C++ Program to Reverse Number Using While Loop

You can reverse a number in C++ using a while loop by extracting digits with num % 10, building the reversed number with rev = rev * 10 + digit, and reducing the original number with num /= 10 until it becomes zero.

Examples

Input1234

Output4321

Input1000

Output1

Input7

Output7

How to Think About It

To reverse a number, think of peeling off the last digit repeatedly using the remainder operator %. Then, build the reversed number by shifting the current reversed number left (multiply by 10) and adding the peeled digit. Keep doing this until the original number is zero.

Algorithm

1

Get the input number.

2

Initialize reversed number as 0.

3

While the input number is greater than 0:

4

Extract the last digit using remainder operator.

5

Add this digit to reversed number after shifting reversed number left by one digit.

6

Remove the last digit from input number by dividing it by 10.

7

After loop ends, return the reversed number.

Code

cpp

#include <iostream>
using namespace std;

int main() {
    int num, rev = 0;
    cout << "Enter a number: ";
    cin >> num;
    int temp = num;
    while (temp > 0) {
        int digit = temp % 10;
        rev = rev * 10 + digit;
        temp /= 10;
    }
    cout << "Reversed number: " << rev << endl;
    return 0;
}

Output

Enter a number: 1234
Reversed number: 4321

Dry Run

Let's trace reversing the number 1234 through the code.

1

Initial values

num = 1234, rev = 0, temp = 1234

2

First iteration

digit = 1234 % 10 = 4; rev = 0 * 10 + 4 = 4; temp = 1234 / 10 = 123

3

Second iteration

digit = 123 % 10 = 3; rev = 4 * 10 + 3 = 43; temp = 123 / 10 = 12

4

Third iteration

digit = 12 % 10 = 2; rev = 43 * 10 + 2 = 432; temp = 12 / 10 = 1

5

Fourth iteration

digit = 1 % 10 = 1; rev = 432 * 10 + 1 = 4321; temp = 1 / 10 = 0

6

Loop ends

temp is 0, reversed number is 4321

temp	digit	rev
1234	4	4
123	3	43
12	2	432
1	1	4321

Why This Works

Step 1: Extract last digit

Using temp % 10 gets the last digit of the number.

Step 2: Build reversed number

Multiply current reversed number by 10 to shift digits left, then add the extracted digit.

Step 3: Remove last digit

Divide the number by 10 using temp /= 10 to discard the last digit.

Alternative Approaches

Using recursion

cpp

#include <iostream>
using namespace std;

int reverse(int num, int rev = 0) {
    if (num == 0) return rev;
    return reverse(num / 10, rev * 10 + num % 10);
}

int main() {
    int num;
    cout << "Enter a number: ";
    cin >> num;
    cout << "Reversed number: " << reverse(num) << endl;
    return 0;
}

Recursion is elegant but uses extra call stack memory compared to the while loop.

Using string conversion

cpp

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main() {
    string numStr;
    cout << "Enter a number: ";
    cin >> numStr;
    reverse(numStr.begin(), numStr.end());
    cout << "Reversed number: " << numStr << endl;
    return 0;
}

This method is simple but converts number to string, which may be less efficient.

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

Time Complexity

The loop runs once for each digit in the number, so time complexity is O(d), where d is the number of digits.

Space Complexity

Only a few integer variables are used, so space complexity is O(1), constant space.

Which Approach is Fastest?

The while loop approach is fastest and most memory efficient compared to recursion or string conversion.

Approach	Time	Space	Best For
While loop	O(d)	O(1)	Efficient and simple for numeric reversal
Recursion	O(d)	O(d)	Elegant but uses call stack memory
String conversion	O(d)	O(d)	Simple code, less efficient for large numbers