Bit Manipulation Basics AND OR XOR NOT Left Right Shift in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the AND (&) bitwise operator do between two bits?

The AND operator returns 1 only if both bits are 1; otherwise, it returns 0.
Example: 1 & 1 = 1, 1 & 0 = 0, 0 & 0 = 0.

beginner

Explain the OR (|) bitwise operator with an example.

The OR operator returns 1 if at least one of the bits is 1; otherwise, it returns 0.
Example: 1 | 0 = 1, 0 | 0 = 0.

beginner

What is the result of XOR (^) between two bits?

XOR returns 1 if the bits are different, and 0 if they are the same.
Example: 1 ^ 0 = 1, 1 ^ 1 = 0.

beginner

What does the NOT (~) operator do to a bit?

The NOT operator flips the bit: 0 becomes 1, and 1 becomes 0.
Example: ~0 = 1, ~1 = 0 (in single bit view).

beginner

What happens when you perform a left shift (<<) on a number?

Left shift moves all bits to the left by the given number of positions.
Each shift to the left multiplies the number by 2.
Example: 3 << 1 = 6 (binary 11 becomes 110).

What is the result of 5 & 3 in binary? (5 = 0101, 3 = 0011)

A 0111

B 0001

C 0010

D 0100

5 & 3 = 0101 & 0011 = 0001 (only the last bit is 1 in both).

Which operator flips all bits of a number?

A AND (&)

B OR (|)

C XOR (^)

D NOT (~)

The NOT (~) operator flips every bit from 0 to 1 and 1 to 0.

What does the expression 6 ^ 3 evaluate to? (6 = 0110, 3 = 0011)

A 0101

B 0111

C 0011

D 0001

6 ^ 3 = 0110 ^ 0011 = 0101 (bits differ in positions 2 and 0).

What is the effect of right shifting a number by 1 (>> 1)?

A Flips all bits

B Multiplies the number by 2

C Divides the number by 2

D Sets all bits to zero

Right shift moves bits right, effectively dividing the number by 2 (dropping remainder).

Which bitwise operator returns 1 only when exactly one bit is 1?

A XOR (^)

B AND (&)

C OR (|)

D NOT (~)

XOR returns 1 only when bits differ (exactly one bit is 1).

Describe how the AND, OR, XOR, and NOT bitwise operators work with simple examples.

Think about how each operator treats pairs of bits.

Explain what happens when you left shift and right shift a binary number by 1 position.

Imagine moving digits in a number left or right.

Practice

1. What does the bitwise AND operator (&) do when applied to two bits?

easy

A. Returns 0 only if both bits are 0

B. Returns 1 if either bit is 1

C. Flips the bits from 0 to 1 and 1 to 0

D. Returns 1 only if both bits are 1

Solution

1. Step 1: Understand bitwise AND operation

   The AND operator compares each bit of two numbers and returns 1 only if both bits are 1.

2. Step 2: Apply to example bits

   For bits 1 & 1, result is 1; for 1 & 0 or 0 & 1 or 0 & 0, result is 0.

3. Final Answer:

   Returns 1 only if both bits are 1 -> Option D

4. Quick Check:

   AND = 1 only if both bits 1 [OK]

Hint: AND returns 1 only when both bits are 1 [OK]

Common Mistakes:

• Confusing AND with OR operator
• Thinking AND flips bits
• Assuming AND returns 1 if either bit is 1

2. Which of the following is the correct syntax to perform a left shift by 2 on an integer variable x in C#?

easy

A. x >> 2;

B. x << 2;

C. x <<< 2;

D. x >> 2 <<;

Solution

1. Step 1: Recall left shift operator syntax in C#

   The left shift operator is << and used as x << n to shift bits left by n places.

2. Step 2: Identify correct syntax among options

   x << 2; uses x << 2; which is correct. x >> 2; is right shift, C uses invalid operator, D is invalid syntax.

3. Final Answer:

   x << 2; -> Option B

4. Quick Check:

   Left shift syntax is x << n [OK]

Hint: Use << for left shift, >> for right shift [OK]

Common Mistakes:

• Using >>> which is invalid in C#
• Confusing left shift with right shift
• Adding extra symbols after shift operator

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

int a = 5; // binary 0101
int b = 3; // binary 0011
int c = a ^ b;
Console.WriteLine(c);

medium

A. 2

B. 7

C. 6

D. 8

Solution

1. Step 1: Convert numbers to binary and apply XOR

   5 in binary is 0101, 3 is 0011. XOR (^) returns 1 if bits differ, else 0.

2. Step 2: Calculate XOR bit by bit

   0101 ^ 0011 = 0110 (binary), which is 6 in decimal.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   5 ^ 3 = 6 [OK]

Hint: XOR sets bits where inputs differ [OK]

Common Mistakes:

• Confusing XOR with OR or AND
• Calculating decimal sum instead of XOR
• Misreading binary values

4. Identify the error in the following C# code snippet that attempts to invert bits of an integer num:

int num = 12;
int inverted = ~num();
Console.WriteLine(inverted);

medium

A. Using parentheses after variable name num()

B. Using bitwise NOT operator incorrectly

C. Missing semicolon after int num = 12

D. Console.WriteLine cannot print integers

Solution

1. Step 1: Check usage of variable num

   The code uses num() which looks like a function call but num is an integer variable, so parentheses are invalid.

2. Step 2: Confirm bitwise NOT operator usage

   The bitwise NOT operator ~ is correctly used before the variable, so no error there.

3. Final Answer:

   Using parentheses after variable name num() -> Option A

4. Quick Check:

   Variables don't use () [OK]

Hint: Variables are not functions; no () after variable names [OK]

Common Mistakes:

• Thinking ~num() is valid syntax
• Assuming missing semicolon error
• Believing Console.WriteLine can't print integers

5. Given an integer n = 8, which expression correctly checks if n is a power of two using bit manipulation in C#?

hard

A. (n & (n - 1)) == 0

B. (n | (n - 1)) == 0

C. (n ^ (n - 1)) == 0

D. (~n & (n - 1)) == 0

Solution

1. Step 1: Understand the power of two property

   A number is a power of two if it has exactly one bit set to 1 in binary.

2. Step 2: Apply bitwise trick

   For such numbers, n & (n - 1) clears the lowest set bit. If result is 0, n was a power of two.

3. Final Answer:

   (n & (n - 1)) == 0 -> Option A

4. Quick Check:

   Power of two check uses n & (n-1) == 0 [OK]

Hint: Power of two if n & (n-1) equals zero [OK]

Common Mistakes:

• Using OR or XOR instead of AND
• Forgetting to compare result to zero
• Misunderstanding bitwise NOT usage