Question

Assume that registers \(s0 and \)s1 hold the values 0x80000000 and 0xD0000000, respectively.

1. What is the value of \(t0 for the following assembly code?

add \)t0,\(s0,\)s1

2. Is the result in \(t0 the desired result, or has there been an overflow?

3. For the contents of registers \)s0 and \(s1 as specified above, what is the value of \)t0 for the following assembly code?

sub \(t0,\)s0,\(s1

4. Is the result in \)t0 the desired result, or has there been an overflow?

5. For the contents of registers \(s0 and \)s1 as specified above, What is the value of \(t0 for the following assembly code?

add \)t0,\(s0,\)s1

add \(t0,\)t0,\(s0

6. Is the result in \)t0 the desired result, or has there been an overflow?

Short answer

The results of each instruction:

1. The value of $t0=50000000
2. Yes, it has overflow.
3. The value of $t0=B0000000
4. No, there is no overflow.
5. The value of $t0=D0000000
6. Yes, there is an overflow

Step-by-step solution

Determine the MIPS instruction number formats:

In MIPS assembly language, instructions use hexadecimal numbers. The machine will convert them into binary digits represented as bits. The rightmost bit is referred to as the Least Significant Bit. The left-most bit is referred to as the Most Significant Bit.

MIPS takes the 32-bit inputs, which means all the input data will be considered in 32 bits format.

Determine the value of $t0

Given that the values of registers $s0 and $s1 :

$$\begin{gathered} \$s0=0\times 80000000 \\ \$s1=0\times D0000000 \end{gathered}$$

1.Given instruction : add $t0,$s0,$s1

Add instruction will add the $s0 and $s1 . the result will be stored in $t0.

The value of $t0 for the given assembly code is as follows:

Hexadecimal values will be converted into binary values.

$$\begin{gathered} \$s0=1000000000000000 \\ \$s1=1101000000000000 \\ \$s0+\$s1=\$t0=0101000000000000 \end{gathered}$$

Now, the value of $t0 will be converted to hexadecimal.

$\$t0=0\times 50000000$

Determine whether overflow occurred or not

Given that the values of registers $s0 and $s1:

2. Given instruction: add $t0,$s0,$s1

$$\begin{gathered} \$s0=1000000000000000 \\ \$s1=1101000000000000 \\ \$s0+\$s1=\$t0=0101000000000000 \end{gathered}$$

In the given instruction while performing addition, the overflow will occur at the left-most bit.

In the above addition, the leftmost bits of $s0 and $s1 are 1. While adding 1+1 in binary addition the value is 10. In 10,1 is neglect due to overflow only 0 is considered.

So the value of $t0 is considered as 0101 0000 0000 0000 due to overflow.

Determine the value of $t0.

Given that the values of registers $s0 and $s1:

$$\begin{gathered} \$s0=0\times 80000000 \\ \$s1=0\times D0000000 \end{gathered}$$

3. Given instruction: sub $t0,$s0,$s1

Sub instruction will subtract the $s0 and $s1 . the result will be stored in $t0.

The value of $t0 for the given assembly code is as follows:

Hexadecimal values will be converted into binary values.

$$\begin{gathered} \$s0=1000000000000000 \\ \$s1=1101000000000000 \end{gathered}$$

$\$s0-\$s1=\$t0=1011000000000000$

Now, the value of $t0 will be converted to hexadecimal.

$\$t0=0\times B0000000$

Determine whether overflow occurred or not.

Given that the values of registers $s0 and $s1 :

$$\begin{gathered} \$s0=0\times 80000000 \\ \$s1=0\times D0000000 \end{gathered}$$

4. Given instruction: sub $t0,$s0,$s1

In the given instruction while performing addition, the overflow will occur at the left-most bit.

$$\begin{gathered} \$s0=1000000000000000 \\ \$s1=1101000000000000 \end{gathered}$$

$\$s0-\$s1=\$t0\Rightarrow 1011000000000000$

In the above subtraction, the leftmost bits of $s0 and $s1 are 1 and 0 when getting borrowed. While subtracting 1 and 0 the result is 1. So there is no overflow occurred.

Determine the value of $t0.

Given that the values of registers $s0 and $s1 :

5. Given instruction:

$$\begin{gathered} \$s0=0\times 80000000 \\ \$s1=0\times D0000000 \end{gathered}$$

add $t0, $s0, $s1

add $t0,$t0,$s0

add instruction will add the $s0 and $s1 . the result will be stored in $t0.

The value of $t0 for the given assembly code is as follows:

Hexadecimal values will be converted into binary values.

$$\begin{gathered} \$s0=1000000000000000 \\ \$s1=1101000000000000 \end{gathered}$$

$\$s0+\$s1=\$t0=0101000000000000$

Now, the value of $t0 will be converted to hexadecimal.

$\$t0=0\times B0000000$

The next instruction will add the values of $t0 and $s0 as follows:

$$\begin{gathered} \$t0=0101000000000000 \\ \$s0=1000000000000000 \end{gathered}$$

$\$t0+\$s0=1101000000000000$

The value of $t0 can be obtained by converting the binary to hexadecimal

$\$t0=0\times D0000000$

Determine whether overflow occurred or not

Given that the values of registers $s0 and $s1 :

$$\begin{gathered} \$s0=0\times 80000000 \\ \$s1=0\times D0000000 \end{gathered}$$

6. Given instruction:

add $t0, $s0, $s1

add $t0,$t0,$s0

add instruction will add the $s0 and $s1 . the result will be stored in $t0.

The value of $t0 for the given assembly code is as follows:

Hexadecimal values will be converted into binary values.

$$\begin{gathered} \$s0=1000000000000000 \\ \$s1=1101000000000000 \end{gathered}$$

$\$s0+\$s1=\$t0=0101000000000000$

In the above addition, the leftmost bits of $s0 and $s1 are 1. While adding 1+1 in binary addition the value is 10. In 10,1 is neglect due to overflow only 0 is considered.

So the value of $t0 is considered as 0101 0000 0000 0000 due to overflow.