CS 3721: Programming Languages Lab
Lab #02: Recursion over Lists in Scheme
Due date: February 5, 3:30pm. At the beginning of the next recitation.
Goals of this lab:
•How to conditionally evaluate expressions using cond ;What lambda is ;How to create
and use an anonymous function with lambda ;How to use define and lambda to create
a named function ;How to build a simple function that recurses over a list
1. Conditionals in Scheme. We will talk about two control-flow operators, if, and cond.
cond syntax:
(cond < clause1> < clause2> ...)
Each < clause > should have the form:
(< test > < expression >)
and the last < clause > may be an “else clause” which has the form:
(else < expression >)
Semantics: Acond expression is evaluated by evaluating the < test > expressions of
successive < clause >s in order until one of them evaluates to a true value. When
a< test > evaluates to a true value, then the corresponding < expression > in its
< clause > is evaluated, and the result of that < expression > is returned as the result
of the entire cond expression.
What does the following cond expression evaluate to? Fill in the blank.
(cond
((< 3 1) 3)
((> 4 2) (> 2 4)) => (evaluates to) _______________
((= 5 5) ’(= 5 5))
(else ’else)
)
Write your own example expression using the cond operator, and then write down what
that expression evaluates to.
if syntax:
(if < test > < consequent > < alternate >) , or
(if < test > < consequent >)
Semantics: An if expression is evaluated as follows: first, < test > is evaluated. If it
yields a true value, then < consequent > is evaluated and its value is returned. Otherwise
< alternate > is evaluated and its value is returned. Write an example expression using
the if operator, and then write down what that expression evaluates to.
2. Function expressions. Scheme computation is based on functions and recursive calls
instead of on assignment and iterative loops.
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