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# C++ Program to Check Whether a Number is Palindrome or Not Program to Check Whether a Number is Palindrome or Not and reverses an integer (entered by the user) using while loop. Then, if statement is used to check whether the reversed number is equal to the original number or not.

To understand this example to Check Whether a Number is a Palindrome or Not, you should have the knowledge of following C++ programming topics:

• C++ while and do…while Loop
• C++ if, if…else and Nested if…else

This program takes an integer from the user and that integer is reversed.

If the reversed integer is equal to the integer entered by the user then, that number is a palindrome if not that number is not a palindrome.

#### Program to Check Palindrome Number

```#include <iostream>
using namespace std;

int main()
{
int n, num, digit, rev = 0;

cout << "Enter a positive number: ";
cin >> num;

n = num;

do
{
digit = num % 10;
rev = (rev * 10) + digit;
num = num / 10;
} while (num != 0);

cout << " The reverse of the number is: " << rev << endl;

if (n == rev)
cout << " The number is a palindrome";
else
cout << " The number is not a palindrome";

return 0;
}```

Output

```Enter a positive number: 12321
The reverse of the number is: 12321
The number is a palindrome
```
```Enter a positive number: 12331
The reverse of the number is: 13321
The number is not a palindrome
``` In the above program, the user is asked to enter a positive number which is stored in the variable num.

The number is then saved into another variable n to check it when the original number has been reversed.

Inside the do…while loop, last digit of the number is separated using the code.`digit = num % 10;` This digit is then added to the rev variable.

Before adding the digit to rev, we first need to multiply the current data in the revvariable by 10 in order to add the digit to the nth place in the number.

For example: in the number 123, 3 is in the zeroth place, 2 in the oneth place and 1 in the hundredth place.

So, to add another number 4 after 123, we need to shift the current numbers to the left, so now 1 is in the thousandth place, 2 in the oneth place, 3 is in the onethplace and 4 in the zeroth place.

This is done easily by multiplying 123 by 10 which gives 1230 and adding the number 4, which gives 1234. The same is done in the code above.

When the do while loop finally ends, we have a reversed number in rev. This number is then compared to the original number n.

If the numbers are equal, the original number is a palindrome, otherwise, it’s not.

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