PLI Tutorial 10
Scheme interpretation and compilation
(Draft)

Some of these questions are revision.

1. Write a Scheme function (subst exp old new) to replace every occurrence
   of the symbol old in the list exp by the value new.  The function
   should return a new list and not change the old list.  E.g.,
   (subst '(a (b a c) d) 'a '(e f)) ==> (((e f) (b (e f) d)

2. Add let- and let*- expressions to the final Scheme interpreter, 
   eval6.s.

   Note that let-expressions are shorthand for procedure application:
   (let ((x1 e1) ... (xm em)) exp1 ... expn)
   ==>
   ((lambda (x1 ... xm) exp1 ... expn) e1 ... em)

3. Add (begin exp1 ... expn), n >=0, to the final Scheme interpreter, 
   eval6.s.  This form evaluates each expi in turn and returns the value 
   of the final one.  

4. Add do-expressions to the final Scheme interpreter, eval6.s.

   The syntax of a do-expression, a generalised for-loop, is as follows:
   (do ((v1 init1 incr1) ... (vm initm incrm)) ; m >= 0
       (termination-condition val1 ... valn)   ; valn is returned, n >= 0
     exp1 ... expp)                            ; p >= 0

   See the function (factors n) in the examples file for examples.

   Hint: Note that do-expressions are shorthand for tail-recursive
   procedure application:
   (do (...) (...) exp1 ... expp)
   ==>
   (define (newproc v1 ... vm)
     (if termination-condition
         (begin val1 ... valn)
         (begin exp1 ... expp (newproc incr1 ... incrm))))
   (newproc init1 ... initm) 

5. Construct an example of a circular environment.

I need to write some questions about continuations and compilation...