[+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 with the same equality procedure and hash function as a given
hash table. The `hashtable-copy` function does this, but it also copies
the associations, which is not wanted in this case.
However, in order to create a most-efficient hash table whose equality
predicate is `eq(v)?`, you need to invoke `make-eq(v)-hashtable` rather
than the general `make-hashtable`. But although it is possible to
retrieve the equality predicate using `hashtable-equivalence-function`,
and although R6RS guarantees that this procedure returns `eq(v)?`
in the relevant cases, you cannot test whether that is what you have!
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?)
(make-hashtable-eqv))
(else
(make-hashtable equiv?))))
Code like this actually appears in the R6RS implementation of SRFI 69 at
<https://code.launchpad.net/~scheme-libraries-team/scheme-libraries/srfi>,
which suggests that in fact none of the R6RS implementations are making
use of this optimization, at least in this kind of case.
--
Only do what only you can do. John Cowan <co...@ccil.org>
--Edsger W. Dijkstra's advice
to a student in search of a thesis
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