round() for float in C++

Question

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??

Answer 1

The C++03 standard relies on the C90 standard for what the standard calls the Standard C Library which is covered in the draft C++03 standard (closest publicly available draft standard to C++03 is N1804) section 1.2 Normative references:

The library described in clause 7 of ISO/IEC 9899:1990 and clause 7 of ISO/IEC 9899/Amd.1:1995 is hereinafter called the Standard C Library.¹⁾

If we go to the C documentation for round, lround, llround on cppreference we can see that round and related functions are part of C99 and thus won't be available in C++03 or prior.

In C++11 this changes since C++11 relies on the C99 draft standard for C standard library and therefore provides std::round and for integral return types std::lround, std::llround :

#include <iostream>
#include <cmath>

int main()
{
    std::cout << std::round( 0.4 ) << " " << std::lround( 0.4 ) << " " << std::llround( 0.4 ) << std::endl ;
    std::cout << std::round( 0.5 ) << " " << std::lround( 0.5 ) << " " << std::llround( 0.5 ) << std::endl ;
    std::cout << std::round( 0.6 ) << " " << std::lround( 0.6 ) << " " << std::llround( 0.6 ) << std::endl ;
}

Another option also from C99 would be std::trunc which:

Computes nearest integer not greater in magnitude than arg.

#include <iostream>
#include <cmath>

int main()
{
    std::cout << std::trunc( 0.4 ) << std::endl ;
    std::cout << std::trunc( 0.9 ) << std::endl ;
    std::cout << std::trunc( 1.1 ) << std::endl ;

}

If you need to support non C++11 applications your best bet would be to use boost round, iround, lround, llround or boost trunc.

Rolling your own version of round is hard

Rolling your own is probably not worth the effort as Harder than it looks: rounding float to nearest integer, part 1, Rounding float to nearest integer, part 2 and Rounding float to nearest integer, part 3 explain:

For example a common roll your implementation using std::floor and adding 0.5 does not work for all inputs:

double myround(double d)
{
  return std::floor(d + 0.5);
}

One input this will fail for is 0.49999999999999994, (see it live).

Another common implementation involves casting a floating point type to an integral type, which can invoke undefined behavior in the case where the integral part can not be represented in the destination type. We can see this from the draft C++ standard section 4.9 Floating-integral conversions which says (emphasis mine):

A prvalue of a floating point type can be converted to a prvalue of an integer type. The conversion truncates; that is, the fractional part is discarded. The behavior is undefined if the truncated value cannot be represented in the destination type.[...]

For example:

float myround(float f)
{
  return static_cast<float>( static_cast<unsigned int>( f ) ) ;
}

Given std::numeric_limits<unsigned int>::max() is 4294967295 then the following call:

myround( 4294967296.5f ) 

will cause overflow, (see it live).

We can see how difficult this really is by looking at this answer to Concise way to implement round() in C? which referencing newlibs version of single precision float round. It is a very long function for something which seems simple. It seems unlikely that anyone without intimate knowledge of floating point implementations could correctly implement this function:

float roundf(x)
{
  int signbit;
  __uint32_t w;
  /* Most significant word, least significant word. */
  int exponent_less_127;

  GET_FLOAT_WORD(w, x);

  /* Extract sign bit. */
  signbit = w & 0x80000000;

  /* Extract exponent field. */
  exponent_less_127 = (int)((w & 0x7f800000) >> 23) - 127;

  if (exponent_less_127 < 23)
    {
      if (exponent_less_127 < 0)
        {
          w &= 0x80000000;
          if (exponent_less_127 == -1)
            /* Result is +1.0 or -1.0. */
            w |= ((__uint32_t)127 << 23);
        }
      else
        {
          unsigned int exponent_mask = 0x007fffff >> exponent_less_127;
          if ((w & exponent_mask) == 0)
            /* x has an integral value. */
            return x;

          w += 0x00400000 >> exponent_less_127;
          w &= ~exponent_mask;
        }
    }
  else
    {
      if (exponent_less_127 == 128)
        /* x is NaN or infinite. */
        return x + x;
      else
        return x;
    }
  SET_FLOAT_WORD(x, w);
  return x;
}

On the other hand if none of the other solutions are usable newlib could potentially be an option since it is a well tested implementation.

Answer 2

Beware of floor(x+0.5). Here is what can happen for odd numbers in range [2^52,2^53]:

-bash-3.2$ cat >test-round.c <<END

#include <math.h>
#include <stdio.h>

int main() {
    double x=5000000000000001.0;
    double y=round(x);
    double z=floor(x+0.5);
    printf("      x     =%f\n",x);
    printf("round(x)    =%f\n",y);
    printf("floor(x+0.5)=%f\n",z);
    return 0;
}
END

-bash-3.2$ gcc test-round.c
-bash-3.2$ ./a.out
      x     =5000000000000001.000000
round(x)    =5000000000000001.000000
floor(x+0.5)=5000000000000002.000000

This is http://bugs.squeak.org/view.php?id=7134. Use a solution like the one of @konik.

My own robust version would be something like:

double round(double x)
{
    double truncated,roundedFraction;
    double fraction = modf(x, &truncated);
    modf(2.0*fraction, &roundedFraction);
    return truncated + roundedFraction;
}

Another reason to avoid floor(x+0.5) is given here.

Answer 3

Based on Kalaxy's response, the following is a templated solution that rounds any floating point number to the nearest integer type based on natural rounding. It also throws an error in debug mode if the value is out of range of the integer type, thereby serving roughly as a viable library function.

    // round a floating point number to the nearest integer
    template <typename Arg>
    int Round(Arg arg)
    {
#ifndef NDEBUG
        // check that the argument can be rounded given the return type:
        if (
            (Arg)std::numeric_limits<int>::max() < arg + (Arg) 0.5) ||
            (Arg)std::numeric_limits<int>::lowest() > arg - (Arg) 0.5)
            )
        {
            throw std::overflow_error("out of bounds");
        }
#endif

        return (arg > (Arg) 0.0) ? (int)(r + (Arg) 0.5) : (int)(r - (Arg) 0.5);
    }

Answer 4

It may be worth noting that if you wanted an integer result from the rounding you don't need to pass it through either ceil or floor. I.e.,

int round_int( double r ) {
    return (r > 0.0) ? (r + 0.5) : (r - 0.5); 
}

Answer 5

It's usually implemented as floor(value + 0.5).

Edit: and it's probably not called round since there are at least three rounding algorithms I know of: round to zero, round to closest integer, and banker's rounding. You are asking for round to closest integer.

Answer 6

Boost offers a simple set of rounding functions.

#include <boost/math/special_functions/round.hpp>

double a = boost::math::round(1.5); // Yields 2.0
int b = boost::math::iround(1.5); // Yields 2 as an integer

For more information, see the Boost documentation.

Edit: Since C++11, there are std::round, std::lround, and std::llround.