Chapter 2.3, Problem 2.27PP

Program Plan Intro

Detecting overflow of unsigned addition:

The “overflow” occurs in arithmetic operation when the result of whole integer cannot fit within the word size range of the given data type.

• Consider “x” and “y” in the range $\text{0}\le \text{x, y}\le {\text{UMax}}_{\text{w}}$ and also assume sum, “s” is given below

${\text{s = x +}}_{\text{w}}^{\text{u}}\text{y}$

• From the above equation, “s” represents sum. “${\text{+}}_{\text{w}}^{\text{u}}$” represent unsigned addition with “w” bits.
• Then the computation of sum “s” overflowed if and only if “s < x”.

Example:

The example for detecting overflow of unsigned addition is given below:

Assume “x” is “9”, “y” is “12” and “w” is “4”.

Then, $\begin{array}{c}{\text{s = 9 +}}_{\text{4}}^{\text{u}}\text{12}\\ \text{=5}\end{array}$

Now it show that overflow occurred, since “5 < 9”.

From the above addition, observe that $\text{x}\text{+}\text{y }\ge \text{x}$.

• Hence if “s” is not in overflow, it will surely take “s ≥ x”.
• Otherwise that is if “s” is in overflow, then “${\text{s = x + y – 2}}^{\text{w}}$”.

Addition using binary is given below

$\begin{array}{l}\text{9}=>\text{1001}\\ \text{12}=>\text{1100}\\ \overline{\underset{\_}{21}}=>\underset{\_}{\overline{\text{10101}}}\end{array}$

From the above binary addition, the sum is “21” which containing a 5-bit representation “10101”.

• However, if removed the high-order bit, the result will obtain “0101” that is decimal value of “5”.
• This value is equivalent with “21 mode 16= 5”.