CODEWITHSUNDEEP
c++
Sai Suman
Sai Suman

Reputation: 47

Number Reversal in c++

When I pass an odd digit number > 1 to the function, I am getting 1 less than the actual result I should be getting and the code is working perfectly fine for an even digit number, I am not able to figure out why.

int rev_number(int num){
int arr [10];
int i=0;
int j=0;
int rev_num=0;
while(num > 0){
    arr[i] = num%10;
    num = num/10;
    i++;
}
i--;
while(i>=0){
    rev_num += arr[j] * pow(10,i);
        j++;
        i--;
    }
    return rev_num;
}

Upvotes: 0

Views: 149

Answers (4)

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

All code examples of the function showed here are invalid because the function does not process valid integer 0. So I decided to show a correct function realization.:)

using System;

namespace ReverseNumberExercise
{
    class Program
    {
        static int ReverseNumber(int x)
        {
            int y = 0;

            do
            {
                long digit;
                x = ( int )Math.DivRem( x, 10L, out digit );
                y = y * 10 + ( int )digit;
            } while ( x != 0 );

            return (y);
        }

        static void Main()
        {
            while (true)
            {
                Console.Write("Enter an integral number: ");

                string input = Console.ReadLine();

                if (string.IsNullOrEmpty(input)) break;

                int x;

                if (!int.TryParse(input, out x))
                {
                    Console.WriteLine("Invalid number. Try again.");
                }

                Console.WriteLine("Reversed number is: {0}", ReverseNumber(x));

                Console.WriteLine();
            }

            Console.WriteLine("\nGood bye!");
        }
    }
}

Upvotes: -1

Mats Petersson
Mats Petersson

Reputation: 129374

Since pow deals with floating point values, it's integer representation may get truncated, so certain values may end up being lower than expected.

A better solution is to use integer math, such as this:

while(i>=0){
rev_num *= 10;
rev_num += arr[j];
    j++;
    i--;
}

There are several other ways to express the same thing, and storing the values in an array isn't strictly necessary, you could just multiply the modulo value back in.

However, my favourite "reverse a number" is to read it in as a string, and then use the standard reverse function.

So, something like this:

#include <string>
#include <algorithm>
#include <iostream>

int main()
{
   std::string s;
   std::cin >> s;
   std::reverse(s.begin(), s.end());
   std::cout << s << std::endl;
}

Upvotes: 1

Maxime Ch&#233;ramy
Maxime Ch&#233;ramy

Reputation: 18831

You should avoid using pow which return a double while you only want to manipulate integers. For example, imagine if the result of arr[j] * pow(10,i) is 753.99..., the cast would only put 753 in rev_num! In some cases, that will work, but not always. It's a fifty-fifty chance to work.

Try replacing:

rev_num += arr[j] * pow(10,i);

to

rev_num = rev_num * 10 + arr[j];

Have you tried to round or to add .5? Does that also fix your problem?

Upvotes: 3

Akshat Singhal
Akshat Singhal

Reputation: 1801

How about the following simple code 

int reverseNumber(int num) {
     int reverse = 0;
     while(num > 0) {
          reverse = reverse*10 + num%10;
          num = num/10;
     }
return reverse;
}

Upvotes: 3

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