关于 eq 的 Scheme 和 R5RS 问题

Question

Can you explain why the first one is false and the second one is true?

And how this works? Thanks.

(eq? '(1 2 3) '(1 2 3)) ;False
(eq? '() '()) ;True

Answer 1

There's only one empty list, so all uses of () refer to that list, and it's eq? to itself. The Scheme Specification description of the storage model says:

Notwithstanding this, it is understood that the empty list cannot be newly allocated, because it is a unique object.

and the specification of eqv? (which is referenced by the eq? description) says that two objects are equivalent if

obj ₁ and obj ₂ are both the empty list

But when you create a non-empty list, it creates a fresh one every time, and they're not eq? to each other even if they contain the same elements.

Answer 2

Quoted from TSPL3 :

[..] Two objects are considered identical if they are represented internally by the same pointer value [..] The empty list () is identical to itself wherever it appears. [..] Two pairs, vectors, or strings created by different applications of cons, vector, string, etc., are distinct.

If you write instead

(let ((x '(1 2 3)))
  (eq? x x))

it will be #t .