double to string without scientific notation or trailing zeros, efficiently

Question

This routine is called a zillion times to create large csv files full of numbers. Is there a more efficient way to to this?

    static std::string dbl2str(double d)
    {
        std::stringstream ss;
        ss << std::fixed << std::setprecision(10) << d;              //convert double to string w fixed notation, hi precision
        std::string s = ss.str();                                    //output to std::string
        s.erase(s.find_last_not_of('0') + 1, std::string::npos);     //remove trailing 000s    (123.1200 => 123.12,  123.000 => 123.)
        return (s[s.size()-1] == '.') ? s.substr(0, s.size()-1) : s; //remove dangling decimal (123. => 123)
    }

Answer 1

Efficient in terms of speed or brevity?

char buf[64];
sprintf(buf, "%-.*G", 16, 1.0);
cout << buf << endl;

Displays "1". Formats up to significant 16 digits, with no trailing zeros, before reverting to scientific notation.