CODEWITHSUNDEEP
c++mathintegerlong-integerlargenumber
chasep255
chasep255

Reputation: 12185

Adding Integers without longs

I am writing a class to model big integers. I am storing my data as unsigned ints in a vector pointer called data. What this function does is it adds n to the current big integer. What seems to be slowing down my performance is having to use long long values. Do any of you know a way around this. I currently have to make sum a long long or else it will overflow.

void Integer::u_add(const Integer & n)
{
    std::vector<unsigned int> & tRef = *data;
    std::vector<unsigned int> & nRef = *n.data;
    const int thisSize = tRef.size();
    const int nSize = nRef.size();
    int carry = 0;
    for(int i = 0; i < nSize || carry; ++i)
    {
        bool readThis = i < thisSize;
        long long sum = (readThis ? (long long)tRef[i] + carry : (long long)carry) + (i < nSize ? nRef[i] : 0);
        if(readThis)
            tRef[i] = sum % BASE; //Base is 2^32
        else
            tRef.push_back(sum % BASE);
        carry = (sum >= BASE ? 1 : 0);
    }
}

Also just wondering if there is any benefit to using references to the pointers over just using the pointers themselves? I mean should I use tRef[i] or (*data)[i] to access the data.

Upvotes: 0

Views: 92

Answers (1)

Peter Raynham
Peter Raynham

Reputation: 657

Instead of using base 2^32, use base 2^30. Then when you add two values the largest sum will be 2^31-1 which fits into an ordinary long (signed or unsigned).

Or better still, use base 10^9 (roughly equal to 2^30), then you don't need a lot of effort to print the large numbers in decimal format.

If you really need to work in base 2^32, you could try a kludge like the following, provided unsigned ints don't throw overflow exceptions:

sum = term1 + term2
carry = 0
if (sum < term1 || sum < term2)
    carry = 1

Upvotes: 1

Related Questions