Binary Arithmetic: Addition and Binary Shifts (Logical)

This lesson covers binary addition and logical binary shifts, both required for OCR J277 Section 2.6. You must be able to add two 8-bit binary numbers and understand the effect of left and right shifts.

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Binary Addition Rules

Binary addition follows four simple rules:

Addition	Result	Carry
0 + 0	0	0
0 + 1	1	0
1 + 0	1	0
1 + 1	0	1 (carry 1 to the next column)

There is one additional rule when a carry is involved:

Addition	Result	Carry
1 + 1 + 1 (carry)	1	1

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Worked Example: Binary Addition

Add 01101010 + 00110101

Work from right to left, just like denary addition:

Row	b8	b7	b6	b5	b4	b3	b2	b1	Decimal
Carry	1	1	1	1	1
	0	1	1	0	1	0	1	0	106
+	0	0	1	1	0	1	0	1	53
Sum	1	0	0	1	1	1	1	1	159

Check: 106 + 53 = 159. Converting 10011111 to denary: 128 + 16 + 8 + 4 + 2 + 1 = 159. Correct!

OCR Exam Tip: Always show your carry digits above the calculation. This is how examiners verify your working, and you may get method marks even if the final answer is wrong.

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Overflow

Overflow occurs when the result of a binary addition is too large to be stored in the available number of bits. For 8-bit numbers, the maximum value is 255 (11111111).

Example of overflow:

Row	ov	b8	b7	b6	b5	b4	b3	b2	b1	Decimal
		1	1	0	0	0	0	0	0	192
+		1	0	0	0	0	0	0	0	128
Sum	1	0	1	0	0	0	0	0	0	320 (9 bits!)

The result requires 9 bits, but only 8 bits are available. The extra bit is lost, and the stored value would be 01000000 (64), which is incorrect. This is an overflow error.

OCR Exam Tip: If the exam asks you to "explain what happens" when overflow occurs, state that the result exceeds the maximum value that can be stored in the available bits, causing an incorrect value to be stored.

flowchart TD
    A[Two 8-bit binary numbers] --> B[Add column by column right to left]
    B --> C{1 + 1?}
    C -->|Yes| D[Write 0, carry 1]
    C -->|No| E[Write sum, carry 0]
    D --> F[Continue to next column]
    E --> F
    F --> G{Result > 8 bits?}
    G -->|Yes| H[Overflow - extra bit lost]
    G -->|No| I[Valid 8-bit result]
    I --> J[Optional logical shift]
    J --> K[Left shift = x2 per position]
    J --> L[Right shift = integer divide by 2]
    K --> M[New 8-bit value, 0s fill the right]
    L --> N[New 8-bit value, 0s fill the left]

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Logical Binary Shifts

A binary shift moves all the bits in a binary number to the left or right by a specified number of positions. Bits shifted off the end are lost, and empty positions are filled with 0s.

Left Shift

A left shift moves all bits to the left. Each left shift by 1 position multiplies the value by 2.

Example: Left shift 00010110 (22) by 1 position

Step	b8	b7	b6	b5	b4	b3	b2	b1	Decimal
Before	0	0	0	1	0	1	1	0	22
After	0	0	1	0	1	1	0	0	44

The leftmost bit (0) is lost, and a 0 is added on the right. The value doubled from 22 to 44.

Left shift by 2 positions multiplies by 4 (2 x 2):

Step	b8	b7	b6	b5	b4	b3	b2	b1	Decimal
Before	0	0	0	1	0	1	1	0	22
After	0	1	0	1	1	0	0	0	88

22 x 4 = 88. Correct!

Right Shift

A right shift moves all bits to the right. Each right shift by 1 position performs integer division by 2 (the result is rounded down).

Example: Right shift 00010110 (22) by 1 position

Step	b8	b7	b6	b5	b4	b3	b2	b1	Decimal
Before	0	0	0	1	0	1	1	0	22
After	0	0	0	0	1	0	1	1	11

The rightmost bit (0) is lost, and a 0 is added on the left. The value halved from 22 to 11.

Right shift 00010111 (23) by 1 position:

Step	b8	b7	b6	b5	b4	b3	b2	b1	Decimal
Before	0	0	0	1	0	1	1	1	23
After	0	0	0	0	1	0	1	1	11

23 / 2 = 11.5, but since we lose the rightmost bit, the result is 11 (integer division, rounded down).

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Summary of Shift Effects

Shift	Direction	Positions	Effect on value
Left shift by 1	Left	1	Multiply by 2
Left shift by 2	Left	2	Multiply by 4
Left shift by 3	Left	3	Multiply by 8
Right shift by 1	Right	1	Integer divide by 2
Right shift by 2	Right	2	Integer divide by 4
Right shift by 3	Right	3	Integer divide by 8

OCR Exam Tip: If a shift question asks for the denary value before and after, convert both binary numbers to denary and verify that the multiplication/division is correct.

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Overflow with Left Shifts

Left shifts can also cause overflow if significant bits (1s) are shifted off the left end:

Step	b8	b7	b6	b5	b4	b3	b2	b1	Decimal	Note
Before	1	1	0	0	0	0	0	0	192
Left 1	1	0	0	0	0	0	0	0	128	lost a 1 bit!

192 x 2 = 384, which cannot fit in 8 bits. The result is 128, which is incorrect.

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Practice