Why not impl<T> Borrow<T> for NonNan<T> ?

[changhe3]

It seems we can satisfy the following behavior as long as we exclude NAN:

In particular Eq, Ord and Hash must be equivalent for borrowed and owned values: x.borrow() == y.borrow() should give the same result as x == y.

[mbrubeck]

Fixed in bfabcc9. However, this has limited use because:

1. It is implemented only for NotNan<f32> and NotNan<f64>. Implementing it for a generic T would be incorrect because T could implement Hash in a way that is inconsistent with NotNan::hash.

2. This isn't useful for HashMap::get or BTreeMap::get, because f32 and f64 do not implement Hash or Ord themselves.

[changhe3]

I think this is useful for the extremely rare cases where you use float as keys for range queries in data structures like btreemultimap.

[mbrubeck]

Unfortunately, I think that fails for the same reason. BTreeMultiMap::<K, V>::range has the following signature:

    pub fn range<T: ?Sized, R>(&self, range: R) -> MultiRange<'_, K, V>
    where
        T: Ord,
        K: Borrow<T>,
        R: RangeBounds<T>,

When K = NotNan<f32>, R = Range<f32>, T = f32, this fails the T: Ord bound.