Rounding off floating numbers in cpp

Question

For a particular question, I need to perform calculations on a floating number, round it off to 2 digits after the decimal place, and assign it to a variable for comparison purposes. I tried to find a solution to this but all I keep finding is how to print those rounded numbers (using printf or setprecision) instead of assigning them to a variable. Please help.

Answer 1

round it off to 2 digits after the decimal place, and assign it to a variable for comparison purposes

To avoid errors that creep in when using binary floating point in a decimal problem, consider alternatives.

Direct approach has corner errors due to double rounding and overflow. These errors may be tolerable for OP larger goals

// Errors: 
// 1) x*100.0, round(x*100.0)/100.0 inexact.
// Select `x` values near "dddd.dd5" form an inexact product `x*100.0`
// and may lead to a double rounding error and then incorrect result when comparing.
// 2) x*100.0 may overflow.
int compare_hundredth1(double x, double ref) {
  x = round(x*100.0)/100.0;
  return (x > ref) - (x < ref);
}

We can do better.

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When a wider floating point type exist:

int compare_hundredth2(double x, double ref) {
  auto x_rounded = math::round(x*100.0L);
  auto ref_rounded = ref*100.0L;
  return (x_rounded > ref_rounded) - (x_rounded < ref_rounded);
}

To use the same width floating point type takes more work:

All finite large values of x, ref are whole numbers and need no rounding to the nearest 0.01.

int compare_hundredth3(double x, double ref) {
  double x_whole; 
  auto x_fraction = modf(x, &x_whole);
  // If rounding needed ...
  if (x_fraction != 0.0) {
    if (x - 0.01 > ref) return 1;  // x much more than ref
    if (x + 0.01 < ref) return -1; // x much less than ref
    // x, ref nearly the same
    double ref_whole; 
    auto ref_fraction = modf(x, &ref_whole);
    x -= ref_whole;
    auto x100 = (x - ref_whole)*100; // subtraction expected to be exact here.
    auto ref100 = ref_fraction*100;
    return (x100 > ref100) - (x100 < ref100);
  }
  return (x > ref) - (x < ref);
}

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The above assume ref is without error. If this is not so, consider using a scaled ref.

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Note: The above sets aside not-a-number concerns.

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More clean-up later.

Answer 2

I usually do something like that:

#include <cmath> // 'std::floor'
#include <limits> // 'std::numeric_limits'
// Round value to granularity
template<typename T> inline T round(const T x, const T gran)
   {
    //static_assert(gran!=0);
    return gran * std::floor( x/gran + std::numeric_limits<T>::round_error() );
   }

double rounded_to_cent = round(1.23456, 0.01); // Gives something near 1.23

Be sure you know how floating point types work though.

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Addendum: I know that this topic has already been extensively covered in other questions, but let me put this small paragraph here. Given a real number, you can represent it with -almost- arbitrary accuracy with a (base10) literal like 1.2345, that's a string that you can type with your keyboard. When you store that value in a floating point type, let's say a double, you -almost- always loose accuracy because probably your number won't have an exact representation in the finite set of the numbers representable by that type. Nowadays double uses 64 bits, so it has 2^64 symbols to represent the not numerable infinity of real numbers: that's a H2O molecule in an infinity of infinite oceans. The representation error is relative to the value; for example in a IEEE 754 double, over 2^53 not all the integer values can be represented. So when someone tells that the "result is wrong" they're technically right; the "acceptable" result is application dependent.

Answer 3

Here's an example with a custom function that rounds up the floating number f to n decimal places. Basically, it multiplies the floating number by 10 to the power of N to separate the decimal places, then uses roundf to round the decimal places up or down, and finally divides back the floating number by 10 to the power of N (N is the amount of decimal places). Works for C and C++:

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

float my_round(float f, unsigned int n)
{
    float p = powf(10.0f, (float)n);
    f *= p;
    f = roundf(f);
    f /= p;
    return f;
}

int main()
{
    float f = 0.78901f;
    printf("%f\n", f);

    f = my_round(f, 2); /* Round with 2 decimal places */
    printf("%f\n", f);
    return 0;
}

Output:

0.789010
0.790000