CODEWITHSUNDEEP
c++stringsumcharacterascii
RonthaDon
RonthaDon

Reputation: 23

How to get the sum of two strings of numbers - C++

How do I get the sum of two strings that are numbers? For example:

string num1 = "12";
string num2 = "4";

string sum = num1 + num2;

So string sum will equal "16"

How do I do this in c++?

I tried doing it with ascii characters but it does not work I cannot convert the string to an integer as well

Upvotes: 1

Views: 27431

Answers (3)

Giriteja Bille
Giriteja Bille

Reputation: 166

The answer to your question can be written like this

    string addStrings(string num1, string num2) {
        string res="";
        int n=num1.size();
        int m=num2.size();
        int carry=0;
        int j;
        for(int i=n-1,j=m-1;i>=0 || j>=0;i--,
            j--){
            int a;
            if(i>=0){
                a=((int)(num1[i])-48);
            }
            else {
                a=0;
            }
            int b; 
            if(j>=0){
                b=((int)(num2[j])-48);
            }
            else{
                b=0;
            }
            cout<<num1[i]<<" "<<num2[i]<<endl;
            cout<<a<<" "<<b<<endl;
            int sum=carry+a+b;
            int u=sum%10;
            res+=u;
            carry=sum/10;
        }
        res+=carry;
        cout<<res<<endl;
        reverse(res.begin(),res.end());
        return res;

    }

I added cout statements to check what's going on inside the function

Upvotes: 1

Wasi Ahmad
Wasi Ahmad

Reputation: 37691

To add large integers using string, you can do something like this.

string doSum(string a, string b)
{
    if(a.size() < b.size())
        swap(a, b);

    int j = a.size()-1;
    for(int i=b.size()-1; i>=0; i--, j--)
        a[j]+=(b[i]-'0');

    for(int i=a.size()-1; i>0; i--)
    {
        if(a[i] > '9')
        {
            int d = a[i]-'0';
            a[i-1] = ((a[i-1]-'0') + d/10) + '0';
            a[i] = (d%10)+'0';
        }
    }
    if(a[0] > '9')
    {
        string k;
        k+=a[0];
        a[0] = ((a[0]-'0')%10)+'0';
        k[0] = ((k[0]-'0')/10)+'0';
        a = k+a;
    }
    return a;
}

int main()
{
    string result = doSum("1234567890", "123789456123");
    cout << result << "\n";
}

Output

125024024013

Reference: See the complete code at Ideone.com

Upvotes: 4

πάντα ῥεῖ
πάντα ῥεῖ

Reputation: 1

The flat answer is 

string sum = std::to_string(std::stoi(num1) + std::stoi(num2));

See the Live Demo.

The broader answer as for your comment is:
You cannot use that technique to do math with big integer values.

That requires using an appropriate 3rd party library.

Upvotes: 4

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