Why typecasting a double to int seems to round it after 16th digit?

Question

//g++  5.4.0

#include <iostream>

int main()
{
    std::cout << "Hello, world!\n";
    std::cout << (int)0.9999999999999999 << std::endl; // 16 digits after decimal
    std::cout << (int)0.99999999999999999 << std::endl; // 17 digits after decimal
}

Output:

Hello, world!
0
1

Why does this happen?

Possible duplicate of [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken) — SABER, Oct 23 '18 at 05:22
see my answer to [round() for float in C++](https://stackoverflow.com/a/24348037/1708801) — Shafik Yaghmour, Oct 23 '18 at 07:16
Try writing out those values as doubles, i.e., without the cast. That should make it all clear. — Pete Becker, Oct 23 '18 at 12:05

score 3 · Answer 1 · 2018-10-29T14:05:46.403

3

The most accurate representation of 0.99999999999999999 is 1.0.¹⁾
The most accurate representation of 0.9999999999999999 is 0.999999999999999888977697537484.

1) In 64-bit double precision IEEE754 floating point representation.

Since there is no rounding but truncation, one gives 1 and the other gives 0 when converted to an integer type.

edited Oct 29 '18 at 14:05

answered Oct 23 '18 at 06:05

score 1 · Answer 2 · 2018-10-23T07:25:27.883

1

The most accurate floating point representation of the value 0.99999999999999999 (17 digits after the decimal) is exactly 1.0.

The most accurate floating point representation of the value 0.9999999999999999 (16 digits after the decimal) is less than 1.0.

The conversion to int truncates one to 0 and the other to 1.

edited Oct 23 '18 at 07:25

answered Oct 23 '18 at 05:33

Why typecasting a double to int seems to round it after 16th digit?

2 Answers2