Multitasking and Context Switching; Pipelining 2. Pipeline Task Assume a 5-stage

Question

Multitasking and Context Switching; Pipelining 2. Pipeline Task Assume a 5-stage pipelined processor using the standard 5-stages (IF, ID, EX, MEM, WB) that uses no forwarding or stalling circuitry. Rather, you will use the compiler to add no-ops to the code to ensure correct execution. Note: You can assume that if the processor reads and writes to the same register in a given cycle, the value read out will be the new value that is written in that cycle Write your own short program that re-writes the code below including the no-ops that are needed to protect against hazard conditions add $3, $2, $3 lw $4, 100($3) sub $7, $6, $2 xor $6, $4, $3 The first thing your program needs to do is to print out your family (last name) and your student ID. Create strings and use the print_string syscall to display this information to the MARS Run 1/0 window. Include as few no-ops as possible. Do not change the order of the code statements. To receive full credit, be sure to document the hazard condition in your code comments Deliverables:

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Code:-

#This is a simple program to check the address calculation
#to avoid data hazards between registers using "nop" instruction
#nop instruction generate delay to the execution of instructions
#but not harmful
#Variable declaration section
.data
     #Write your lastname variable
     lastName: .asciiz "LastName"
     #Write your student ID
     sId: .asciiz " YourId"
     #an array of 100 size , word boundary means 4 byte
     arr: .word 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
#Main Program
.text
     #Display your last name
     la $a0,lastName
     li $v0,4
     syscall
     #Display your Id
     la $a0,sId
     li $v0,4
     syscall
     #add value of register 2 and 6 with 0
     li $2,0
     li $6,0
     #load address of the array in $3 register
     la $3,arr
     # $3 using next instruction , so avoiding data hazard put NOP
     nop
     nop    
     nop
     #add starting address of array with 0 and store again in $3
     add $3,$2,$3
    #$4 getting value of $3 address(Base address)+offset contains element, so here also data hazard happening
     nop
     nop
     nop
     #load value of 100th offset space , $4=26
     lw $4,100($3)
     #$7=0-0=0
     sub $7,$6,$2
     #again $6 using next instruction for storing , so avoid data hazard nop instruction using
     nop
     nop
     nop
     #xoring 26 with base address of array and store the result in $6(Means another address)
     xor $6,$4,$3
    #End of the program
     li $v0,10
     syscall

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