Why Set Clear Toggle a Specific Bit in DSA Go?

The Big Idea

Discover how a single magic finger can flip any switch instantly without disturbing the rest!

The Scenario

Imagine you have a row of light switches representing different features in a device. You want to turn one specific feature on, off, or flip its state without disturbing the others.

The Problem

Manually checking and changing each switch one by one is slow and prone to mistakes. You might accidentally change the wrong switch or forget to update the right one.

The Solution

Using bit operations lets you directly set, clear, or toggle a specific switch (bit) quickly and safely without affecting others. This is like having a magic finger that flips only the switch you want.

Before vs After

✗ Before

flags := 0
// To turn on feature 3
flags = flags + 8  // manually adding 2^3
// To turn off feature 3
flags = flags - 8  // manually subtracting 2^3

✓ After

flags := 0
// Set bit 3
flags |= 1 << 3
// Clear bit 3
flags &^= 1 << 3
// Toggle bit 3
flags ^= 1 << 3

What It Enables

This lets you control individual features or options efficiently in programs, saving time and avoiding errors.

Real Life Example

In a game, you can quickly turn on or off abilities like "invisibility" or "speed boost" by toggling bits representing each ability.

Key Takeaways

Manual changes to bits are slow and error-prone.

Bit operations let you change one bit without affecting others.

This makes feature control fast, safe, and clear.

Practice

1. What does the operation num |= (1 << 3) do to the number num?

easy

A. Sets the 4th bit of num to 1

B. Clears the 4th bit of num to 0

C. Toggles the 4th bit of num

D. Checks if the 4th bit of num is set

Solution

1. Step 1: Understand bit shifting

   The expression 1 << 3 shifts 1 left by 3 bits, resulting in binary 00001000, which targets the 4th bit.

2. Step 2: Apply bitwise OR to set bit

   Using |= with this mask sets the 4th bit of num to 1 without changing other bits.

3. Final Answer:

   Sets the 4th bit of num to 1 -> Option A

4. Quick Check:

   Set bit = |= (1 << position) [OK]

Hint: Use OR with shifted 1 to set a bit [OK]

Common Mistakes:

• Confusing set with clear operation
• Using wrong bit position (off by one)
• Thinking it toggles instead of sets

2. Which of the following Go code snippets correctly clears the 2nd bit (bit index 1) of an integer variable num?

easy

A. num &= (1 << 1)

B. num |= (1 << 1)

C. num &= ^(1 << 1)

D. num ^= (1 << 1)

Solution

1. Step 1: Identify clearing operation

   To clear a bit, we AND num with the inverse mask that has 0 at the target bit and 1 elsewhere.

2. Step 2: Use bitwise AND with negated mask

   ^(1 << 1) creates a mask with all bits 1 except the 2nd bit 0. ANDing clears that bit.

3. Final Answer:

   num &= ^(1 << 1) -> Option C

4. Quick Check:

   Clear bit = &= ^(1 << position) [OK]

Hint: Use AND with negated shifted 1 to clear bit [OK]

Common Mistakes:

• Using OR instead of AND to clear
• Not negating the mask
• Confusing bit index counting

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

package main
import "fmt"
func main() {
    num := 10 // binary 1010
    num ^= (1 << 1)
    fmt.Println(num)
}

medium

A. 10

B. 8

C. 12

D. 14

Solution

1. Step 1: Understand initial value and bit positions

   Number 10 in binary is 1010. Bit positions (0-based) from right: bit0=0, bit1=1, bit2=0, bit3=1.

2. Step 2: Toggle 2nd bit (bit index 1) using XOR

   XOR with (1 << 1) toggles bit1. Bit1 is 1, toggling it changes to 0. New binary: 1000 which is 8.

3. Final Answer:

   8 -> Option B

4. Quick Check:

   Toggle bit = ^= (1 << position) [OK]

Hint: XOR toggles bit: 1->0 or 0->1 [OK]

Common Mistakes:

• Confusing toggle with set or clear
• Misreading bit positions
• Calculating decimal incorrectly

4. The following Go code is intended to clear the 3rd bit (bit index 2) of num. What is the error?

num := 15
num &= (1 << 2)

medium

A. It sets the 3rd bit instead of clearing it

B. No error, it correctly clears the 3rd bit

C. It toggles the 3rd bit instead of clearing

D. It clears all bits except the 3rd bit

Solution

1. Step 1: Analyze the operation performed

   The code does num &= (1 << 2), which ANDs num with 00000100 (only 3rd bit set).

2. Step 2: Understand effect on bits

   This clears all bits except the 3rd bit, effectively keeping only the 3rd bit if it was set, not clearing it.

3. Final Answer:

   It clears all bits except the 3rd bit -> Option D

4. Quick Check:

   Clear bit = &= ^(1 << position) [OK]

Hint: Use AND with negated mask to clear bit [OK]

Common Mistakes:

• Using AND with mask instead of negated mask
• Confusing clear with set or toggle
• Assuming &= (1 << n) clears bit

5. You have an integer num initially 0. You want to toggle the 1st bit, set the 3rd bit, and clear the 2nd bit in this order. Which sequence of operations achieves this correctly in Go?

hard

A. num ^= (1 << 0); num |= (1 << 2); num &= ^(1 << 1)

B. num |= (1 << 0); num ^= (1 << 2); num &= ^(1 << 1)

C. num &= ^(1 << 1); num ^= (1 << 0); num |= (1 << 2)

D. num |= (1 << 2); num ^= (1 << 0); num &= ^(1 << 1)

Solution

1. Step 1: Toggle 1st bit (bit 0)

   Use XOR with (1 << 0) to toggle the 1st bit: num ^= (1 << 0).

2. Step 2: Set 3rd bit (bit 2)

   Use OR with (1 << 2) to set the 3rd bit: num |= (1 << 2).

3. Step 3: Clear 2nd bit (bit 1)

   Use AND with negated mask to clear 2nd bit: num &= ^(1 << 1).

4. Final Answer:

   num ^= (1 << 0); num |= (1 << 2); num &= ^(1 << 1) -> Option A

5. Quick Check:

   Toggle, Set, Clear in order with ^=, |=, &= ^ [OK]

Hint: Toggle ^=, Set |=, Clear &= ^ in correct order [OK]

Common Mistakes:

• Mixing order of operations
• Using wrong operators for each action
• Not negating mask when clearing