C++ Program to Toggle Bits of a Number

To toggle bits in C++, use the bitwise XOR operator ^ with a mask. For example, result = num ^ mask; toggles bits where mask has 1s.

Examples

Inputnum = 5, mask = 1

Outputresult = 4

Inputnum = 10, mask = 7

Outputresult = 13

Inputnum = 0, mask = 255

Outputresult = 255

How to Think About It

To toggle bits, think of flipping bits from 0 to 1 or 1 to 0. Using the XOR operator ^ with a mask that has 1s in positions to toggle will flip those bits in the number, leaving others unchanged.

Algorithm

1

Get the input number and the mask indicating which bits to toggle.

2

Apply the XOR operator between the number and the mask.

3

Store the result as the toggled number.

4

Return or print the toggled number.

Code

cpp

#include <iostream>
using namespace std;

int main() {
    int num = 10; // original number
    int mask = 7; // bits to toggle
    int result = num ^ mask; // toggle bits using XOR
    cout << "Original number: " << num << endl;
    cout << "Mask: " << mask << endl;
    cout << "Number after toggling bits: " << result << endl;
    return 0;
}

Output

Original number: 10
Mask: 7
Number after toggling bits: 13

Dry Run

Let's trace toggling bits of number 10 with mask 7 through the code

1

Initial values

num = 10 (binary 1010), mask = 7 (binary 0111)

2

Apply XOR

result = num ^ mask = 1010 ^ 0111 = 1101 (binary)

3

Convert result

1101 binary = 13 decimal

num (binary)	mask (binary)	result (binary)	result (decimal)
1010	0111	1101	13

Why This Works

Step 1: Using XOR to toggle bits

The XOR operator ^ flips bits where the mask has 1s and keeps bits unchanged where the mask has 0s.

Step 2: Mask defines toggle positions

The mask is a number with 1s in bit positions you want to toggle and 0s elsewhere.

Step 3: Result is the toggled number

Applying num ^ mask produces a new number with specified bits flipped.

Alternative Approaches

Toggle single bit using shift

cpp

#include <iostream>
using namespace std;

int main() {
    int num = 10;
    int bit_pos = 1; // toggle bit at position 1 (0-based)
    int result = num ^ (1 << bit_pos);
    cout << "Original number: " << num << endl;
    cout << "Number after toggling bit " << bit_pos << ": " << result << endl;
    return 0;
}

This toggles only one bit at a time by shifting 1 to the desired position.

Toggle all bits using NOT operator

cpp

#include <iostream>
using namespace std;

int main() {
    unsigned int num = 10;
    unsigned int result = ~num;
    cout << "Original number: " << num << endl;
    cout << "Number after toggling all bits: " << result << endl;
    return 0;
}

This flips all bits of the number, not selective bits.

Complexity: O(1) time, O(1) space

Time Complexity

Bitwise XOR is a single CPU operation, so toggling bits runs in constant time.

Space Complexity

No extra memory is needed besides variables to hold numbers, so space is constant.

Which Approach is Fastest?

Using XOR with a mask is the fastest and simplest way to toggle bits compared to looping or other methods.

Approach	Time	Space	Best For
XOR with mask	O(1)	O(1)	Toggling specific bits quickly
Shift and XOR single bit	O(1)	O(1)	Toggling one bit at a time
NOT operator	O(1)	O(1)	Flipping all bits