Chapter 2, Problem 2.19.3E

Explanation of Solution

The value of $t2:

The given is:

$t0 = 0xAAAAAAAA

$t1 = 0x12345678

srl $t2, $t0, 3

andi $t2, $t2, 0xFFEF

• The “srl” instruction is used to shift one position right to the bit. Here the register value “$t0” is shifted 3 positions right to the bit and finally store the result in the “$t2”.
• The “andi” instruction is used to “AND” the constant value to the register “$t2” and store it in the result in the “$t2”.

Step1: Convert the hexadecimal value into the binary value

$\begin{array}{l}0\text{xAAAAAAAA }=1010\text{ 1010 1010 1010 1010 1010 1010 1010} \\ 0\text{x12345678 }=0001\text{ 0010 0011 0100 0101 0110 0111 1000}\end{array}$

Step2: Right shift the “$t0” value

$0001\text{ 01}010\text{101 0101 0101 0101 0101 0101}$

Here the last three bits removed and “0s” are added in the first 3 position.

Step3: convert the hexadecimal value “0xFFEF” into binary value