[r6rs-discuss] [Formal] Unclear how equality predicates behave on NaN

[Per Bothner]

Submitter's name: Per Bothner
Submitter's email address: per at bothner.com
Type of issue: Defect
Priority: Major
R6RS component: Arithmetic
Version of the report: 5.91
One-sentence summary of the issue: Unclear how equality predicates 
behave on NaN

9.10.2 Numerical operations p 39
Arithmetic operations
= explicitly specifies the behavior for +inf.0 and -inf.0, but not
for +nan.0.  Implicitly, if any operand is +nan.0, then the result
should be #f, and this follows IEEE-754.  However, it should be
stated explicitly.
(The same is also true for <, >, <= and >=.)

However, in that case the specification of eqv? (9.6 p 35) is wrong.
It says that (eqv? x y) implies that (= x y) for numbers.  This is
wrong: (eqv? +nan.0 +nan.0) must return #t.  This follows because
we really need (let ((x +nan.0)) (eq? x x)) to be #t.

If x and y are flonums of the same resolution/implementation,
then a reasonable definition of (eqv? x y) is that it is true
if and only y are bit-for-bit equal.
-- 
	--Per Bothner
per at bothner.com   http://per.bothner.com/