Double precision issues when converting it to a large integer

Precision is the number of digits in a number. Scale is the number of digits to the right of the decimal point in a number. For example, the number 123.45 has a precision of 5 and a scale of 2.

I need to convert a double with a maximum scale of 7(i.e. it may have 7 digits after the decimal point) to a __int128. However, given a number, I don't know in advance, the actual scale the number has.

#include <iostream> #include "json.hpp" using json = nlohmann::json; #include <string> static std::ostream& operator<<(std::ostream& o, const __int128& x) { if (x == std::numeric_limits<__int128>::min()) return o << "-170141183460469231731687303715884105728"; if (x < 0) return o << "-" << -x; if (x < 10) return o << (char)(x + '0'); return o << x / 10 << (char)(x % 10 + '0'); } int main() { std::string str = R"({"time": [0.143]})"; std::cout << "input: " << str << std::endl; json j = json::parse(str); std::cout << "output: " << j.dump(4) << std::endl; double d = j["time"][0].get<double>(); __int128_t d_128_bad = d * 10000000; __int128_t d_128_good = __int128(d * 1000) * 10000; std::cout << std::setprecision(16) << std::defaultfloat << d << std::endl; std::cout << "d_128_bad: " << d_128_bad << std::endl; std::cout << "d_128_good: " << d_128_good << std::endl; } 

Output:

input: {"time": [0.143]} output: { "time": [ 0.143 ] } 0.143 d_128_bad: 1429999 d_128_good: 1430000 

As you can see, the converted double is not the expected 1430000 instead it is 1429999. I know the reason is that a float point number can not be represented exactly. The problem can be solved if I know the number of digit after the decimal point.

For example,

I can instead use __int128_t(d * 1000) * 10000. However, I don't know the scale of a given number which might have a maximum of scale 7.

Question> Is there a possible solution for this? Also, I need to do this conversion very fast.