Reverse a number or digits

tootoo · #1 06-Apr-2007, 09:08

Write program that reverse a number digits. The program will ask the user to enter a number (one variable) consist from three digits then the program reverse the digits of the number.
Note: you must define the number as integer. To solve the problem use the / and the %

The program output might be like this

Enter any number consist from three digits = 123
The reverse of this number= 321

cable_guy_67 · #2 06-Apr-2007, 09:27

You will probably want to post the code you are using. You will be more likely to get help if you show your work first. Members are much more likely to jump in and help than to do your work for you.

Mark

tootoo · #3 06-Apr-2007, 10:50

CPP / C++ / C Code:

#include <iostream>


int main ()
{
short x;

std::cout<<"enter a number";
std::cin>>x;

    short a=(x%10);
    short b=(?%?)?10;  
    short c=(x/?);

std::cout<<?<<?<<?<<std::endl;
   
    std::cin.ignore();
	std::cin.get();

    
    return 0;
}

davekw7x · #4 06-Apr-2007, 11:04

Quote:

Originally Posted by tootoo

CPP / C++ / C Code:

#include <iostream>
.
.

Posting a hint that you got from another forum is really bad form. Why don't you try it yourself?

Show something that you wrote. If it doesn't compile, then tell us exactly what the messages were. If it compiles but doesn't give the answers you expected, then tell us what you did get and what you didn't understand about it.

Regards,

Dave

cable_guy_67 · #5 06-Apr-2007, 12:46

Taking the code you posted and commenting out all the lines with a question mark gives me...

CPP / C++ / C Code:

#include <iostream>

int main ()
{
  short x;
  
  std::cout<<"enter a number";
  std::cin>>x;
  
  short a=(x%10);
//   short b=(?%?)?10;
//   short c=(x/?);

//   std::cout<<?<<?<<?<<std::endl;

  std::cin.ignore();
  std::cin.get();

  
  return 0;
}

Since all that happens when you do this is you are prompted for a number then have to hit a key to quit. Add a cout with the value you got for a and see if it works.

CPP / C++ / C Code:

  std::cout << "a = " << a << std::endl;

It seems you got the least significant digit, Bravo! Now, you are going to want to use your short (int) and dividing by 10 to your advantage. Once you do that (and get the right answer) you just use the same method you used on a to find the new least significant digit.

Rinse, Lather, Repeat

Then just structure your output statement with the proper variables.

At least that seems to be the assignment to me.

Mark

surya.k · #6 22-Jul-2007, 08:08

wht they said is really good,
a standard logic to start working on it 4 u,..

CPP / C++ / C Code:

int r=0;
while(x!=0)
 {
  d=x%10;
  r=(r*10)+d;
  x=x/10;
 }
std::cout<<r;

regrards
surya

#7 29-Jul-2007, 20:34

Quote:

Originally Posted by tootoo

Write program that reverse a number digits. The program will ask the user to enter a number (one variable) consist from three digits then the program reverse the digits of the number.
Note: you must define the number as integer. To solve the problem use the / and the %

The program output might be like this

Enter any number consist from three digits = 123
The reverse of this number= 321

Maybe something like:

CPP / C++ / C Code:

#include <iostream>

using namespace std;

int main()
{
    cout << "Enter a positive number: ";
    int number;
    cin >> number;
    if( number > 0 )
    {
        do
        {
            cout << (char)( '0' + (number % 10) );
            number /= 10;
        } while( number != 0 );
        cout << endl;
    }
    else
    {
        cout << "Number not positive!" << endl;
    }
    return 0;
}

Output:

Code:

Enter a positive number: 123
321

Enter a positive number: 448737
737844

Enter a positive number: -123
Number not positive!

:davis:

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