Variable Definitions

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Variable Definitions

Programs consist not only of function definitions but also variable definitions, but these weren't included in our first grammar.

Here is the grammar rule for variable definitions:

The shape of a variable definition is similar to that of a function definition. It starts with a ``('', followed by the keyword define, followed by a variable, followed by an expression, and closed by a right parenthesis ``)'' that matches the very first one. The keyword define distinguishes variable definitions from expressions, but not from function definitions. For that, a reader must look at the second component of the definition.

Next we must understand what a variable definition means. A variable definition like

(define RADIUS 5)

has a plain meaning. It says that wherever we encounter RADIUS during an evaluation, we may replace it with 5.

When DrScheme encounters a definition with a proper expression on the right-hand side, it must evaluate that expression immediately. For example, the right-hand side of the definition

(define DIAMETER (* 2 RADIUS))

is the expression (* 2 RADIUS). Its value is 10 because RADIUS stands for 5. Hence we can act as if we had written

(define DIAMETER 10)

In short, when DrScheme encounters a variable definition, it determines the value of the right-hand side. For that step, it uses all those definitions that precede the current definition but not those that follow. Once DrScheme has a value for the right-hand side, it remembers that the name on the left-hand side stands for this value. Whenever we evaluate an expression, every occurrence of the defined variable is replaced by its value.

(define RADIUS 10)

(define DIAMETER (* 2 RADIUS))

;; area : number -> number 
;; to compute the area of a disk with radius r 
(define (area r) 
  (* 3.14 (* r r)))

(define AREA-OF-RADIUS (area RADIUS))

Figure: An example of variable definitions

Consider the sequence of definitions in figure . As DrScheme steps through this sequence of definitions, it first determines that RADIUS stands for 10, DIAMETER for 20, and area is the name of a function. Finally, it evaluates (area RADIUS) to 314.0 and associates AREA-OF-RADIUS with that value.

Exercises

Exercise 8.6.1

Make up five examples of variable definitions. Use constants and expressions on the right-hand side. Solution

Exercise 8.6.2

Evaluate the following sequence of definitions

(define RADIUS 10)

(define DIAMETER (* 2 RADIUS))

(define CIRCUMFERENCE (* 3.14 DIAMETER))

by hand. Solution

Exercise 8.6.3

Evaluate the following sequence of definitions

(define PRICE 5)

(define SALES-TAX (* .08 PRICE))

(define TOTAL (+ PRICE SALES-TAX))

by hand. Solution

Next: Structure Definitions Up: Intermezzo 1 Syntax and Semantics Previous: Boolean Expressions

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