I'm reading the book called the little schemer.

Before read that, i've finished reading first three chapter of SICP.

My question is that why second argument to cons must be a list.

However, (cons a b) works for all values a and b and

(car (cons a b)) = a

(cdr (cons a b)) = b

有帮助吗?

解决方案

The second argument to cons is not necessarily a list. It's a list only if you're, well, building a list (proper or otherwise). It's perfectly valid if the cdr part of a cons cell is not a list, for example, when building an association list:

(define lookup-table (list (cons 'x 10) (cons 'y 20) (cons 'z 30)))
(assoc 'z lookup-table)
=> '(z . 30)

其他提示

Not all implementations of Lisp allow a non-list as the second argument to cons. For example, see https://scheme.cs61a.org/

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