Left shift and right shift behavior in Embedded C

Introduction

Left and right shifts move bits in a number to the left or right. This helps multiply or divide numbers quickly in low-level programming.

When you want to multiply a number by powers of two quickly.

When you need to divide a number by powers of two without using division.

When working with hardware registers to set or clear specific bits.

When optimizing code for speed in embedded systems.

When manipulating flags or masks in bitwise operations.

Syntax

Embedded C

result = value << n;  // Left shift value by n bits
result = value >> n;  // Right shift value by n bits

Left shift (<<) moves bits to the left, adding zeros on the right.

Right shift (>>) moves bits to the right. For unsigned values, zeros are added on the left. For signed values, behavior may depend on the compiler (logical vs arithmetic shift).

Examples

Left shift by 1 doubles the number (5 * 2 = 10).

Embedded C

unsigned int x = 5;  // binary 0000 0101
unsigned int y = x << 1;  // y = 10 (0000 1010)

Right shift by 2 divides the number by 4 (20 / 4 = 5).

Embedded C

unsigned int x = 20;  // binary 0001 0100
unsigned int y = x >> 2;  // y = 5 (0000 0101)

Right shift on signed negative numbers may keep the sign bit to preserve the sign.

Embedded C

int x = -8;  // binary depends on system
int y = x >> 1;  // may keep sign bit (arithmetic shift)

Sample Program

This program shows how left and right shifts work on unsigned and signed integers. It prints the original values and the results after shifting.

Embedded C

#include <stdio.h>

int main() {
    unsigned int a = 3;  // binary 0000 0011
    unsigned int left_shift = a << 2;  // shift left by 2 bits
    unsigned int right_shift = a >> 1; // shift right by 1 bit

    printf("Original a = %u\n", a);
    printf("a << 2 = %u\n", left_shift);
    printf("a >> 1 = %u\n", right_shift);

    int b = -16;  // signed integer
    int b_right_shift = b >> 2;  // right shift signed
    printf("Original b = %d\n", b);
    printf("b >> 2 = %d\n", b_right_shift);

    return 0;
}

Important Notes

Left shifting a number by n bits multiplies it by 2^n, but be careful of overflow.

Right shifting unsigned numbers fills with zeros on the left (logical shift).

Right shifting signed numbers may fill with the sign bit (arithmetic shift), preserving the sign.

Summary

Left shift (<<) moves bits left, multiplying by powers of two.

Right shift (>>) moves bits right, dividing by powers of two for unsigned numbers.

Signed right shift may keep the sign bit to preserve negative values.

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