SICP 4.1.2中definition-variable的(caadr exp)返回值咨询

Let's break this down clearly—your question cuts to a key detail of how Scheme parses definitions, so it's a great one to unpack.

First, let's recap the two valid define forms covered in SICP 4.1.2:

• Variable definition: (define ⟨var⟩ ⟨value⟩)
  Here, (cadr exp) directly grabs the symbol ⟨var⟩ (for example, in (define x 5), (cadr exp) returns x).
• Procedure definition: (define (⟨var⟩ ⟨param₁⟩ … ⟨paramₙ⟩) ⟨body⟩)
  This is syntactic sugar for (define ⟨var⟩ (lambda (⟨param₁⟩ … ⟨paramₙ⟩) ⟨body⟩)). Unlike variable definitions, (cadr exp) here isn't a single symbol—it's the list (⟨var⟩ ⟨param₁⟩ … ⟨paramₙ⟩).

Now, (caadr exp) is shorthand for (car (car (cdr exp)))—let's use a concrete example to see what it does:
Take the procedure definition (define (add a b) (+ a b)):

1. (cdr exp) gives ((add a b) (+ a b))
2. (car (cdr exp)) (aka cadr exp) gives (add a b)
3. (car (cadr exp)) (aka caadr exp) gives add—a symbol, not null or '().

Why it never returns null or '() #

In valid Scheme syntax (per SICP's rules), a procedure definition's (⟨var⟩ ⟨param₁⟩ …) part must start with a symbol (the procedure name). That list can't be empty, and its first element can't be null—so (caadr exp) will always resolve to a valid symbol (the name of the procedure being defined).

Even for edge cases like a procedure with no parameters: (define (foo) 42)—(cadr exp) is (foo), so (caadr exp) still returns foo, not an empty value.

内容的提问来源于stack exchange，提问作者Wizard