Left shift and right shift behavior in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of left shift on unsigned integer

What is the output of this C code snippet when compiled and run on a typical 32-bit system?

Embedded C

unsigned int x = 1;
unsigned int y = x << 3;
printf("%u", y);

A 8

B 16

C 0

D 1

💡 Hint

Left shifting by 3 bits multiplies the number by 2^3.

Left shifting 1 by 3 bits moves the bit pattern 3 places to the left, resulting in 8.

❓ Predict Output

intermediate

Right shift on signed negative integer

What is the output of this C code snippet on a system where right shift of signed integers is arithmetic?

Embedded C

int x = -16;
int y = x >> 2;
printf("%d", y);

A 0

B 1073741820

C -1

D -4

💡 Hint

Right shift on signed negative integers usually fills with sign bit (arithmetic shift).

Arithmetic right shift divides by powers of two rounding toward negative infinity, so -16 >> 2 is -4.

❓ Predict Output

advanced

Behavior of left shift causing overflow

What is the output of this C code snippet on a 32-bit unsigned integer?

Embedded C

unsigned int x = 0x80000000;
unsigned int y = x << 1;
printf("%u", y);

A 4294967295

B 0

C 2147483648

D 1

💡 Hint

Left shifting the highest bit moves it out of the 32-bit range.

Shifting 0x80000000 left by 1 bit discards the highest bit, resulting in 0.

❓ Predict Output

advanced

Right shift on unsigned integer with large shift count

What is the output of this C code snippet on a 32-bit unsigned integer?

Embedded C

unsigned int x = 1;
unsigned int y = x >> 33;
printf("%u", y);

A 0

B 1

C Undefined behavior

D 2

💡 Hint

Shifting by more than or equal to the bit width results in zero for unsigned types.

Right shifting 1 by 33 bits on a 32-bit unsigned integer results in 0 because bits shifted out are lost.

🧠 Conceptual

expert

Difference between logical and arithmetic right shift

Which statement correctly describes the difference between logical and arithmetic right shift in C?

A Logical right shift fills with zeros; arithmetic right shift fills with the sign bit.

B Logical right shift fills with the sign bit; arithmetic right shift fills with zeros.

C Both logical and arithmetic right shifts fill with zeros regardless of sign.

D Both logical and arithmetic right shifts fill with the sign bit regardless of sign.

💡 Hint

Think about how negative numbers behave when shifted right.

Logical right shift always fills with zeros; arithmetic right shift preserves the sign by filling with the sign bit.

Practice

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

easy

A. Divides the number by powers of two

B. Inverts all bits of the number

C. Multiplies the number by powers of two

D. Leaves the number unchanged

Solution

1. Step 1: Understand left shift operation

   The left shift operator moves bits to the left, adding zeros on the right.

2. Step 2: Effect on unsigned integer value

   Shifting left by 1 bit multiplies the number by 2, shifting by n bits multiplies by 2^n.

3. Final Answer:

   Multiplies the number by powers of two -> Option C

4. Quick Check:

   Left shift = multiply by 2^n [OK]

Hint: Left shift multiplies by 2 for each shift [OK]

Common Mistakes:

• Thinking left shift divides the number
• Confusing left shift with bitwise NOT
• Assuming left shift leaves number unchanged

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

easy

A. x >>3

B. x << 3;

C. x >>> 3;

D. x >> 3;

Solution

1. Step 1: Identify right shift operator syntax

   In embedded C, right shift is done using >> operator with spaces optional.

2. Step 2: Check options for correct syntax

   The correct syntax is x >> 3; using the right shift operator >>, shift amount 3, and semicolon to form a valid statement. Lacking semicolon, using left shift <<, or invalid operator like >>> are incorrect.

3. Final Answer:

   x >> 3; -> Option D

4. Quick Check:

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

Hint: Right shift uses >> operator with semicolon [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 code snippet?

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

medium

A. 16

B. 8

C. 1

D. 32

Solution

1. Step 1: Understand initial value and shift

   Variable x is 4 (binary 100). Left shifting by 2 bits moves bits left twice.

2. Step 2: Calculate shifted value

   4 << 2 = 4 * 2^2 = 4 * 4 = 16.

3. Final Answer:

   16 -> Option A

4. Quick Check:

   4 << 2 = 16 [OK]

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

Common Mistakes:

• Confusing left shift with right shift
• Calculating 4 << 2 as 8 instead of 16
• Ignoring unsigned type behavior

4. Consider this code snippet:

int x = -16;
int y = x >> 2;
printf("%d", y);

What is the likely output assuming arithmetic right shift?

medium

A. -4

B. 4

C. -1

D. 0

Solution

1. Step 1: Understand signed right shift behavior

   For signed integers, right shift usually preserves the sign bit (arithmetic shift).

2. Step 2: Calculate shifted value

   -16 in binary (two's complement) is shifted right by 2 bits, dividing by 4 but keeping sign, so -16 >> 2 = -4.

3. Final Answer:

   -4 -> Option A

4. Quick Check:

   Signed right shift divides and keeps sign [OK]

Hint: Signed right shift divides and keeps sign bit [OK]

Common Mistakes:

• Assuming right shift always fills with zeros
• Confusing signed and unsigned shift results
• Expecting positive result from negative input

5. You want to multiply an unsigned 8-bit integer val by 8 using bit shifts without overflow. Which code snippet is correct?

unsigned char val = 20;
unsigned char result = val << 3;

What should you check before shifting to avoid overflow?

hard

A. Ensure val <= 8 before shifting

B. Ensure val <= 31 before shifting

C. Ensure val <= 20 before shifting

D. No check needed, shifting never overflows

Solution

1. Step 1: Understand shift and overflow risk

   Shifting left by 3 multiplies by 8. For 8-bit unsigned, max value is 255.

2. Step 2: Calculate max safe value

   To avoid overflow: val * 8 <= 255 -> val <= 31.875, so val must be ≤ 31.

3. Final Answer:

   Ensure val <= 31 before shifting -> Option B

4. Quick Check:

   Max val before shift = 255 / 8 = 31 [OK]

Hint: Check max value before shift to prevent overflow [OK]

Common Mistakes:

• Ignoring overflow risk on left shift
• Assuming shifting never causes overflow
• Using wrong max value for 8-bit unsigned