Scheme console printing

Question

Just started with Scheme. I'm having problem with printing on console. A simple list printing example:

 (define factorial
   (lambda (n)
     (cond 
       ((= 0 n) 1)
       (#t (* n (factorial (- n 1)))))))

I want to print n, every time the function is called. I figured that I can't do that within the same function? Do I need to call another function just so I can print?

Solution

Printing in Scheme works by calling display (and possibly, newline). Since you want to call it sequentially before/after something else (which, in a functional (or in the case of Scheme, functional-ish) language only makes sense for the called functions side-effects), you would normally need to use begin, which evaluates its arguments in turn and then returns the value of the last subexpression. However, lambda implicitly contains such a begin-expression.

So in your case, it would go like this:

 (lambda (n)
   (display n) (newline)
   (cond [...]))

Two remarks:

1. You can use (define (factorial n) [...]) as a shorthand for (define factorial (lambda (n) [...])).
2. The way you implement factorial forbids tail call-optimization, therefore the program will use quite a bit of stack space for larger values of n. Rewriting it into a optimizable form using an accumulator is possible, though.

If you only want to print n once, when the user calls the function, you will indeed need to write a wrapper, like this:

 (define (factorial n)
   (display n) (newline)
   (inner-factorial n))

And then rename your function to inner-factorial.

Comments

[Paul Richter]

You don't need a begin within a lambda body.

[eric17859]

Oh thx! But how about printing after the condition? I'm not working on factorial actually, but it's easier explaining with it. Something like print only if the condition is true, i.e. print if n!=0.

[fnl]

I fixed the part about begin. The body of a cond-clause may also contain multiple expressions, so you can easily just call display there.

[eric17859]

Sry if I wasn't clear. I mean, if n=5, I would print 1 2 3 4 5

[fnl]

I assume you still want to extend factorial to do this? Then it would go like: (define factorial (lambda (n) (cond ((= 0 n) 1) (#t (let ((result (* n (factorial (- n 1))))) (display n) (newline) result)))))

[eric17859]

thanks a lot! Can you tell me why you need to create a variable? why it doesn't work with (display n) (newline) (* n (factorial (- n 1)))

[fnl]

That would also work, but achieve a different effect: Because the current n is now printed before the recursive call, the output would be reversed.

[eric17859]

Hm.. it didn't work. procedure application: expected procedure, given: #<void>; arguments were: #<void> 1.. Ah, I don't get these error msgs.

[fnl]

Having diverged quite a bit from the original question, I recommend you ask a separate one.

[dyoo]

The original answer had a few typos. I've tried to edit the answer to be correct. e.g. (define factorial (n) ...) is syntactically incorrect: it should be (define (factorial n) ...).