<div class="gmail_quote">2009/4/5 David Van Horn <<a href="mailto:dvanhorn@ccs.neu.edu">dvanhorn@ccs.neu.edu</a>> <blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"> <div class="im">Jon Loldrup wrote: <blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"> Well it sure does evaluate successfully now, and testing the first 4 expressions on page 4 in "The Reasoned Schemer" makes them all return the same value as the ones in that book. Only thing is, the book doesn't mention the definitions of fail and succeed (only that they are longs for #u and #s), so I'm not sure whether this solution is some crude hack or the official way to do it. </blockquote> </div> The book gives definitions of #s and #u, both in English and Scheme. See the code at the end of the book. The definitions you gave are equivalent, so this is no crude hack. </blockquote><div> That doesn't make sense in my mind: If #u and #s are actually defined how come I can't use them and have to use the longs "fail" and "succeed"? eg.: the line (run #f (q) #u (== #t q)) results in Module Language: invalid module text read: bad syntax `#u' mvh Jon Loldrup </div></div>