Left shift and right shift in C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the left shift operator (<<) do in C?

It moves the bits of a number to the left by a specified number of positions, effectively multiplying the number by 2 for each shift.

beginner

What is the effect of the right shift operator (>>) in C?

It moves the bits of a number to the right by a specified number of positions, effectively dividing the number by 2 for each shift (for unsigned numbers).

beginner

Explain what happens when you do: int x = 4; int y = x << 1;

The value of x (4) is shifted left by 1 bit, which multiplies it by 2. So y becomes 8.

intermediate

What is a potential risk when using left shift on signed integers?

Shifting bits into the sign bit can cause undefined behavior or unexpected negative values.

intermediate

How does right shift behave differently for signed and unsigned integers in C?

For unsigned integers, right shift fills with zeros (logical shift). For signed integers, it may fill with the sign bit (arithmetic shift), preserving the sign.

What is the result of 3 << 2 in C?

A 12

B 6

C 1

D 8

3 in binary is 00000011. Shifting left by 2 bits gives 00001100, which is 12.

What does the right shift operator (>>) do to the bits of an unsigned integer?

A Fills left bits with zeros

B Fills left bits with ones

C Reverses the bits

D Swaps the bits

Right shift on unsigned integers fills the left bits with zeros (logical shift).

Which operator would you use to multiply an integer by 8 using bit shifts?

A << 2

B >> 3

C << 3

D >> 2

Shifting left by 3 bits multiplies the number by 2^3 = 8.

What happens if you left shift a signed integer into its sign bit?

A No change

B Always zero

C Always positive

D Undefined behavior or negative value

Shifting into the sign bit can cause undefined behavior or change the sign unexpectedly.

If int x = -16; what might x >> 2 do?

A Divide by 4 and fill with zeros

B Divide by 4 and keep the sign

C Multiply by 4

D Cause a compile error

Right shift on signed integers usually performs arithmetic shift, preserving the sign.

Explain how left shift and right shift operators work in C and give a simple example for each.

Think about how shifting bits changes the number's value.

What are the differences in behavior of right shift on signed versus unsigned integers in C?

Consider how negative numbers are handled.

Practice

1. What does the left shift operator << do to an integer in C?

easy

A. Moves bits to the left, multiplying the number by powers of two

B. Moves bits to the right, dividing the number by powers of two

C. Reverses the bits of the number

D. Sets all bits to zero

Solution

1. Step 1: Understand left shift operation

   The left shift operator moves all bits in the number to the left by the specified count.

2. Step 2: Effect on the number's value

   Each left shift multiplies the number by 2, so shifting left by n multiplies by 2^n.

3. Final Answer:

   Moves bits to the left, multiplying the number by powers of two -> Option A

4. Quick Check:

   Left shift = multiply by 2^n [OK]

Hint: Left shift multiplies by 2 to the power of shift count [OK]

Common Mistakes:

• Confusing left shift with right shift
• Thinking left shift divides the number
• Assuming left shift reverses bits

2. Which of the following is the correct syntax to shift the integer variable x right by 3 bits in C?

easy

A. x >>> 3;

B. x >> 3;

C. x << 3;

D. x >>3

Solution

1. Step 1: Identify the right shift operator in C

   The right shift operator is >> and requires spaces or no spaces around it.

2. Step 2: Check syntax correctness

   x >> 3; is correct syntax. x << 3; is left shift, x >>> 3; is invalid in C, and x >>3 is missing a semicolon.

3. Final Answer:

   x >> 3; -> Option B

4. Quick Check:

   Right shift syntax = x >> n; [OK]

Hint: Right shift uses >> with semicolon to end statement [OK]

Common Mistakes:

• Using >>> which is not valid in C
• Confusing left shift << with right shift >>
• Omitting semicolon at end

3. What is the output of the following C code?

int x = 4;
int y = x << 2;
printf("%d", y);

medium

A. 0

B. 1

C. 8

D. 16

Solution

1. Step 1: Understand the initial value of x

   x is 4, which in binary is 00000100.

2. Step 2: Apply left shift by 2 bits

   Shifting left by 2 bits multiplies 4 by 2^2 = 4 * 4 = 16.

3. Final Answer:

   16 -> Option D

4. Quick Check:

   4 << 2 = 16 [OK]

Hint: Left shift by n multiplies by 2^n [OK]

Common Mistakes:

• Calculating 4 << 2 as 8 instead of 16
• Confusing left shift with right shift
• Ignoring the shift count

4. Find the error in this C code snippet:

int a = 32;
int b = a >> 2
printf("%d", b);

medium

A. Wrong operator used for right shift

B. Incorrect printf format specifier

C. Missing semicolon after right shift operation

D. Variable b is not declared

Solution

1. Step 1: Check syntax line by line

   The line int b = a >> 2 is missing a semicolon at the end.

2. Step 2: Verify other parts

   The operator >> is correct, printf uses %d for int, and b is declared properly.

3. Final Answer:

   Missing semicolon after right shift operation -> Option C

4. Quick Check:

   Every statement needs semicolon [OK]

Hint: Check for missing semicolons after shift operations [OK]

Common Mistakes:

• Assuming >> is invalid operator
• Thinking printf format is wrong
• Ignoring missing semicolon errors

5. Given int x = 1;, what is the value of y after this operation?

int y = (x << 3) >> 1;

hard

A. 4

B. 8

C. 16

D. 2

Solution

1. Step 1: Calculate left shift (x << 3)

   Left shifting 1 by 3 bits multiplies 1 by 2^3 = 8.

2. Step 2: Calculate right shift of result by 1

   Right shifting 8 by 1 bit divides 8 by 2^1 = 2, so 8 / 2 = 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   (1 << 3) >> 1 = 8 >> 1 = 4 [OK]

Hint: Left shift then right shift: multiply then divide by powers of two [OK]

Common Mistakes:

• Calculating (1 << 3) as 16
• Ignoring right shift effect
• Mixing up shift directions