Printing in Base 10-16 with letters - c++

Question

I am stuck on a project where I have to print out any number in any base from 10-16. The problem is that in those bases, you have to add a letter to the front, which I don't really understand how to do with recursion. Can anyone help me?
int conversionFunction(int num, int base)
{
if (num == 0)
return 0;
int x = num % base;
num /= base;
if (x < 0)
num = num + 1;
conversionFunction(num, base);
if (x < 0){
cout << x+(base * -1);
}
else{
cout << x;
return x;
}
}
If I do 246 in base 16, I get 156. I know that the actual answer should be F6. 15 translates to F when converting. But how would I do that?

Answer 1

Something like
static const char* digits = "0123456789abcdef";
and
cout << digits[num % base];
is a nice way. static just means that digits is has global lifetime but is scoped to your function (basically, you won't have to recreate it over and over every time you enter your function).

Answer 2

You seem to be stuck just on the problem of converting between bases. I can think of two ways to do it:
Divide by decreasing powers of the radix, from n-1 to 0, where n is the largest power. That requires you to know the largest value that you might have to convert. Each division gives you a digit in the place that corresponds to that power. Using your example, you could decide to go up to four digits, so you'd have:
246 / 16^^3 = 0
246 / 16^^2 = 0
246 / 16^^1 = F
6 / 16^^0 = 6
So the answer is 0x00F6.
Use modulo arithmetic with increasing powers of the radix, from 1 to n. Again, each operation gives you a digit in the place that corresponds to the power of the radix. Using the same example:
246 mod 16^^1 = 6
240 mod 16^^2 = F
So again, you've got 0xF6.

Answer 3

Here's a version with comments in the code using a similar approach as in okovkos answer and Calebs second solution. It starts with the least significant digit and extracts until num is zero. It supports conversions in the range (INTMAX_MIN, INTMAX_MAX] using a base in the range [2, 36].
#include <iostream>
#include <string>
#include <cstdint> // std::intmax_t, std::uintmax_t
std::string itos(
std::intmax_t num, // number to convert, range: (INTMAX_MIN, INTMAX_MAX]
const int base=10, // base, range: [2, 36]
const std::string& prefix="", // user defined prefix
bool add_plus=false) // add plus sign for positive numbers
{
static const std::string digits = "0123456789abcdefghijklmnopqrstuvwxyz";
if(base>36 || base<2) return ""; // erroneous base
std::string rv; // the return value we'll create
if(num) {
bool negative = false;
if(num<0) {
if(num==INTMAX_MIN) return ""; // the ONE std::intmax_t number you can't use
// make it positive for the calculation
num = -num;
negative = true;
}
std::uintmax_t x;
while(num) {
x = num % base; // extract least significant digits index
rv.insert(rv.begin(), digits[x]); // insert digit first
num -= x; // reduce num with the extracted value
num /= base; // divide num down for next extraction
}
// the below two inserts could be moved to just before the
// return if you want to add the prefix for the value zero too
// insert prefix
rv.insert(0, prefix);
// insert minus sign if negative or plus if desired
if(negative) rv.insert(rv.begin(), '-');
else if(add_plus) rv.insert(rv.begin(), '+');
} else rv = "0"; // special case
return rv;
}
int main() {
std::cout << "bin " << itos(255, 2, "0b") << "\n";
std::cout << "oct " << itos(255, 8, "0") << "\n";
std::cout << "dec " << itos(255, 10, "", true) << "\n";
std::cout << "hex " << itos(-INTMAX_MAX, 16, "0x") << "\n";
std::cout << "hex " << itos(INTMAX_MAX, 16, "0x") << "\n";
}
Possible output:
bin 0b11111111
oct 0377
dec +255
hex -0x7fffffffffffffff
hex 0x7fffffffffffffff