At Fri, 03 Aug 2012 14:11:20 +0100, Tim Brown wrote: > > (for ([j (in-range 10)] [i (in-cycle (in-range 3) (in-value (displayln > ".")))]) (displayln i)) > stdout: ".\n0\n1\n2\n#<void>\n0\n1\n2\n#<void>\n0\n1\n" > > Which suggests that the in-value is evaluated once printing a dot and > returning #<void> -- then returning void (which is printed out with the > (displayln i) clause.
More precisely, evaluation of the `in-value' call is evaluated once, printing a dot and returning a sequence that contains #<void>. > The documentation for in-cycle says: > "where each seq is initiated afresh in each iteration" > > Can (in-value ..) not be "initiated afresh"? > Why not? > Is there a bug in the documentation of (in-cycle)? > Or should (in-value) be documented as "cannot be initiated afresh"? I think you are confusing "initiating a sequence" with "evaluating an expression". Evaluation of an `in-cycle' call evaluates all of its arguments to sequence values. The documentation for `in-cycle' doesn't say that explicitly, but `in-cycle' is documented as a procedure, so `(in-cycle ....)' is a procedure call, and that's the way procedure calls work. So, as you say, `(in-value (displayln "."))' is evaluated once to the sequence that contains just #<void>, while `(in-range 3)' is evaluated to the sequence that contains 1, 2, and 3. Those are put together into a cycle sequence that is the result of `(in-cycle (in-range 3) (in-value (displayln ".")))'. As the cycle sequence's elements are requested, the sequence's implementation initiates each of the given number and #<void> sequences once per cycle. That means it restarts sequences of 1, 2, 3, and #<void>. It doesn't (and can't) re-evaluate the expressions that produced the sequences. Hope that helps, Matthew ____________________ Racket Users list: http://lists.racket-lang.org/users
