dec-to-base

/*
Date: 14 APR 2011
*/

#include <iostream>
#include <string.h>
using namespace std;

char* dectobase(int val,
    int base, // 2 for binary, 16 for hex.
    const char digits[] = "0123456789ABCDEF", //can be overridden, if you want base 27.
    int i = 0 //for office use only
    )
{
    int r = val % base; //remainder, this one will be converted and dropped onto our stack
    int v = ((val - r) / base); //this will be recursed on
    char* ret = '\0';
    if ((base > strlen(digits)) || (base < 2)) // base 1 = you fail
        return ret;

    if (v == 0) //base case, most-significant digit.
    {
        ret = new char [i]; //is this right?
        i = 0; //reset, use i as index on way back out
        ret[0] = digits[r];
        return ret;
    }
    else
    {
        ret = dectobase(v, base, digits, ++i);
        ret[i++] = digits[r];
        return ret;
    }
}

int main()
{
    int val = 0;
    cout << "val:";
    cin >> val;
    cout << "hex:" << dectobase(val, 16) << endl;
    cout << "bin:" << dectobase(val, 2) << endl;
    cout << "oct:" << dectobase(val, 8) << endl;
    cout << "wtf:" << dectobase(val, 10, ")!@#$%^&*(") << endl;
    return 0;
}