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MIPS assembly language question

If $s0 has the base address of array A, and A has the following elements: 9, 6, 4, 24, 33, 0, 1; What is the value of $t1 after the instructions are executed? addi $t1, $zero, 20 add $t0, $t1, $S0 lw $t1, (Sto)

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Answer #1

Solution :

addi $t1, $zero, 20
This instruction will add 20 on zero register and assign result to $t1 register.
hence $t1 will have value 20.

add $t0, $t1, $s0
This instruction will add content of $t1 and $s0 and result will be stored in $t0.
hence $t0 will store value as address of A + 20.

lw $t1, ($t0)
this will load value from memory address (address of A + 20) and store it in $t1.


one integer requries 4 bytes
hence address (address of A+20) will point to 5th element in the array A.
Hence $t1 will have 33.

if you have any doubts then you can ask in comment section if you find the solution helpful then upvote the answer. Thank you.

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