Some new learner of C# faces this type of problem. Don''t worry use this code once.
using System;
class ReverseNumber
{
public static void Main()
{
int num,rem,i,counter=0,temp;
// num : Contains the actual number inputted via the console
// rem : remainder of the number ''num'' when divided by 10
// i : loop variable
// counter : determines the no. of digits in the inputted number ''num''
// temp : temporary variable used to save the value of ''num'' (Explained further)
Console.Write("Enter an integer number (Not more than 9 digits) : ");
num = int.Parse(Console.ReadLine());
temp = num;
// Here we are saving ''num'' in ''temp'' coz its value after determining the no. of digits will loose.
// So after its work is done, ''num'' will contain value = 0
// The value of ''num'' is resetted to its original value later from ''temp'' variable
// Determine the no. of digits in the inputted number
while(num > 0)
{
rem = num % 10;
num = num / 10;
if (num <= 0)
{
break;
}
else
{
counter = counter + 1;
}
}
Console.WriteLine("Number of digits are = " + (counter+1));
Console.Write("The reversed digits are : ");
rem = 0;
// resetting the value of remainder ''rem''
num = temp;
// resetting the lost value of ''num'' from ''temp''
// *** Determine the reversed of inputted digits ***
// Funda :
// 1) Divide the number by 10 & determine the remainder. (Save the remainder in ''rem'')
// This will give us the last digit in the actual inputted number.
// 2) Write the number so obtained into the console
// 3) Divide the same number by 10 & get the quotient this time.
// Since division is between the integers, we will get the new number, deprived of the last digit.
// Then again goto step 1) & continue until & unless the counter is equal to ''i'' (coz thats the loop varibale)
for(i = 0; i<=counter; i++)
{
rem = num % 10;
Console.Write(rem);
num = num / 10;
}
Console.ReadLine();
}
}