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Re: [Scheme-reports] EQV? on numbers should be based on operational equivalence

On 05/08/2012 12:06 PM, John Cowan wrote:
> Andrew Robbins scripsit:
>>> I believe that they are not: no R7RS (or R6RS, for that matter) standard
>>> procedure can distinguish between one NaN and another.
>> I beg to differ. Consider the functions:
> [neat hack using bytevectors snipped]
> Right enough.  I suppose such bytevector hacks will have to be removed
> from the definition of "operationally equivalent" if it is to be adopted.
> R3RS already excludes eq? and eqv? as well as functions defined in
> terms of them, such as {mem,ass}{q,v}.

Why?  If two number written out to bytevectors (or binary files in general)
have different bit-patterns, they're not operationally equivalent, and
they're not eqv?.  What is the problem?
	--Per Bothner
per@x   http://per.bothner.com/

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