Question

Assume that \(s0 holds the value 128_ten.

1. For the instruction add \)t0, \(s0, \)s1, what is the range(s) of values for \(s1 that would result in overflow?
2. For the instruction sub \)t0, \(s0, \)s1, what is the range(s) of values for \(s1 that would result in overflow?
3. For the instruction sub \)t0, \(s1, \)s0, what is the range(s) of values for $s1 that would result in overflow?

Short answer

Ranges of values:

1. $x>2^{31}-129andx<-2^{31}-128$
   
2. $x<-2^{31}+129andx>2^{31}+128$
   
3. $x<-2^{31}+128andx>2^{31}+127$
   

Step-by-step solution

Define signed overflow.

In the MIPS instructions, the arithmetic operations are performed using signed operands.

In addition instructions add and addi, overflow occurs when the result gets an opposite sign of the operands.

In the subtraction instruction sub, overflow occurs when the result gets the opposite sign of the subtrahend.

Determine the range of overflow for the add instruction.

1.The given instruction add $t0, $s0, $s1.

$s0 holds the decimal value .The given instruction will add the values of $s0 and $s1 registers and store the result at $t0.

The overflow occurs when the result exceeds$-2^{31}\le x\le 2^{31}$

Here, x is greater than the limit.

$$\begin{gathered} 128+x>2^{31}-1 \\ x>2^{31}-1-128 \\ x>2^{32}-129 \end{gathered}$$

Here, x is lesser than the limit.

$$\begin{gathered} 128+x<-2^{31} \\ x<-2^{31}-128 \end{gathered}$$

Therefore, the range for overflow of add $t0, $s0, $s1 instruction is

$x>2^{31}-129andx<-2^{31}-128$(Impossible)

Determine the range of overflow for the sub instruction.

2.The given instruction sub $t0, $s0, $s1.

$s0 holds the decimal value. The given instruction will subtract $s1 from $s0 and hold the result in $t0.

The overflow occurs when the result exceeds

Here, x is greater than the limit.

$$\begin{gathered} 128-x>2^{31}-1 \\ x<-2^{31}+1+128 \\ x<-2^{31}+129 \end{gathered}$$

Here, x is lesser than the limit.

$$\begin{gathered} 128-x<-2^{31} \\ x>2^{31}+128 \end{gathered}$$

Therefore, the range for overflow of sub $t0, $s0, $s1 instruction is

$x<-2^{31}+129andx>2^{31}+128$(Impossible)

Determine the range of overflow for the sub instruction

3.The given instruction sub $t0, $s1, $s0.

$s0 holds the decimal value ${128}_{ten}$. The given instruction will subtract $s0 from $s1 and hold the result in $t0.

The overflow occurs when the result exceeds$-2^{31}\le x<2^{31}$

Here, x is greater than the limit.

$$\begin{gathered} x-128<-2^{31} \\ x<-2^{31}+128 \end{gathered}$$

Here, x is lesser than the limit.

$$\begin{gathered} x-128<2^{31}-1 \\ x<-2^{31}-1+128 \\ x<-2^{31}+127 \end{gathered}$$

Therefore, the range for overflow of sub $t0, $s0, $s1 instruction is

$x<-2^{31}+128andx>2^{31}+127$(Impossible)