The correct answer is c. addi $2, $1, 10.

Here's the step by step explanation:

The addi instruction in MIPS assembly language adds a register and a constant. The format is addi $destination, $source, constant. The result is stored in the destination register.

a. addi $1, $2, 10: This adds the value in $2 (6) to 10 and stores the result in $1. This is not the correct answer because we want the result in $2.

b. addi $1, $2, 0x10: This adds the value in $2 (6) to 0x10 (16 in hexadecimal) and stores the result in $1. Again, this is not the correct answer because we want the result in $2.

c. addi $2, $1, 10: This adds the value in $1 (10) to 10 and stores the result in $2. This gives us the desired result of 26 in $2.

d. addi $2, $1, 0x10: This adds the value in $1 (10) to 0x10 (16 in hexadecimal) and stores the result in $2. This would result in a value of 26 in $2 if the value in $1 was 10, but the value in $1 is not 10, so this is not the correct answer.

Question

The correct answer is c. addi $2, $1, 10.

Here's the step by step explanation:

The addi instruction in MIPS assembly language adds a register and a constant. The format is addi $destination, $source, constant. The result is stored in the destination register.

a. addi $1, $2, 10: This adds the value in $2 (6) to 10 and stores the result in $1. This is not the correct answer because we want the result in $2.

b. addi $1, $2, 0x10: This adds the value in $2 (6) to 0x10 (16 in hexadecimal) and stores the result in $1. Again, this is not the correct answer because we want the result in $2.

c. addi $2, $1, 10: This adds the value in $1 (10) to 10 and stores the result in $2. This gives us the desired result of 26 in $2.

d. addi $2, $1, 0x10: This adds the value in $1 (10) to 0x10 (16 in hexadecimal) and stores the result in $2. This would result in a value of 26 in $2 if the value in $1 was 10, but the value in $1 is not 10, so this is not the correct answer.

Knowee AI · Accepted Answer

The correct answer is c. addi $2, $1, 10.

Here's the step by step explanation:

The addi instruction in MIPS assembly language adds a register and a constant. The format is addi $destination, $source, constant. The result is stored in the destination register.

a. addi $1, $2, 10: This adds the value in $2 (6) to 10 and stores the result in $1. This is not the correct answer because we want the result in $2.

b. addi $1, $2, 0x10: This adds the value in $2 (6) to 0x10 (16 in hexadecimal) and stores the result in $1. Again, this is not the correct answer because we want the result in $2.

c. addi $2, $1, 10: This adds the value in $1 (10) to 10 and stores the result in $2. This gives us the desired result of 26 in $2.

d. addi $2, $1, 0x10: This adds the value in $1 (10) to 0x10 (16 in hexadecimal) and stores the result in $2. This would result in a value of 26 in $2 if the value in $1 was 10, but the value in $1 is not 10, so this is not the correct answer.

Given that general purpose register $1 has the value 10 in it, and $2 has the value 6, whichone of the following lines results in the value 26 being stored in $2?a. addi $1, $2, 10b. addi $1, $2, 0x10c. addi $2, $1, 10d. addi $2, $1, 0x10

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