Chapter 2, Problem 2.19.2E

Explanation of Solution

The value of $t2:

The given is:

$t0 = 0xAAAAAAAA

$t1 = 0x12345678

sll $t2, $t0, 4

andi $t2, $t2, -1

• The “sll” instruction is used to shift one position left to the bit. Here the register value “$t0” is shifted 4 positions left to the bit and finally store the result in the “$t2”.
• The “andi” instruction is used to “AND” the constant value to the register “$t2” and store the result in the “$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: Left shift the “$t0” value

$1010\text{ 1010 1010 1010 1010 1010 1010 0000}$

Step2:

Here the first four bits removed and “0s” are added in the last 4 position.

Step3: Take 2’s complement for “-1” value.

$\begin{array}{l}1\text{ = 0000 0000 0000 0000 0000 0000 0000 }0001\\ \text{interchange given binary number:}\\ 1111\text{ 1111 1111 1111 1111 1111 1111 }1110\\ \\ \text{add "1" to the interchange binary number}\\ 1111\text{ 1111 1111 1111 1111 1111 1111 }1110\\ \text{ �}\end{array}$