In previous labs, you've seen that it's possible to define complex
expressions from simpler expressions. For example, we might write the
val explicitly as a rational number''.
(/ (inexact->exact (numerator val)) (inexact->exact (denominator val)))
What happens when we want to try this expression using different values? One possibility is to redefine grade and then reexecute the expression.
> (define val (sqrt 2)) > (/ (inexact->exact (numerator val)) (inexact->exact (denominator val))) 1592262918131443/1125899906842624 > (define val 5) > (/ (inexact->exact (numerator val)) (inexact->exact (denominator val))) 5 > (define val (exp 1)) 765128314358509/281474976710656
But this seems cumbersome. It would be nice to simply give a name to this expression and use that name. Unfortunately, our current way of naming doesn't quite work.
> (define val (sqrt 2)) > (define frac (/ (inexact->exact (numerator val)) (inexact->exact (denominator val))) > frac 1592262918131443/1125899906842624 > (define val 5) > frac 1592262918131443/1125899906842624
What's going on? Scheme evaluated the expression that accompnaies
frac once, when it was first defined. Hence,
since the expression had a particular value once, it retains that
What we'd really like to do is to say that ``
is a procedure that takes a value as an input and returns
an appropriate fraction''. You know that Scheme has procedures, since
you've used lots of built-in procedures, including
list. But can you
define your own procedures? Yes.
define to give names to procedures, just as you use it
to give names for values. The values just look different. The general
form of a procedure is
(lambda (formal-parameters) expression)
For example, we might write our
frac procedure as
(define frac (lambda (val) (/ (inexact->exact (numerator val)) (inexact->exact (denominator val)))))
valid procedure can now be called as if it were a built-in
> (frac (sqrt 2)) 1592262918131443/1125899906842624 > (frac 5) 5 > (frac 22/7) 22/7
We can define procedures for anything we already know how to do in Scheme.
For example, here is a simple
(define square (lambda (n) (* n n)))
We can test it.
> (square 2) 4 > (square -4) 16 > (square square) *: expects type <number> as 1st argument, given: #<procedure:square>; other arguments were: #<procedure:square> > (square 'a) *: expects type <number> as 1st argument, given: a; other arguments were: a
Convention in Scheme (and all programming languages) is that we carefully document what our procedures do, including input values, output values, and assumptions. We use comments provide information to the reader of our program (that is, to people instead of the computer). In Scheme, comments begin with a semicolon and end with the end of the line.
;;; Samuel A. Rebelsky ;;; Department of Mathematics and Computer Science ;;; Grinnell College ;;; email@example.com ;;; Procedue: ;;; square ;;; Parameters: ;;; val, a number ;;; Purpose: ;;; Compute val*val ;;; Produces: ;;; The result of the computation ;;; Preconditions: ;;; val must be a number ;;; Postconditions: ;;; The result is the same "type" of number as val (e.g., if ;;; val is an integer, so is the result; if val is exact, ;;; so is the result). ;;; Citations: ;;; Based on code created by John David Stone dated March 17, 2000 ;;; and contained in the Web page ;;; http://www.math.grin.edu/~stone/courses/scheme/procedure-definitions.xhtml ;;; Changes to ;;; Parameter names ;;; Formatting ;;; Comments (define square (lambda (value) (* value value)))
At times, we'll want to write procedures that take more than one parameter. Such procedures look just like procedures with one parameter, except that you can list more parameters between the parentheses.
(lambda (param1, param2 ... paramn) expression )
For example, here is a simple procedure that finds the average of two nubmers
;;; Samuel A. Rebelsky ;;; Department of Mathematics and Computer Science ;;; Grinnell College ;;; firstname.lastname@example.org ;;; Procedure: ;;; pairave ;;; Parameters: ;;; val1, an exact number ;;; val2, an exact number ;;; Purpose: ;;; Compute the average of two numbers. ;;; Produces: ;;; The average of those two numbers. ;;; Preconditions: ;;; Both val1 and val2 are exact numbers. ;;; Postconditions: ;;; The result is an exact number. The result is equidistant ;;; from val1 and val2. (define pairave (lambda (val1 val2) (/ (+ val1 val2) 2)))
Monday, 4 September 2000
Wednesday, 31 January 2001
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