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Andrew Wilcox <awilcox@xxxxxxxxxxxxxxxxx> writes: > Two Scheme implementations returning results flagged as exact, from > the same computation, performed on the same inputs, will return > equal values. I believe this is unimplementable sanely. Assume, for the sake of argument, that (= (expt (sqrt 2) 2) 2) is an example of an expression which is #t and #f on different implementations. Of course with your proposal these are inexact booleans. Now consider this: (define (test) (let ((l '())) (if (= (expt (sqrt 2) 2) 2) (set! l (cons #f l))) (length l))) Does this return an inexact 0 or 1? If yes, how would you implement this? -- __("< Marcin Kowalczyk \__/ qrczak@xxxxxxxxxx ^^ http://qrnik.knm.org.pl/~qrczak/