OCR J277 · Component 1 · Topic 1.2.4d

Binary Shifts

Left Shifts · Right Shifts · Multiplying and Dividing in Binary · Issues with Shifts · Exam Practice

CSZone OCR GCSE Computer Science J277

Learning Objectives

By the end of this video you will be able to...

Explain what is meant by a binary shift

Perform a left binary shift and state its effect — multiplying by a power of 2

Perform a right binary shift and state its effect — dividing by a power of 2

Explain the issues that can occur when shifting binary digits, including the loss of data

Apply all of this to exam-style questions

⚡ Builds on the place value method from 1.2.4a — the same 8-bit columns: 128, 64, 32, 16, 8, 4, 2, 1

Binary Shifts

What is a binary shift?

DEFINITION

A binary shift moves every bit in a binary number a number of places to the left or to the right.

A left shift moves all the bits towards the most significant (leftmost) end of the register.

A right shift moves all the bits towards the least significant (rightmost) end of the register.

⚡ Whichever direction the bits move, the spaces left empty at the other end are filled with 0s.

Left Shifts

A left shift multiplies the number

RULE

Shifting a binary number left by one or more places multiplies it by a power of 2.

←1 place = ×2

Shift left by 1 place doubles the value.

←2 places = ×4

Shift left by 2 places quadruples the value.

←3 places = ×8

Shift left by 3 places multiplies the value by 8.

Worked Example — Left Shift

Shifting 00001101 (13) left

	128	64	32	16	8	4	2	1
Start = 13	0	0	0	0	1	1	0	1

←1 place

0

1

0

1

0

= 00011010 = 26 (13 × 2 = 26 ✓) — a new 0 fills the right

←2 places

0

1

0

1

0

= 00110100 = 52 (13 × 4 = 52 ✓)

←3 places

0

1

0

1

0

= 01101000 = 104 (13 × 8 = 104 ✓)

Right Shifts

A right shift divides the number

RULE

Shifting a binary number right by one or more places divides it by a power of 2.

→1 place = ÷2

Shift right by 1 place halves the value.

→2 places = ÷4

Shift right by 2 places divides the value by 4.

→3 places = ÷8

Shift right by 3 places divides the value by 8.

Worked Example — Right Shift

Shifting 01101000 (104) right

	128	64	32	16	8	4	2	1
Start = 104	0	1	1	0	1	0	0	0

→1 place

0

1

0

1

0

= 00110100 = 52 (104 ÷ 2 = 52 ✓) — a new 0 fills the left

→2 places

0

1

0

1

0

= 00011010 = 26 (104 ÷ 4 = 26 ✓)

→3 places

0

1

0

1

= 00001101 = 13 (104 ÷ 8 = 13 ✓)

Issues with Shifts

What happens to bits at the edge?

THE PROBLEM

A register only has a fixed number of bits (e.g. 8). When a shift moves bits off the end of the register, those bits are permanently lost.

Left shift: bits fall off the left (most significant) end. Losing a significant bit makes the result much smaller than expected.

Right shift: bits fall off the right (least significant) end. The result is rounded down, and the shift cannot be reversed to get the original number back.

⚡ Both directions still fill the empty spaces with 0 — but it's the bits pushed off the opposite end that cause the problem.

Data Loss — Left Shift

Shifting 11000000 (192) left by 1

256	128	64	32	16	8	4	2	1
	1	1	0	0	0	0	0	0

←1 place

1

0

The leading 1 (worth 256) is pushed off the end and lost.

RESULT

Only 8 bits remain: 10000000 = 128. We expected 192 × 2 = 384, but only 128 is stored.

⚡ The multiply-by-2 rule breaks down when a significant bit is shifted off the register.

Data Loss — Right Shift

Shifting 00000111 (7) right by 1

128	64	32	16	8	4	2	1	lost
0	0	0	0	0	1	1	1

→1 place

0

1

The trailing 1 (worth 1) is pushed off the right end and lost.

RESULT

Remaining 8 bits: 00000011 = 3. We expected 7 ÷ 2 = 3.5, but binary can't store fractions — the result is rounded down to 3.

⚡ Check it can't be reversed: shifting 00000011 (3) left by 1 gives 6, not 7.

Exam Technique

Identifying a shift from before and after

EXAM-STYLE TASK

A question may give a binary number before and after a shift, and ask you to state the direction and number of places shifted.

	128	64	32	16	8	4	2	1
Before = 3	0	0	0	0	0	0	1	1
After = 24	0	0	0	1	1	0	0	0

Direction: the bits have moved towards the most significant end — this is a left shift.

PLACES

3 → 24 is ×8, and ×8 = 2³, so this is a shift left by 3 places. Exam answer: "Left shift, 3 places".

Exam Practice

Exam-Style Questions — Binary Shifts

Question 1 · 2 marks

Shift the binary number 00001011 left by 2 places. Give your answer in binary and state the effect on the denary value.

ANS

00101100 — denary value is multiplied by 4 (11 × 4 = 44)

Question 2 · 2 marks

Shift the binary number 11100000 right by 3 places. Give your answer in binary and denary.

ANS

00011100 = 28 (224 ÷ 8 = 28, no bits lost)

Common Mistakes

Four mistakes that cost marks in the exam

1

Forgetting to fill empty places with 0. After any shift, the spaces left behind must be filled with 0 — never left blank.

2

Mixing up the direction. Left shift = multiply (value gets bigger). Right shift = divide (value gets smaller).

3

Assuming ×2ⁿ / ÷2ⁿ always holds exactly. If a bit is shifted off the end of the register, it is lost — the result will not be exactly the expected multiple or fraction.

4

Getting the number of places wrong. Shift by 1 place = ×2 or ÷2. Shift by n places = ×2ⁿ or ÷2ⁿ — not ×2 multiplied by n.

Summary

1.2.4d — Binary Shifts

LEFT SHIFT

Multiplies by a power of 2. n places = ×2ⁿ. New 0s fill in on the right.

RIGHT SHIFT

Divides by a power of 2. n places = ÷2ⁿ. New 0s fill in on the left.

DATA LOSS

Bits shifted off the end of the register are permanently lost — left shift loses significant bits, right shift loses precision.

EXAM SKILL

Compare before/after binary to identify direction and number of places shifted.

⚡ Exam phrasing: "Shifting left by n places multiplies the value by 2ⁿ; shifting right by n places divides the value by 2ⁿ. Bits shifted beyond the register are lost."

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That's 1.2.4d done.

Next up: 1.2.4e — Character Encoding

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