Question

I. What value will be in register r2 after execution of the following instructions? Show your work with the register values after executing each instruction.

MOV r2, #0x0

LDR r1, =0xCF

MOVS r1, r1,

LSR #1

ADC r2, r2, #0

MOVS r1, r1,

LSR #1

ADC r2, r2, #0

II. Assume a 32 bit register holds a data of four bytes as B3B2B1B0. Write a sequence of ARM instruction that takes this data as input and swaps the bytes 0 and 3 so the result would be B0B1B2B3.

III. Which ARM processor mode/modes has the fewest available number of registers available? How many and why??

Answer 1

Answer :- i) The ARM assembly code has been written below-

MOV r2, #0x0 ;r2 = 0

LDR r1, =0xCF ;r1 = 0xCF

MOVS r1, r1, LSR #1 ;r1 = 0x67 and carry flag = 1 i.e lsb of oxCF

ADC r2, r2, #0 ;r2 = r2+0+carry = 0x01

MOVS r1, r1, LSR #1 ;r1 = 0x33 and c = 1

ADC r2, r2, #0 ;r2 = 0x01 + 0x0 + c = 0x02.

Answer :- ii) The ARM assembly code has been written below-(assume r1 holds 4 bit data)

AND r2, r1, #0x01 ;r2 = lsb i.e. B0

AND r3, r1, #0x08 ;r3 = msb i.e. B3

AND r1, r1, #0x06 ;r1 = 0 B2 B1 0

LSL r2, r2, #3 ;r2 = B0 0 0 0

LSR r3, r3, #3 ;r3 = 0 0 0 B3

ORR r2, r2, r3 ;r2 = B0 0 0 B3

ORR r1, r1, r2 ;r1 = B0 B2 B1 B3