Reputation: 237

reverse the position of integer digits?

i have to reverse the position of integer like this

input = 12345

output = 54321

i made this but it gives wrong output e.g 5432

#include <iostream>
using namespace std;

int main(){
 int num,i=10;   
 cin>>num;   

 do{
    cout<< (num%i)/ (i/10);
    i *=10;
   }while(num/i!=0);

 return 0;
}

Upvotes: 6

Answers (13)

Neel Salpe

Reputation: 1303

public class TestDS {
    
    public static void main(String[] args) {
        System.out.println(recursiveReverse(234));
               System.out.println(recursiveReverse(234 ,0));
    }
    
    public static int reverse(int number){
        int reversedNumber = 0;
        int temp = 0; 
    
        while(number > 0){
            //use modulus operator to strip off the last digit
            temp = number%10;
 
            //create the reversed number
            reversedNumber = reversedNumber * 10 + temp;
            number = number/10;
        }

        return reversedNumber;
    }
    
    private static int reversenumber =0;
    public static int recursiveReverse(int number){
        
        if(number <= 0){
            return reversenumber;
        }
        
        reversenumber = reversenumber*10+(number%10);
        number =number/10;
        
        return recursiveReverse(number);
    }

    public static int recursiveReverse(int number , int reversenumber){
        if(number <= 0){
            return reversenumber;
        }
        
        reversenumber = reversenumber*10+(number%10);
        number =number/10;
        
        return recursiveReverse(number,reversenumber);
    }
}

Upvotes: -2

Marinaldo de Jesus naldo dj

Reputation: 1

#include <iostream>
using namespace std;
int main() {
   int num, i = 10, sign; // Adiciona uma variável para o sinal
   cin >> num;

   // Armazena o sinal e trabalha com o valor absoluto
   sign = (num < 0) ? -1 : 1;
   num = abs(num);

   // Imprime o sinal antes da reversão (se necessário)
   if (sign == -1) {
      cout << "-";
   }

  // Reverte e imprime os dígitos
  do {
      cout << (num % i) / (i / 10);
      i *= 10;
  } while (num / i != 0);

  cout << (num % i) / (i / 10);

  return 0;
 }

Upvotes: -1

Amith Chinthaka

Reputation: 1051

template <typename T>
T reverse(T n, size_t nBits = sizeof(T) * 8)
{
    T reverse = 0;
    auto mask = 1;

    for (auto i = 0; i < nBits; ++i)
    {
        if (n & mask)
        {
            reverse |= (1 << (nBits - i - 1));
        }
        mask <<= 1;
    }

    return reverse;
}

This will reverse bits in any signed or unsigned integer (short, byte, int, long ...). You can provide additional parameter nBits to frame the bits while reversing.

i. e. 7 in 8 bit = 00000111 -> 11100000 7 in 4 bit = 0111 -> 1110

Upvotes: 0

Nauman Tauseef

Reputation: 1

I have done this simply but this is applicable upto 5 digit numbers but hope it helps

 #include<iostream>
    using namespace std;
void main()
{ 
    int a,b,c,d,e,f,g,h,i,j;
    cin>>a;
    b=a%10;
    c=a/10;
    d=c%10;
    e=a/100;
    f=e%10;
    g=a/1000;
    h=g%10;
    i=a/10000;
    j=i%10;
    cout<<b<<d<<f<<h<<j;
}`

Upvotes: -1

Saad Mirza

Reputation: 1177

int a,b,c,d=0;
cout<<"plz enter the number"<<endl;
cin>>a;
b=a;
do
{
    c=a%10;
    d=(d*10)+c; 
    a=a/10;
}
while(a!=0);
cout<<"The reverse of the number"<<d<<endl;
if(b==d)
{
    cout<<"The entered number is palindom"<<endl;
}
else
{
    cout<<"The entered number is not palindom"<<endl;
}

}

Upvotes: 0

Yiannis Hadjipolycarpou

Reputation: 1

The following code works for a two-digit number only.

#include<iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    cout << (n%10) << (n/10);    
return 0;
}

Upvotes: -1

rerun

Reputation: 25505

int _tmain(int argc, _TCHAR* argv[])
{
int x = 1234;
int out = 0;
while (x != 0)
{
    int Res = x % (10 );
    x /= 10;
    out *= 10;
    out +=  Res;
}
cout << out;


}

Upvotes: -1

Jerry Coffin

Reputation: 490108

If I were doing it, I'd (probably) start by creating the new value as an int, and then print out that value. I think this should simplify the code a bit. As pseudocode, it'd look something like:

output = 0;

while (input !=0)
    output *= 10
    output += input % 10
    input /= 10
}
print output

The other obvious possibility would be to convert to a string first, then print the string out in reverse:

std::stringstream buffer;

buffer << input;

cout << std::string(buffer.str().rbegin(), buffer.str().rend());

Upvotes: 0

Elalfer

Reputation: 5338

Here is a solution

    int num = 12345;
    int new_num = 0;
    while(num > 0)
    {
            new_num = new_num*10 + (num % 10);
            num = num/10;
    }
    cout << new_num << endl;

Upvotes: 12