Notes: Wednesday 22 January 2003
| Schedule |
- Now: Problem Set 1
- 24 January: Read SICP, 1.2
- 27 January: Read GEB, Little Harmonic Labyrinth and Chapter 5
- 29 January (needed for PS2): Read SICP, 1.3
- 3 February: Problem Set 2
Expression ::= Primitive
Evaluation Rule 1: Primitives. If the expression is a primitive, it is self-evaluating.
Expression ::= Name
Evaluation Rule 2: Names. If the expression is a name, it evaluates to the value associated with that name.
Expression ::= (Expression ExpressionList)
ExpressionList ::=
ExpressionList ::= Expression ExpressionList
Evaluation Rule 3: Application. If the expression is an application:
(a) Evaluate all the subexpressions of the combination (in any order)
(b) Apply the value of the first subexpression to the values of all the other subexpressions.Application Rule 1: Primitives. If the procedure to apply is a primitive, just do it.
Application Rule 2: Compound Procedures. If the procedure is a compound procedure, evaluate the body of the procedure with each formal parameter replaced by the corresponding actual argument expression value.
Expression ::= (lambda (Parameters) Expression)
Parameters ::=
Parameters ::= Name Parameters
Eval 4-lambda. Lambda expressions self-evaluate. (Do not do anything until it is applied.)Expression ::= (define Name Expression)
Eval 4-define. If the expression is (define Name Expression) associate the Expression with Name (for Evaluation Rule 2).Expression ::= (if Expression0 Expression1 Expression2)
Eval 4-if. If the expression is (if Expression0 Expression1 Expression2) evaluate Expression0. If it evaluates to #f, the value of the if expression is the value of Expression2. Otherwise, the value of the if expression is the value of Expression1.Expression ::= (begin ExpressionList Expression)
Eval 4-begin. If the expression is (begin ExpressionList Expression) evaluate all the sub-expressions in ExpressionList in order from left to right. Then evaluate Expression. The value of the begin expression is the value of Expression.
The common syntax for defining a procedure is actually syntactic sugar (an easier way to write something that means exactly the same thing) for a lambda expression. For example,
(define (square x) (* x x))
is just a short way of expressing:
(define square (lambda (x) (begin (* x x))))
((lambda (x y) (+ x y)) 100 (* 10 10)).
You will need more space for this. Once you feel confident you can evaluate any Scheme expression following the Rules of Evaluation, you should stop.
Evaluation Rule 3a does not say in what order the subexpressions should be evaluated in. For example, we could evaluate them left to right, or right to left, or in any other order. This is like the MIU-system Rule 3 that does not say which occurance of III should be replaced. Does it ever matter in which order the subexpressions of an application are evaluated? (Tough question, but try to think of a Scheme expression where it would make a difference.)
- Why Verb-Initial Languages are Not Frequent [PDF], Andre Gr¨ning, May 2002. Interesting paper that speculates on why most natural languages are SOV or SVO using results from computer simulations.
- Revised5 Report on the Algorithmic Language Scheme [PDF] - this is the official definition of the Scheme language
- C++ Standard Core Language Active Issues (389 issues about C++ that the language experts can't agree on)
"Somehow it seems to fill my head with ideas - only I don't exactly know what they are!"
The best book on programming for the layman is Alice in
Wonderland;
Alice, in Alice and Wonderland by Lewis Carroll after hearing The Jaberwocky.
but that's because it's the best book on anything for the layman.
