如何安全地比较 std::complex<double> 零和一些精度 Epsilon？ </double>

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Current answer is wrong and don't tell me how to compare is complex double near to zero

I need to safely check is my complex number a zero (or very similar to it). How can I do it for floating point numbers?

Can I use something like:

std::complex<double> a;
if(std::abs(std::real(a)) <= std::numerical_limits<double>::epsilon() && std::abs(std::imag(a)) <= std::numerical_limits<double>::epsilon())
{
//...
}

I will divide values by a and don't want to get the INF as result.

Answer 1

I need to check is my complex number a zero. How can I do it for floating point numbers?

You can compare it with a floating point literal with value of 0. You cannot use an integer literal. Example:

a == 0.0

Can I use something like...

What you've shown doesn't compare whether the complex number is zero; it compares whether the complex number is near zero.

Whether that is a good way to compare if the number is near zero depends on use case. For example, epsilon is not necessarily the best threshold of "near". Another thing that you might consider is whether you should compare std::abs(a) to the threshold instead of comparing the components separately ie whether you should use euclidean distance instead of manhattan distance.

Answer 2

Have you tried std::fpclassify from <cmath> ?

if (std::fpclassify(a.real()) == FP_ZERO) {}