Notes: Friday 27 February 2004

Schedule

• Friday, 5 March: Problem Set 5
• Before 15 March: Read rest of GEB part I (Chapters 2-4 and 6-9, in addition to Chapters 1 and 5 you have already read).

Notes

Places, Frames and Envrionments

• Names identify places. (define x expr) and (lambda (x) expr) both create places named x.
• Frames hold names. All names must be inside a frame.
• An environment is a pointer to a frame.
• All frames except the outermost frame have pointers to a parent frame.
• The global environment is a pointer to the outermost frame. The outermost frame contains all the Scheme primitives.

(define nest 
  (lambda (x)
    (lambda (x)
      (+ x x))))

Compound Application Rule

1. Construct a new frame, enclosed in the environment of this procedure (that is, the parent of this frame is the frame this procedure's environment points to)
2. Make places in that frame with the names of each parameter
3. Put the values of the parameters in those places
4. Evaluate the body in the new environment

Why must expressions be evaluated in an environment?

How do we find the value associated with a name in an environment?