Question

Using the Scheme functional language AND do not use set! (or any procedure with !)

Question 5 [4 marks] The following program can be used to determine if a given interpreter is using applicative-order or normal- order evaluation: (define (test x y) (if (= x 0) y)) (test 0 (1 3 0)) a. What will be the behaviour of this code on an interpreter that uses applicative-order evaluation? Explain why. b. What behaviour will be observed with an interpreter that uses normal-order evaluation? Explain why.

Answer 1

Answers:

a.

In an applicative-order evaluation, the function arguments are evaluated first before they passed to the function for the operation.

for example:
we have a function (define (add a b) (+ a b))
this statement needs to be evaluated, (add (add 1 2) 3)

It will be evaluated as follows

(add (add 1 2) 3)
(add (+ 1 2) 3)
(add 3 3)
(+ 3 3)
6

First the parameter (add 1 2) was evaluated and then the outer expression was evaluated.

In this program, the function is defined as below,
(define (test x y) (if (= x 0) x y))
The function calling,
(test 0 (/ 3 0))
The function test is called with two parameters,
parameter 1: 0
parameter 2: (/ 3 0)

On evaluation of this program (/ 3 0) will give error because division by zero is not possible. The correct evaluation of the statement (test 0 (/ 3 0)) is as below,

Expression	Description
(test 0 (/ 3 0))	the function is called. Since it is an applicative order evaluation, all the parametes should be evaluated
(/ 0 3)	this will be produce an error.

Hence an interpreter using applicative-order evlauation will throw an error for the given code. Since Scheme is an application-order evaluation, it will throw an error.

b.

In a normal-order evaluation, the evaluation of an expression happens from left to right and the function arguments will be not evaluated until they are necessary.

for example:
we have a function (define (add a b) a+b)
this statement needs to be evaluated, (add (add 1 2) 3)

It will be evaluated as follows

(add (add 1 2) 3)
(+ (add 1 2) 3)
(+ (+ 1 2) 3)
(+ 3 3)
6

Here the execution happened from left to right. First the outer add was evaluated and then the first parameter (add 1 2) was evaluated and at last the whole expression was evaluated.

For the given program,
(define (test x y) (if (= x 0) x y))

(test 0 (/ 3 0))

Evaluation:

Expression	Description
(test 0 (/ 3 0))
(if (= 0 0) 0 (/ 3 0))	function test is evaluated
(if true 0 (/ 3 0))	(= 0 0) is evaluated to be true and now the if condition should be evaluated
0	since the condition is true, 0 is returned. The expression (/ 3 0) is never evaluated

In an interpreter using normal order evaluation, the code will given an output 0. There will be no errors.