1. ## stacks to convert from decimal to hexadecimal

Pardon me chaps, We are using stacks to convert a number, lets say 103, for instance to binary octal and hexadecimal. I am a bit perplexed by the whole situation and I was wondering if you blokes could perhaps share a little bit of your insight with me.I feel as though I've mastered the binary portion of the program.
Code:
```#include <iostream>
#include <iomanip>
using namespace std;

template <class T, int n>
class stack
{
private: T elt[n];
int counter;
public:
void clearstack() {  counter = -1;                            }
bool emptystack() {  return counter==-1?true:false;            }
bool fullstack()  {  return counter==n -1?true:false;          }
void push(T x)    {  counter++; elt[counter]=x;                }
T pop()           {  T x; x=elt[counter]; counter--;return x;  }
};

void main()
{
stack <int,7> s;

int m,n,p;
s.clearstack();

cout<<"Enter a number to convert: ";

cin>>m;

while( !  s.fullstack() )
{

n = m % 2;

s.push(n);

m =	m / 2;

}

while(! s.emptystack())
{
p = s.pop();

cout<<p;
}
cout<<endl;
}```
So if you could help me out with the whole hexadecimal portion of this program I would be very appreciative thank you very kindly.
(english enough for you sean)

2. Would you please edit your post and start using English in what we like to call, "the right way". I can't even get past your first sentence.

3. punctuation always helps

But I think what you're looking for could be found with a good board search here, or even a....Google search *gasp*

4. > void main()
Read the FAQs, watch the Avatar, you know its wrong.

> I feel as though I've mastered the binary portion of the program.
> n = m % 2;
So what's the big jump to %8 for octal and %16 for hex ?