[x1f604]

I think the libstdc++ implementation does indeed have the comparisons ordered in the way that you describe. I stepped into the std::clamp() call in gdb and got this:

    ┌─/usr/include/c++/12/bits/stl_algo.h──────────────────────────────────────────────────────────────────────────────────────
    │     3617     \*  @pre `_Tp` is LessThanComparable and `(__hi < __lo)` is false.
    │     3618     \*/
    │     3619    template<typename _Tp>
    │     3620      constexpr const _Tp&
    │     3621      clamp(const _Tp& __val, const _Tp& __lo, const _Tp& __hi)
    │     3622      {
    │     3623        __glibcxx_assert(!(__hi < __lo));
    │  >  3624        return std::min(std::max(__val, __lo), __hi);
    │     3625      }
    │     3626

[cmovq]

Thanks for sharing. I don't know if the C++ standard mandates one behavior or another, it really depends on how you want clamp to behave if the value is NaN. std::clamp returns NaN, while the reverse order returns the min value.

[cornstalks]

From §25.8.9 Bounded value [alg.clamp]:

> 2 Preconditions: `bool(comp(proj(hi), proj(lo)))` is false. For the first form, type `T` meets the Cpp17LessThanComparable requirements (Table 26).

> 3 Returns: `lo` if `bool(comp(proj(v), proj(lo)))` is true, `hi` if `bool(comp(proj(hi), proj(v)))` is true, otherwise `v`.

> 4 [Note: If NaN is avoided, `T` can be a floating-point type. — end note]

From Table 26:

> `<` is a strict weak ordering relation (25.8)

[x1f604]

Based on my reading of cppreference, it is required to return negative zero when you do std::clamp(-0.0f, +0.0f, +0.0f) because when v compares equal to lo and hi the function is required to return v, which the official std::clamp does but my incorrect clamp doesn't.