 R0, [R1, R4, LSL #2]!
; R0=MEM[0x1008], R1=0x1000 ; R0=MEM[0x1000], R1=0x0FF8 ; R1=0x100C, R0=MEM[0x100C] ; R0=MEM[0x1008], R1=0x1000 ; R0=MEM[0x1000], R1=0x1008 ; R1=0x1008, R0=MEM[0x1008] ; R0=MEM[0x1040], R1=0x1000 ; R0=MEM[0x1000], R1=0x1004 ; R1=0x1020, R0=MEM[0x1020]
Q2 Given an array b at memory address 0x40001000 containing 64 32-bit integers b[0] to b[63], write ARM assembly language instructions for the following pseudo code statements. Assume i and j are 32-bit unsigned integers stored in memory locations 0x40000000 and 0x40000004 respectively.
(i) R0 = b[7]
LDR R1, =0x40001000
LDR R0, [R1, #7*4]
(ii) R0 = b[i]
LDR R1, =0x40001000 LDR R2, =0x40000000 LDR R2, [R2]
LDR R0, [R1, R2, LSL #2]
(iii) i = b[i] + b[j]
LDR R1, =0x40001000 LDR R2, =0x40000000 LDR R3, [R2], #4
LDR R0, [R1, R3, LSL #2] LDR R3, [R2], #-4
LDR R3, [R1, R3, LSL #2] ADD R0, R0, R3
STR R0, [R2]
;R1->b
; R0 = b[7] (MEM[b + 28])
;R1->b
;R2->i
;R2=i
; R0 = b[i] (MEM[b + i*4])
; R1 -> b
; R2 -> i
; R3 = i AND R2 -> j
; R0 = b[i] (MEM[b + i*4]) ; R3 = j AND R2 -> i
; R3 = b[j] (MEM[b + j*4]) ; R0 = b[i] + b[j]
; i = R0 (b[i] + b[j])
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CS1021 Tutorial 7 2018 jones@scss.tcd.ie
(iv) b[i] = b[10] + b[j]
LDR R1, =0x40001000 LDR R0, [R1, #10*4] LDR R2, =0x40000004 LDR R3, [R2], #-4
LDR R3, [R1, R3, LSL #2] ADD R0, R0, R3
LDR R3, [R2]
STR R0, [R1, R3, LSL #2]
; R1 -> b
; R0 = b[10]
; R2 -> j
; R3 = j AND R2 -> i
; R3 = b[j] (MEM[b + j*4])
; R0 = b[10] + b[j]
; R3 = j
; b[i] (MEM[b + i*4]) = R0 (b[10] + b[j])
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CS1021 Tutorial 7 2018 jones@scss.tcd.ie
Q3 In a Scrabble® like game, players form words and each word is awarded a score that is the sum of the points for each letter in the word. English language editions of Scrabble contain 100 letter tiles with the following letter points and letter distribution:
2 blank tiles (scoring 0 points)
1 point:
2 points: 3 points: 4 points: 5 points: 8 points: 10 points:
E×12, A×9, I×9, O×8, N×6, R×6, T×6, L×4, S×4, U×4 D×4, G×3
B×2, C×2, M×2, P×2
F×2, H×2, V×2, W×2, Y×2
K×1
J×1, X×1 Q×1, Z×1
For example, the word “MAZE” would have a score of 15 (3 + 1 + 10 + 1).
Write an ARM assembly language program that will compute the word score for a NUL terminated string containing UPPER CASE alphabetic characters and spaces (for blanks). The word is stored in memory at the address contained in R1. The score for each letter is stored in flash memory as a sequence (or table or array) of 26 byte values. The first byte is the score for “A”, the second byte is the score for “B”, and so on (use the DCB assembler directive to create this table). Your program should calculate the word score in R0.
LDR R0, =0
LDR R1, =MAZE LDR R4, =POINTS
; score = 0
; R1 -> word
; R4 -> points table
; get ch AND R1 = R1 + 1 ; if ch == 0
; finished
; if ch = ‘ ‘
; ignore as points == 0
; index from ‘A’
; get points for letter
; add to score
; next ch
; A=1, B=3, C=3, D=2, E=1
; F=4, G=2, H=4, I=1, J=8
; K=5, L=1, M=3, N=1, O=1 ; P=3, Q=10, R=1, S=1, T=1 ; U=1, V=4, W= 4, X=8, Y=4 ; Z=10
L0 LDRB
CMP R2,
BEQ L1 CMP R2, BEQ L0 SUB R2, LDRB R2, ADD R0, B L0
L1 … …
;
; points for each letter ;
POINTS
DCB DCB DCB DCB DCB DCB
;
; test word
;
MAZE DCB
R2, [R1], #1 #0
#0x20
R2, #0x41 [R4, R2] R0, R2
1, 3, 3, 2, 1 4, 2, 4, 1, 8 5, 1, 3, 1, 1 3, 10, 1, 1, 1 1, 4, 4, 8, 4 10
“MAZE”, 0, 0
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