Reverse a Number in C++

Reversing a number — turning 1234 into 4321 — is one of the most common beginner exercises, and for good reason. It teaches you the two-step pattern of pulling digits off a number one at a time, which shows up everywhere from digit sums to palindrome checks.

The Two Operations You Need

To reverse a number, you process it digit by digit using two operators:

• number % 10 gives you the last digit (the remainder after dividing by 10).
• number / 10 removes the last digit (integer division drops the fraction).

For example, with 1234: 1234 % 10 is 4, and 1234 / 10 is 123. Do that repeatedly and you visit every digit from right to left.

#include <iostream>

int main() {
    int number = 1234;
    std::cout << "Last digit: " << number % 10 << "\n";  // 4
    std::cout << "Rest: " << number / 10 << "\n";         // 123
    return 0;
}

Building the Reversed Number

The reversal itself uses a while loop. Each pass, we take the last digit and append it to a reversed value. To “append,” we shift reversed one place to the left by multiplying by 10, then add the new digit:

#include <iostream>

int main() {
    int number = 1234;
    int reversed = 0;

    while (number != 0) {
        int digit = number % 10;      // grab last digit
        reversed = reversed * 10 + digit;  // shift left, add digit
        number = number / 10;         // drop last digit
    }

    std::cout << "Reversed: " << reversed << "\n";  // 4321
    return 0;
}

Trace it: reversed goes 0 -> 4 -> 43 -> 432 -> 4321, while number shrinks 1234 -> 123 -> 12 -> 1 -> 0. When number hits 0, the loop ends and reversed holds the answer.

Handling Negative Numbers

If the user might enter a negative number like -123, the cleanest approach is to remember the sign, reverse the absolute value, then put the sign back:

#include <iostream>
#include <cstdlib>   // for std::abs

int main() {
    int number;
    std::cout << "Enter a number: ";
    std::cin >> number;

    int sign = (number < 0) ? -1 : 1;
    number = std::abs(number);

    int reversed = 0;
    while (number != 0) {
        reversed = reversed * 10 + number % 10;
        number /= 10;
    }

    std::cout << "Reversed: " << sign * reversed << "\n";
    return 0;
}

Reversing the absolute value keeps the loop simple, and reapplying sign at the end gives -123 -> -321.

A Note on Overflow

Reversing a large number can produce a value too big for an int. For example, reversing 1000000009 overflows a 32-bit int. For beginner exercises with small inputs this rarely bites, but if you want to be safe, use a wider type like long long for the reversed variable:

long long reversed = 0;

This is a good habit: whenever a calculation can grow past the input’s size, give the result room to breathe.

Where This Pattern Leads

The “peel a digit, shrink the number” loop is a building block. Swap the reversed = reversed * 10 + digit line for sum += digit and you get the sum of digits. Compare the reversed number to the original and you get a palindrome check. Master this one loop and a dozen classic problems open up.

Related Articles

• Reverse a String in C++ — the string version of the same idea
• C++ Modulo Operator (%) — how digit extraction works
• Integer Division in C++ — why number / 10 drops a digit
• Palindrome Program in C++ — reverse and compare
• C++ Loops Tutorial — while loops in depth

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