Am Fr., 28. Okt. 2022 um 22:55 Uhr schrieb Shiro Kawai <shiro.ka...@gmail.com>: > > I assume the same applies to the parameterization, that is, the following > should run in a constant space. Correct? > > (define p (make-parameter 0)) > (let loop () > (parameterize ((p 1)) > (loop)))
Yes, this is the expectation (and a reason why parameterize possesses a tail-context guarantee. The idea is that a parameter is conceptually the same as an implicit argument. An a loop like (let loop ((p 0)) (loop 1)) also runs in bounded space. I should check whether the sample implementation is correct in this regard.
