shortest funtion c++ of rounding a float number

Question

Which is the shortest form of this function(without changing the name of this)?

int NearestInteger(float N) 
{
    return round(N);
}

@Vasiu: do your requirements differ from the functionality in answers [here](http://stackoverflow.com/questions/485525/round-for-float-in-c?rq=1)? — Tony Delroy, Jul 21 '15 at 10:04
Isn't it exactly the way you wrote it? Why do you think there's a shorter version? Why do you think you need one? — Lightness Races in Orbit, Jul 21 '15 at 11:29

score 2 · Answer 1 · edited Jun 08 '21 at 05:47

2

You can use std::round function if you have C++11.

Otherwise, you may need to use one of them depends on your requirement std::floor or std::ceil

edited Jun 08 '21 at 05:47

Glorfindel

answered Jul 21 '15 at 09:51

Steephen

score 1 · Answer 2 · edited May 23 '17 at 12:14

1

If you don't have C++11, you can do this:

int NearestInteger(float N) {
    return int(floor(N + 0.5f));
}

However, it may work incorrectly for large floats. First of all, if your number does not fit int, any trash can be returned. Second, it may return wrong results for numbers larger than 2^24, because not each integer is exactly representable in single precision float number.

edited May 23 '17 at 12:14

Community

answered Jul 21 '15 at 10:03

stgatilov

1

Excepting *"The behavior is undefined if the truncated value cannot be represented in the destination type."* per 4.9/1 `[conv.fpint]`. – Tony Delroy Jul 21 '15 at 10:08
@TonyD Yes, you are right. The caller has to worry about overflow. – stgatilov Jul 21 '15 at 10:10

2 Answers2