Chapter 2, Problem 2.19.1E

Explanation of Solution

The value of $t2:

The given is:

$t0 = 0xAAAAAAAA

$t1 = 0x12345678

Let consider the given question as:

sll $t2, $t0, 4

or $t2, $t2, $t1

• 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 “or” instruction is used to “OR” the constant value to the register “$t2” and store the result in the “$t2”.

Step1: Convert the hexadecimal value into the binary value.

The corresponding binary value of “A” = 1010

The corresponding binary value of “1” = 0001

The corresponding binary value of “2” = 0010

The corresponding binary value of “3” = 0110

The corresponding binary value of “4” = 0100

The corresponding binary value of “5” = 0101

The corresponding binary value of “6” = 0110

The corresponding binary value of “7” = 0111

The corresponding binary value of “8” = 1000

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

Step2: Left shift the “$t0” value

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

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