Scheme getting last element in list
Solution 1
Your syntax is totally wrong. You have an extra set of parentheses around the body of the function, not enough around the cond
clauses, and your recursive case isn't even within the cond
, so it gets done whether the test succeeds or fails. The following procedure should work:
(define (last_element l)
(cond ((null? (cdr l)) (car l))
(else (last_element (cdr l)))))
Solution 2
Just to add: in professional-level Racket, the last function is a part of the racket/list library.
Solution 3
you can retrieve the last element of a list by calling
(define (lastElem list) (car (reverse list)))
or, recursively using if built-in
(define (last list)
(if (zero? (length (cdr list)))
(car list)
(last (cdr list))))
calccrypto
Updated on July 06, 2020Comments
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calccrypto almost 4 years
Im trying to write a simple scheme function that returns the last element of a list. My function looks like it should work, but I managed to fail on something:
(define (last_element l)( (cond (null? (cdr l)) (car l)) (last_element (cdr l)) )) (last_element '(1 2 3)) should return 3
DrRacket keeps on giving me the errors:
mcdr: contract violation expected: mpair? given: ()
Since
(null? '())
is true, I don't get why this doesn't work.This is a function I think I will need for a homework assignment (writing the function
last-element
is not the assignment), and the instructions say that I cannot use the built-in functionreverse
, so I can't just do(car (reverse l))
How do I fix this function?
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Greg Hendershott over 11 years@calccrypto The glass-half-full version of this answer is that you had nearly all the individual elements correct. Just keep in mind that parentheses aren't the same thing as curly braces; you'll want to break that habit of an "opening
(
" on the first line, and putting)
s on their own line. Also, let DrRacket guide you with the indentation. You'll get the hang of it soon.