Converting many bits to Base 10 - c++

Question

I am building a class in C++ which can be used to store arbitrarily large integers. I am storing them as binary in a vector. I need to be able to print this vector in base 10 so it is easier for a human to understand. I know that I could convert it to an int and then output that int. However, my numbers will be much larger than any primitive types. How can I convert this directly to a string.
Here is my code so far. I am new to C++ so if you have any other suggestions that would be great too. I need help filling in the string toBaseTenString() function.
class BinaryInt
{
private:
bool lastDataUser = true;
vector<bool> * data;
BinaryInt(vector<bool> * pointer)
{
data = pointer;
}
public:
BinaryInt(int n)
{
data = new vector<bool>();
while(n > 0)
{
data->push_back(n % 2);
n = n >> 1;
}
}
BinaryInt(const BinaryInt & from)
{
from.lastDataUser = false;
this->data = from.data;
}
~BinaryInt()
{
if(lastDataUser)
delete data;
}
string toBinaryString();
string toBaseTenString();
static BinaryInt add(BinaryInt a, BinaryInt b);
static BinaryInt mult(BinaryInt a, BinaryInt b);
};
BinaryInt BinaryInt::add(BinaryInt a, BinaryInt b)
{
int aSize = a.data->size();
int bSize = b.data->size();
int newDataSize = max(aSize, bSize);
vector<bool> * newData = new vector<bool>(newDataSize);
bool carry = 0;
for(int i = 0; i < newDataSize; i++)
{
int sum = (i < aSize ? a.data->at(i) : 0) + (i < bSize ? b.data->at(i) : 0) + carry;
(*newData)[i] = sum % 2;
carry = sum >> 1;
}
if(carry)
newData->push_back(carry);
return BinaryInt(newData);
}
string BinaryInt::toBinaryString()
{
stringstream ss;
for(int i = data->size() - 1; i >= 0; i--)
{
ss << (*data)[i];
}
return ss.str();
}
string BinaryInt::toBaseTenString()
{
//Not sure how to do this
}

Answer 1

I know you said in your OP that "my numbers will be much larger than any primitive types", but just hear me out on this.
In the past, I've used std::bitset to work with binary representations of numbers and converting back and forth from various other representations. std::bitset is basically a fancy std::vector with some added functionality. You can read more about it here if it sounds interesting, but here's some small stupid example code to show you how it could work:
std::bitset<8> myByte;
myByte |= 1; // mByte = 00000001
myByte <<= 4; // mByte = 00010000
myByte |= 1; // mByte = 00010001
std::cout << myByte.to_string() << '\n'; // Outputs '00010001'
std::cout << myByte.to_ullong() << '\n'; // Outputs '17'
You can access the bitset by standard array notation as well. By the way, that second conversion I showed (to_ullong) converts to an unsigned long long, which I believe has a max value of 18,446,744,073,709,551,615. If you need larger values than that, good luck!

Answer 2

Just iterate (backwards) your vector<bool> and accumulate the corresponding value when the iterator is true:
int base10(const std::vector<bool> &value)
{
int result = 0;
int bit = 1;
for (vb::const_reverse_iterator b = value.rbegin(), e = value.rend(); b != e; ++b, bit <<= 1)
result += (*b ? bit : 0);
return result;
}
Live demo.
Beware! this code is only a guide, you will need to take care of int overflowing if the value is pretty big.
Hope it helps.