round() for float in C++

2018-12-31 05:43发布

I need a simple floating point rounding function, thus:

double round(double);

round(0.1) = 0
round(-0.1) = 0
round(-0.9) = -1

I can find ceil() and floor() in the math.h - but not round().

Is it present in the standard C++ library under another name, or is it missing??

20条回答
忆尘夕之涩
2楼-- · 2018-12-31 06:27

A certain type of rounding is also implemented in Boost:

#include <iostream>

#include <boost/numeric/conversion/converter.hpp>

template<typename T, typename S> T round2(const S& x) {
  typedef boost::numeric::conversion_traits<T, S> Traits;
  typedef boost::numeric::def_overflow_handler OverflowHandler;
  typedef boost::numeric::RoundEven<typename Traits::source_type> Rounder;
  typedef boost::numeric::converter<T, S, Traits, OverflowHandler, Rounder> Converter;
  return Converter::convert(x);
}

int main() {
  std::cout << round2<int, double>(0.1) << ' ' << round2<int, double>(-0.1) << ' ' << round2<int, double>(-0.9) << std::endl;
}

Note that this works only if you do a to-integer conversion.

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与君花间醉酒
3楼-- · 2018-12-31 06:28

You could round to n digits precision with:

double round( double x )
{
const double sd = 1000; //for accuracy to 3 decimal places
return int(x*sd + (x<0? -0.5 : 0.5))/sd;
}
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