How to reverse a number?

[Guitars]

I want to know how to reverse a number in a column, here is my code:

#include <iostream>
using namespace std;
int main()
{
	int n;
	do
	{
	cin>>n;
	n/10;
	cout<<n%10;
	} while(n>0);
	return 0;
}

[Guitars]

Bump.

[ciphermagi]

Your code makes no sense.

n/10; // This line doesn't do anything.
cout << n % 10; // All this does is print out the remainder of n/10. 

[Stewbond]

Do you mean, if you enter 123 it should output 321?

Try this algorithm:

int Reverse(int num)
{
  //get num digits
  int nb_digits = 0;
  while (num / pow(10,nb_digits) > 0)
    ++nb_digits;

  //add the digits, one at a time:
  int out = 0, iterator = nb_digits;
  while (iterator > 0)
  {
    out += num / pow(10,iterator-1) * pow(10,nb_digits-iterator);
    num %= pow(10,--iterator);
  }
  return out;
}

make sure you have int pow (int, int) available. If it's not in <cmath>, make it:

int pow (int num, int exp)
{
  int out = 1;
  while (exp-- > 0)
    out *= num;
  return out;
}

Edit: I just tested it, it works.

1 2 3 4 5 6 7 8 9 10 11

int main() { while (true) { int in; cout << "Number: "; cin >> in; cout << "Reverse: " << Reverse(in) << endl; } return 0; }

Number : 1 Reverse: 1 Number : 123 Reverse: 321 Number : 164353 Reverse: 353461 Number : ^CPress any key to continue . . .

[Zu007]

#include<conio.h>
#include<iostream.h>
void main(void)
{
clrscr();
long i,num,rem;
cout<<"Enter a number";
cin>>num;
for(i=num;i>0;i=i/10)
{
rem=num%10;
num=num/10;
cout<<rem;
}
getch();
}

[Guitars]

Can you guys explain how it works?

[strongdrink]

If you want to turn a number negative very simply, try this..

int number = 123;

// Negate the number
number -= number * 2;

[oonej]

yeah, it divides by 10 to the power of some number until it reaches its max size (so it knows how large the number is) then it does a similar feet, by finding the remainder of each division to produce its inverse..

[strongdrink]

Another way would be to convert it to a string and then go from there.. although the numeric way is probably much better

[Stewbond]

Here I started off by findign the number of digits. I did this with:

int nb_digits = 0;
while (num / pow(10,nb_digits) > 0) // number / 10^i > 0
  ++nb_digits; // increase i 

Now that I know the number of digits, we can deal with each decimal place separately. If you want to know the 3rd digit you can do:

num %= pow(10,3+1); // Get rid of large digits
num /= pow(10,3); // Get rid of small digits 

We can use a for loop to go through all digits instead of just hard-coding the third digit.

Then we just put this digit in the spot we want it. If we have a 3-digit number and this represents the third digit, then we want to put it in the 1st digit place.
output = num * pow(10,0); //Where 0 is the decimal place we want this to go to