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Editorial Team
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Editorial Team
Asked: May 21, 20262026-05-21T09:09:50+00:00 2026-05-21T09:09:50+00:00

I’m trying to write good round_double function which will round double in specified precision:

  • 0

I’m trying to write good round_double function which will round double in specified precision:

1.

double round_double(double num, int prec)
{
    for (int i = 0; i < abs(prec); ++i)
        if(prec > 0)
            num *= 10.0;
        else
            num /= 10.0;
    double result = (long long)floor(num + 0.5);
    for (int i = 0; i < abs(prec); ++i)
        if(prec > 0)
            result /= 10.0;
        else
            result *= 10.0;
    return result;
}

2.

double round_double(double num, int prec)
{
    double tmp = pow(10.0, prec);
    double result = (long long)floor(num * tmp + 0.5);
    result /= tmp;
    return result;
}

This functions do what I wan’t but they are, in my opinion, not good enough. Because starting from precision = 13 – 14, they returning bad results.

The cause I’m sure that there is possible to write good double_round is that just printing the number via cout in specified precision (say 18) is prints better result than result of my function.

For example this part of code:

int prec = 18;
double num = 10.123456789987654321;
cout << setiosflags(ios::showpoint | ios::fixed)
<< setprecision(prec) << "round_double(" << num << ", " 
<< prec << ") = " << round_double(num, prec) << endl;

Will print round_double(10.123456789987655000, 18) = -9.223372036854776500 for first round_double and round_double(10.123456789987655000, 18) = -9.223372036854776500for second one.

How write good round_double function in c++? Or there is already exists?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-21T09:09:51+00:00Added an answer on May 21, 2026 at 9:09 am

    Don’t cast to long long that is forcing a conversion to an integer with limited range, beyond what 10^13 requires (well 19 for 64-bit with no whole number part). Just calling floor should be enough.

    double round_double(double num, int prec)
    {
        double tmp = pow(10.0, prec);
        double result = floor(num * tmp + 0.5);
        result /= tmp;
        return result;
    }
    

    Note that Mike is also correct, you have a limited range you can represent just in double itself. It isn’t so great if you need clean decimal responses. But the long long is the cause of your totally wacky numbers.

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