Overflow in Binary `Arithmetic`. Discussion with examples.

Overflow Introduction and scenarios.

In terms of computing, two numbers are loaded into registers before performing any arithmetic operations. For a number between 0 to 15 we can use a register of length 4 bits. When two 4 bit numbers are added and the result is greater than 4 bits then we get an overflow condition.

Overflow can occur in either of the two following scenarios:-

Adding two positive numbers.
Or, Adding two negative numbers in signed arithmetic.

Overflow can never occur when adding a positive number to a negative number.

The reason being the result is always smaller than the original larger number. We will discuss the overflow condition with examples below:-

Example1 - Add 2 unsigned binary numbers.
Add two binary numbers 10 and 15 with previous carry = 0.

Sol. Load the values in two registers R1 and R2.

So, R1 = 10 (decimal) = 1010 (in binary A3A2A1A0)
& R2 = 15 (decimal) = 1111 (in binary B3B2B1B0)
Also Refer example on the binary adder discussion page.
Result = (Carry = 1) (Sum = 1001)

In this case Overflow is true and result includes the overflow bit. We can use a separate flip-flop to store the result of overflow condition.

Some more expamples are discussed on fullchipdesign.

Example2 - Add 2 signed number to results in overflow.

Example3 - Add 2 signed number to results in no overflow.

Overflow in Binary Arithmetic. Discussion with examples.

Overflow in Binary `Arithmetic`. Discussion with examples.