LISP中M表达式list[x;y]的功能、对应表达式及理解确认咨询

Great question—diving into McCarthy’s foundational LISP papers is tough work, so kudos for pushing through even when it’s over your current level! Let’s break this down clearly:

1. Confirming list[x;y] is an M-expression #

You’re exactly right—list[x;y] is an M-expression. McCarthy introduced M-expressions as a more human-readable meta-language to describe LISP’s core syntax and semantics, back before S-expressions became the dominant way to write LISP code directly.

2. Is your understanding of list[x;y] as (e1, e2) correct? #

Your intuition is spot-on, with a small refinement:

• In McCarthy’s original framework, list[x;y] denotes a 2-element list (not just a raw ordered pair) where elements are the results of evaluating x and y respectively.
• If you’re thinking of (e1, e2) as a paired set of elements, that’s accurate—but in early LISP terms, this is specifically a list terminated by nil (the empty list). A raw ordered pair (without the nil terminator) would use the cons[x;y] M-expression, corresponding to the S-expression (e1 . e2).

3. What S-expression does list[x;y] correspond to? #

For the general case list[x;y], the equivalent S-expression is (x y)—this is shorthand for the full dotted pair (x . (y . nil)).

Applying this to your specific example list[cadar[e];car[e]]:

• First, convert each M-expression subterm to S-expression syntax: cadar[e] becomes (cadar e), and car[e] becomes (car e).
• Then wrap them in a list as per the list constructor: the final S-expression is ((cadar e) (car e)).

To clarify: this is the syntactic translation of the M-expression to runnable S-expression code. When evaluated, it produces a two-element list where the first element is the result of (cadar e) and the second is the result of (car e).

内容的提问来源于stack exchange，提问作者volf