Sum of three integers stored in the memory (a) Specify memory location using: [ la ] (b) Load immediate integer-1 to register: [ $t1 ] (c) Load immediate integer-2 to register: [ $t2 ] (d) Load Immediate integer-3 to

#a) Specify the memory location [la]

la $s0, 0x0010

#b) Load immediate Integer-1 to register

li $t1, 1

#c) Load immediate Integer-2 to register

li $t2, 2

#d) Load immediate Integer-3 to register

li $t3, 3

#e) Store Integer-1 to memory

sw $t1, ($s0)

#f) Store Integer-2 to memory

sw $t2, 4($s0)

#g) Store Integer-3 to memory

sw $t3, 8($s0)

#h) Load Integer-1 from memory to Register

move $t4,$t1

#i) Load Integer-2 from memory to Register

move $t5,$t2

#j) Load Integer-3 from memory to Register

move $t6,$t3

#k) add the three numbers

add $t7,$t4,$t5

add $t7,$t7,$t6

#l) store the result of the addition again to memory

sw $t7, 12($s0)

#m) Load the result of the addition to register

move $t8,$t7

#sum

li $v0, 1

move $a0,$t8

syscall

Explanation:

Run the above code and you will get your perfect output.

solution:

this situation is called an ip conflict.

data from a server is sent in small bits called packets. in an ip conflict, some packets will end up in one computer while some other packets will end up in another computer.

so…overall none of the computers get the data as the original message or picture being sent by the server cannot be retrieved with only half the packets. so it will show up as an error.

thus the final answer is d.

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