[r6rs-discuss] Comparison procedures' number of arguments

[Ken Dickey]

On Saturday 18 October 2008 18:56:01 John Cowan wrote:
> Derick Eddington scripsit:
> > Why does R6RS specify that predicates like =, <, >=, char<?, fx>=?, etc.
> > accept a minimum of two arguments instead of accepting zero or more?
>
> What would be the "obvious" value of (< 2) or (=)?  I have no idea whether
> #t or #f would win here.

#!NaB  [Not a Boolean].   8^)

But this would seriously torque compiler logic requiring booleans.  [NaN is 
only seen as a win because failure checking can be delayed and save some 
computational time.  Personally I prefer exceptions].

#f is the obvious base case as comparisons with nothing can't be true and in 
Scheme anything not #f is true (even #!Nan).

> (if +nan.0 1 2)
1
> (< +nan.0 1)
#f
> (< 1 +nan.0)
#f

So #f is the obvious, consistent base case.


$0.02,
-KenD

"Do you walk to school or take a lunch?"  ;^)