Binary Shifts (Cambridge (CIE) IGCSE Computer Science): Revision Note
Exam code: 0478 & 0984
Binary Shifts
What is a logical binary shift?
Examiner Tips and Tricks
Cambridge IGCSE 0478 expects you to perform binary shifts on 8-bit values and explain how the result changes (×2, ÷2). Every example here mirrors the real exam format.
A logical binary shift is how a computer system performs basic multiplication and division on non-negative values (0 and positive numbers)
Binary digits are moved left or right a set number of times
A left shift multiplies a binary number by 2 (x2)
A right shift divides a binary number by 2 (/2)
A shift can move more than one place at a time, the principle remains the same
A left shift of 2 places would multiply the original binary number by 4 (x4)
How do you perform a logical left shift of 1?
Here is the binary representation of the denary number 40
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|
0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
To perform a left logical binary shift of 1, we move each bit 1 place to the left
Since the most significant bit is 0, there is no overflow
The 1 column becomes empty so is filled with a 0
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|
|---|---|---|---|---|---|---|---|---|
0 | 1 | 0 | 1 | 0 | 0 | 0 | = 40 | |
0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | = 80 |
The original binary representation of denary 40 (32+8) was multiplied by 2 and became 80 (64+16)
How do you perform a logical left shift of 2?
Here is the binary representation of the denary number 28
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|
0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 |
To perform a left binary shift of 2, we move each bit 2 places to the left
Since the two leftmost bits are 0, nothing important is lost and no overflow occurs
The 1 and 2 column become empty so are filled with a 0
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|
|---|---|---|---|---|---|---|---|---|
0 | 1 | 1 | 1 | 0 | 0 | = 28 | ||
0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | = 112 |
The original binary representation of denary 28 (16+8+4) was multiplied by 4 and became 112 (64+32+16)
Examiner Tips and Tricks
Your textbook might show shifts with longer binary values—but in IGCSE exams, you’ll only ever be asked about 8-bit unsigned integers. That’s why all our examples are capped at 8 bits.
How do you perform a logical right shift of 1?
Here is the binary representation of the denary number 40
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|
0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
To perform a right binary shift of 1, we move each bit 1 place to the right
The bit in the 1 column (LSB) is shifted out and lost
The 128 column becomes empty so is filled with a 0
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|
|---|---|---|---|---|---|---|---|---|
0 | 0 | 1 | 0 | 1 | 0 | 0 | = 40 | |
0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | = 20 |
The original binary representation of denary 40 (32+8) was divided by 2 and became 20 (16+4)
How do you perform a logical right shift of 2?
Here is the binary representation of the denary number 200
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|
1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
To perform a right binary shift of 2, we move each bit 2 places to the right
The bits in the 1 and 2 columns are shifted out and lost
The 128 and 64 columns become empty so are filled with a 0
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|
|---|---|---|---|---|---|---|---|---|
1 | 1 | 0 | 0 | 1 | 0 | = 200 | ||
0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | = 50 |
The original binary representation of denary 200 (128+64+8) was divided by 4 and became 50 (32+16+2)
Overflow in binary shifts
Overflow happens when a 1 is shifted out of the most significant bit (MSB) on the left in a logical left shift
This means important data is lost, which can seriously change the number’s value
Examiner Tips and Tricks
In IGCSE Computer Science, you usually only need to spot an overflow when a 1 is shifted out on the left. You don’t need to worry about underflow in this context!
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