Write LEGv8 assembly language programs for the following C p

Write LEGv8 assembly language programs for the following C program fragments. Please make sure to add any necessary comments. P [20] = P [10] + P [30] + 25; Show the machine code that will be generated by the assembly code in Problem 1. You can write machine code as decimal numbers.

Solution

prob 1

Let the reference of p be X22

LDUR X9,[X22,#80] //index 8 required offset of 80, temporary register X9 gets p[10]

LDUR X10,[X22,#240] //index 30 required offset of 240, temporary reg X10 gets p[30]

ADD X9,X10,X9 //X10+X9 stored in X9, X9 gets p[10]+p[30]

ADD X9,#25 //X9 gets p[10]+p[30]+25

STUR X9,[X22.#160] //stores p[10]+p[30]+25 in p[20] as index 20 requires offset of 160

prob 2

for ADD opcode is 10001011000, opcode size is 11 and instruction is R format

for LDUR opcode is 11111000010, opcode size is 11 and instruction is D format

for STUR opcode is 11111000000, opcode size is 11 and instruction is D format

for D format instruction

where, Rn = base register

op2 = expand opcode field

Rt = destination number

For R format instuction

where, shamt = how many position to shaft

Shift left logical (LSL)

Shift right logical (LSR) \"

Now is the 1st instuction x22[#80] is loaded in X9 , op2 = 0, Rt: X9, Rn: X22

so machine code will be 11111000010  010110000 00 10110 01001 or in decimal 1986 80 0 22 9

in 2nd instuction [X22,#240] is loaded in X10 so machine code will 11111000010 011110000 00 10110 1010

similary for ADD X9,X10,X9 machine code will 10001011000 01001 000000 01010 01001

for ADD X9,#25 machine code: 10001011000 01001 011001 00000 00000

for STUR X9,[X22.#160], op2=0,machine code will be 11111000000 010100000 00 10110 01001

opcode 11 bits

offset 9 bits

op2 2bits

Rn 5 bits

Rt 5bits