Thanks Jack and Mike! You are right. Arguments to procedures will be evaluated before the invocation of the procedure. I thought that because (if) is not an ordinary procedure, and because one can express if in terms of cond (or vice versa) that my procedure is also a non ordinary procedure, which is not the case.
Thanks again On Sunday, January 13, 2019 at 7:29:10 AM UTC+2, Jack Rosenthal wrote: > > On Sat, 12 Jan 2019 at 21:12 -0800, Hassan Shahin wrote: > > When I apply the procedure to 'John it will evaluate to 'symbol. The > idea > > of the procedure is to check the "type" of the given item, which is not > > decided aprior. If I give it 'John I know it is a symbol. > > May be my question should be formulated as this: Since John is not a > pair, > > an empty-list, a number, not a symbol, why the execution doesn't proceed > to > > the else expression no matter what John is, unless you are telling me > that > > the item given to the procedure is evaluated first. > > Perhaps you meant to define John before: > > (define John '(1 2 3)) > (type-of John) => 'pair > > Since type-of is a procedure, this means that when it is applied, it's > arguments will be evaluated during application. Evaluating a symbol > which is not defined will result in an error. > > -- > Jack M. Rosenthal > http://jack.rosenth.al > > A virtual filesystem? I don't know what you are trying to achieve, > but there's probably a better way. > -- Me > > -- You received this message because you are subscribed to the Google Groups "Racket Users" group. To unsubscribe from this group and stop receiving emails from it, send an email to racket-users+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
