in scheme what is the meaning of dot not on cons? why sometimes it is valid and sometime not

Question

what is the meaning of dot, other than just cons notation? dot as I know is just cons notation. so I don't understand the meaning here:

why:

> (equal? . 8)
Exception: invalid syntax (equal? . 8)
Type (debug) to enter the debugger.

but:

> (equal? . ((quote . ((a . (b . (c . ()))))) . ('(a b c))))
#t

what is the meaning of the dot here?

question from:https://stackoverflow.com/questions/65918208/in-scheme-what-is-the-meaning-of-dot-not-on-cons-why-sometimes-it-is-valid-and

Answer 1

The syntax gets parsed and for a Scheme reader '(1 . (2 . ())) gets parsed and creates the same structure as '(1 2). Basically proper lists are singly linked lists where the last cdr points to the empty list.

(somefun . (5)) is the same as (somefun 5) for the reader. When the source code is turned in to a data structure, the interpreter won't know if you wrote the one or the other. Hovever (somefun . 5) doesn't get parsed to a function with one argument since the argument list itself is a number. the parser parsed it into a data structure, but your scheme implementation responsible to compile it or interpret it failed. You may use dotted list in data and some syntax like lambda for rest arguments, but it is very limited. Most of the time using dotted pairs in code fails.

A Scheme system will display structure only one way. eg. if you put in '(1 . (2 . ())) you will se (1 2) back since a dot with a list or empty list cdr gets special list visualization. If you do '(1 . (2 . 3)) it can start using the special rule since the cdr is a pair, but the next won't fly ending it up as (1 2 . 3). If you evaluate '(equal? . ((quote . ((a . (b . (c . ()))))) . ('(a b c)))) you get (equal? '(a b c) '(a b c)) back and that is because it is the simplest visualization of that structure. As code they read to the same structure and evaluate the same.

The most common problem when learning Scheme is to understand the list structure. Eg. how to create ((1 2) (3 4)) using cons and how to access the 4. If you knew it is really ((1 . (2 . ()) . ((3 . (4 . ())))) you see the 4 is after fllowing the cdr, car, cdr, then car or the simplified (cadadr '((1 2) (3 4))). The visualizer will just omit the dot and the extra set of parentheses in the cdr if it has them. eg. '((a) . (b)) will become ((a) b)