[+r6rs-discuss]
It occurs to me that because of the R6RS procedure equivalence rules,
it's impossible to efficiently implement something like `hashtable-map`
(which lifts a function to the domain of hash tables) while being
strictly conformant. To do so, one must be able to create a hash
table
[Warning: sent to both r6rs-discuss and scheme-reports]
John Cowan wrote:
The following code fragment is not portable:
(let ((equiv? (hashtable-equivalence-function some-hash-table)))
(cond
((eqv? equiv? eq?)
(make-hashtable-eq))
((eqv? equiv? eqv?)
