# Lab 5
Name: SUDDULA VINEETH RAGHAVENDRA
Roll No.: CS22B045

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## Question 1

**Code** (if required)

.data
array: .word 1 2 3 4 5 6 7 8 9 10 12
str: .string " "
str1: .string "after updation of the array"
str2: .string "\n"
base: .word 0x10000000

.text
la x21,array
addi x21,x21,40    # we storing the last element address in the x21.
la x22,array
li x3,3
li x2,0
li x11,11
li x25,12
srli x26,x25,2     # here they mentioned factor of 4.
li x9,11
li x5,0
loop:
    beq x2,x3,exit
    addi x2,x2,1
    lw x10,0(x21)
    addi x10,x10,3
    sw x10,0(x21)
    
    lw x10,-4(x21)
    addi x10,x10,3
    sw x10,-4(x21)
    
    lw x10,-8(x21)
    addi x10,x10,3
    sw x10,-8(x21)
    
    lw x10,-12(x21)
    addi x10,x10,3
    sw x10,-12(x21)
    addi x21,x21,-16
    
    j loop
exit:
    li a0,10
    ecall

**Observation:**
The value of IPC is 0.724 when we do the loop unrolling with 4 factor. Without loop unrolling the value of IPC is 0.67. So it is a good optimization.

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## Question 2
**Code** (if required)

# reference code from lab example.
.data
node1: .word 10  #value
       .word 0  #pointer to next node
node2: .word 20
       .word 0
node3: .word 30
       .word 0
node4: .word 40 
       .word 0 
node5: .word 50
       .word 0
node6: .word 60
       .word 0
node7: .word 70 
       .word 0 
node8: .word 80
       .word 0
node9: .word 90
        .word 0
node10:.word 100 
       .word 0 

.text
init:   
la x3,node1
la x4,node2
sw x4,4(x3)
la x3,node3
sw x3,4(x4)
la x4,node4
sw x4,4(x3)
la x3,node5
sw x3,4(x4)
la x4,node6
sw x4,4(x3)
la x3,node7
sw x3,4(x4)
la x4,node8
sw x4,4(x3)
la x3,node9
sw x3,4(x4)
la x4,node10
sw x4,4(x3)

la x3,node1
reverse: #here we used three pointer approach to reverse the actual linked list which is used in DSA.
    li x4,0
    mv x5,x3
    mv x6,x3
    loop:
    beq x5,x0,exit
    lw x6,4(x5)
    sw x4,4(x5)
    mv x4,x5
    mv x5,x6
    j loop    
        
exit:
#By this step the total linked list will be reversed.
    li a7,10
    ecall

**Observation:**
Here we are initializing 10 nodes and here each word will be storing 4 bytes in memory. So at the end each node will be taking 8 bytes in the memory.

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## Question 3

# reference code from lab
.data
node1: .word 10  #value
       .word 0  #pointer to next node
node2: .word 20
       .word 0
node3: .word 30
       .word 0
node4: .word 40 
       .word 0 
node5: .word 50
       .word 0
node6: .word 60
       .word 0
node7: .word 70 
       .word 0 
node8: .word 80
       .word 0
node9: .word 90
        .word 0
node10:.word 100 
       .word 0 

.text
init:   
la x3,node1
la x4,node2
sw x4,4(x3)
la x3,node3
sw x3,4(x4)
la x4,node4
sw x4,4(x3)
la x3,node5
sw x3,4(x4)
la x4,node6
sw x4,4(x3)
la x3,node7
sw x3,4(x4)
la x4,node8
sw x4,4(x3)
la x3,node9
sw x3,4(x4)
la x4,node10
sw x4,4(x3)

allocate_node:
    la x9,node1
    li x10,0
    join:
        beq x9,x0,allocate
        lw x10,4(x9)
        lw x9,4(x10)
        j join
    allocate:
    # here x10 is pointing towards the last node.
        mv x9,x10
        addi x9,x9,8
        sw x9,4(x10)
        li x9,110
        # just we are creating a node with the value of 110.
        sw x9,8(x10)
        sw x0,12(x10)
    # after this step a new node will be allocated.

**Observation:**
By this approach we will be able to create a node by allocating new memory.

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