3. What will be in $8 after execution? addi $8, $0,5 sil_$8, $8,2 // shifting $8...

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3. What will be in $8 after execution? addi $8, $0,5 sil_$8, $8,2 // shifting $8 left twice (logical shift- assume no sign) 4

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Answer 1

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addi $8, $0,5 : This instruction will add 5 to $0 register and then store the result in $8.
$8 $0 + 5
sll $8, $8,2 : This will shift the contents inside the register $8 to the left twice. At present the register is having 101, after the shift operation it will become 10100 (zeros are added to the LSB side after each shift operation). 10100 will be stored inside $8 which is equivalent to 20 in decimal.
$8 $0 << 2

Therefore $8 will contain 20.

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Answer 2

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