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# Accounts Analysis

; 1. Code the following IF-THEN statement:
;
; if (r2 != r7)
; r2 = r2 – r7;
; else
; r2 = r2 + r4;
;
; Your code have to work with any numbers in R2, R7 and R4
; (5, 25(0x19))
;
MOVW R2, #10
MOVW R7, #5
MOVW R4, #15
;
; MOVW R2, #10
; MOVW R7, #10
; MOVW R4, #15
;
;– Approx. 3 instructions —

; 2. Write a program that converts a hexadecimal value between 0x0 and 0xf in register
; R0 into its ASCII representation. Store ASCII representation into R1.
;
; Digits ‘0’ through ‘9’ are represented with the ASCII codes 0x30 to 0x39. The digits
; ‘A’ through ‘F’ are coded as 0x41 through 0x46. (See ‘ascii.pdf’ file)
;
; Test your code with R0 = 15, 0 and 9
; (0x46, 0x30, 0x39)
;
MOV R0, #15
; MOV R0, #0
; MOV R0, #9

;
; 3. Modify the code you did for task #2 to check for valid number (0..15) in R0.
; Store ‘*’ (0x2a) into R1 if R0 is out of valid range. Store ASCII representation
; for valid numbers (from 0 to 15 inclusive).
;
; Test your code with R0 = 255, 15, 0, 9, 12 and 16
; (0x2a, 0x46, 0x30, 0x39, 0x43, 0x2a)
;
MOV R0, #255
; MOV R0, #15
; MOV R0, #0
; MOV R0, #9
; MOV R0, #12
; MOV R0, #16

;
; 4. Write a program that counts number of binary ‘1’s and number of binary ‘0’s in R0.
; Save number of ‘1’s into R1, number of ‘0’s in R2. You may use any other register(s) and
; do not need to preserve the value of R0 register.
;
; Your code have to work with any number in R0
;
; Debug your program with tets values R0 = 0xAAAAAAAA and R0 = 0x55555555. Both should give
; you 16 1’s and 16 0’s
;
; Then run your progarm with R0 = 0x708. Is your result correct? (use online converter to check)

``MOV     R0, #0xAAAAAAAA``

; MOV R0, #0x55555555
; MOVW R0, #0x708

;
; 5. Write a “shift, test and restore” division algorithm (See “Binary Division.pdf” for details).
;
; For i = 1 to 32 do { we’re using 32-bit representations }
; {
; Left Shift the RQ pair
; Subtract the Divisor from R
; If R is positive then
; Set the low order (right most) bit in Q to 1
; Else
; Restore R by Adding back the Divisor
; }
;
; R0 = remainder (R)
; R1 = divident/quotient (Q)
; R2 = divisor (D)
;
; R1,R0 = R1/R2
;
; Do not use UDIV and SDIV ARM commands!
;
; Your code have to work with any valid numbers in R0. You may use any other register(s). Check your
; code with given values first. Then try some another numbers in R1 and R2.

``````MOVW    R1, #163    ; 163/10 = 16, 3
MOVW    R2, #10``````

;
; 6. Write a small program to compare two 64-bit values. Set R0 to 0 if two values
; are equal, set R0 to 1 if two values are not equal. First 64-bit number placed into
; R1(HI bits) and R2(LO bits), second number is in R3(HI bits) and R4(LO bits). Use only
; 4 ARM instructions!
;
; Your code have to work with any numbers in R1-R4
;
; Check your code with all test sets below
;
MOV R1, #0xAAAAAAAA
MOV R2, #0x55555555
MOV R3, #0xAAAAAAAA
MOV R4, #0x55555555
; —-
; MOV R3, #0x55555555
; MOV R4, #0xAAAAAAAA
; —-
; MOV R3, #0xAAAAAAAA
; MOV R4, #0xAAAAAAAA
; —-
; MOV R3, #0x55555555
; MOV R4, #0x55555555

;– Your 4 instructions here —

;
; 7. (Bonus). Write a code that allows you to rotate 64-bit values in registers
; R0 and R1. The code should shift value left…
;
; +——————–+ +——————–+
; +<-! R0 !<–! R1 !<–+ ; ! +——————–+ +——————–+ ! ; ! ! ; +–>———————->————————->+
;
; …then right by one bit.
;
; +——————–+ +——————–+
; +->! R0 !–>! R1 !–>+
; ! +——————–+ +——————–+ !
; ! !
; +<————————<———————–<–+
;
; – Do not use any other registers.
;
; – Make sure that after two shifts the final value is equal to original one.
;
; Tip: Left shift is tricky. You may remember: a << 1 = a * 2; a * 2 = ???
; … and take care of LSB…
;
; Check your code with all 3 test sets below:
;
MOVW R0, #0x0001
MOVT R0, #0x8000
MOVW R1, #0x0001
MOVT R1, #0x8000
; —-
; MOVW R0, #0xAAAA
; MOVT R0, #0x8555
; MOVW R1, #0x0F0F
; MOVT R1, #0xF0F0
; —-
; MOV R0, #0x00000001
; MOV R1, #0x00000001
;
; — left —
;– May be done in 3 instructions —

; — right —
;– May be done in 3 instructions —

; 8. (Bonus++). Write a program that check if the binary pattern 2_11001010 are presented at least once
; somewhere in the number R0. The program should set R1 to 1 if the pattern is found, and clear R1 to 0
; if no pattern in the R0.
;
; Your code have to work with any number in R0
;
; Your code have to preserve (do not change) the value in R0
;
; You may use any other registers
;
; Examples: the number 2_11001101100100101111001010101100 (0xCD92F2AC) has the pattern inside
; ——–
; 2_11001100110011001100110110010100 (0xCCCCCD94) has the pattern inside
; ——–
; 2_11011010101101110011100111010101 (0xDAB739D5) – no pattern
;
MOVW R0, #0xF2AC
MOVT R0, #0xCD92
;—
; MOVW R0, #0xCD94
; MOVT R0, #0xCCCC
;—
; MOVW R0, #0x39D5
; MOVT R0, #0xDAB7

Sample Solution

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