问题
While attempting to generate a list of subsets that exist within two sets, I am running into an issue with cons. A procedure takes in a list named result and attempts to construct a new list out of result and the car of another set. So far, the set is generated with the correct elements, but they are contained within a size N nested list, where N is the number of nestings and the number of elements within the subset for which I am searching.
How can I apply cons with result without creating a nested list?
Example:
;These are two sets that I will be checking
(define s1 '(1 2 3 4))
(define s2 '(1 2 3))
;Result from running these lists through my procedure
(((() . 1) . 2) . 3)
;What I want to have generated
(1 2 3)
I need to be able to call (car list) and receive 1 not ((() . 1) . 2)
回答1:
First of all (((() . 1) . 2) . 3) is not a nested list - it's not a list at all. (((() 1) 2) 3) would be a nested list. (((() . 1) . 2) . 3) is a dotted pair whose first element is also a dotted pair.
So now to explain the behaviour you see: In the lisp family of languages (cons a b) creates a pair containing a as it's car and b as its cdr.
Now a list is either the empty list (()) or a pair whose cdr is also a list (the car might contain anything - if both the cdr and the car are a list, the list is called a nested list).
A pair that is not a list is called a dotted pair. This what you're creating here because you're calling cons with a second argument which is not a list.
Conclusion:
You can not use cons to append to the end of the list. If you need to append to the end of a list, you can use concat with a single-element list as the second argument.
Note however that appending to the end of a list, is an O(n) operation while appending at the front (using cons) is an O(1) operation, so if possible you should change your algorithm, so it only needs to append at the front.
来源：https://stackoverflow.com/questions/4109317/how-to-use-cons-without-generating-nested-lists-in-scheme