c++ - 在比较 float 时使用 epsilon 是否会破坏严格弱排序？

Question

以下类(class)是否打破了严格的弱排序(与常规 std::less 相比(因此忽略边缘情况值，例如 Nan))

struct LessWithEpsilon
{
    static constexpr double epsilon = some_value;
    bool operator() (double lhs, double rhs) const
    {
        return lhs + epsilon < rhs;
    }
};

LessWithEpsilon lessEps{};

Answer 1

来自 https://en.wikipedia.org/wiki/Weak_ordering#Strict_weak_orderings

4. Transitivity of incomparability: For all x,y,z in S, if x is incomparable with y (meaning that neither x < y nor y < x is true) and if y is incomparable with z, then x is incomparable with z.

同样，来自 https://en.cppreference.com/w/cpp/named_req/Compare

If equiv(a, b) == true and equiv(b, c) == true, then equiv(a, c) == true

与 {x, y, z} = {0, epsilon, 2 * epsilon} ，该规则被打破:

!lessEps(x, y) && !lessEps(y, x) && !lessEps(y, z) && !lessEps(z, y)但是 lessEps(x, z) .

equiv(x, y) == true and equiv(y, z) == true但是 equiv(x, z) == false (如 x + epsilon < z)

因此，该类打破了严格弱序。