Martin Ba - 5 months ago 47

C++ Question

**TL;DR** Do the IEEE754 floating point values *including*

`NaN`

`LessThanComparable`

Specifically, question "Why does Release/Debug have a different result for std::min?" got me looking up

`LessThanComparable`

The type must work with < operator and the result should have standard

semantics.

## Requirements

The type T satisfies LessThanComparable if

Given

- a, b, and c, expressions of type T or const T

The following expressions must be valid and have their specified

effects

Establishesstrict weak orderingrelation with the following properties

(...)

I double checked it in the standard and it seems it states basically the same there, and I looked at the Wikipedia def of

Some think that the set of IEEE floating point vales that includes NaN does

`NaN`

For the list given at Wikipedia:

- For all x in S, it is not the case that x < x (irreflexivity).
- For all x, y in S, if x < y then it is not the case that y < x (asymmetry).
`see below`

- For all x, y, z in S, if x < y and y < z then x < z (transitivity).
- For all x, y, z in S, if x is incomparable with y (neither x < y nor y < x hold), and y is incomparable with z, then x is incomparable with z (transitivity of incomparability).

It seems that the strict weak ordering axiom as defined on Wikipedia explicitly calls out possible

`incomparable`

On the other hand, the standard says: (25.5/4)

If we defineas`equiv(a, b)`

, then the`!comp(a, b) && !comp(b, a)`

requirements are that comp and equiv both be transitive relations:

(4.1) —implies`comp(a, b) && comp(b, c)`

`comp(a, c)`

(4.2) —implies`equiv(a, b) && equiv(b, c)`

`equiv(a, c)`

With these definitions,

`equiv(x, NaN)`

`true`

`!comp(a, NaN)==true`

`!comp(Nan, a)==true`

But clearly (4.2) is not satisfied, e.g:

`equiv(3.0, NaN) && equiv(NaN, 7.0) **does not** imply equiv(3.0, 7.0)`

So is what the standard defines not a Strict Weak Ordering, or -- more likely indeed -- am I missing something here?

Answer

Strict weak ordering requires that strongly-ordered equivalence classes exist. This is not true of IEEE754.

The problem isn't that there exist multiple NaN values which are equivalent to each other, but that the entire class of NaNs is unordered with respect to the real line.

The violation of (4.2) causes the test in the fourth bullet point you quoted from Wikipedia to also fail (let `y`

be a NaN).

For an example of incomparability that is allowed in a strict weak ordering, consider sign-magnitude integers. Then:

-4 < -3 < -2 < -1 < { -0, +0 } < +1 < +2 < +3 < +4

Neither `-0 < +0`

nor `+0 < -0`

is true, so the ordering is weak. But the class formed by these equivalent values is strongly ordered with respect to all others.