I am going through the book - The Scheme Programming Language, 4th Edition. In an exercise, there is an expression ((car (list + - * /)) 2 3). Below are the steps in which I believe the expression will be evaluated. Please let me know if my understanding is correct.
The expression (list + - * /) is evaluated to a list of procedures: (+ - * /). (I understand that list always produces a "proper list", could someone please state a few differences between using list and cons?)
The expression (car (+ - * /)) is evaluated to the symbol + which evaluates to a procedure. (I don't really understand how (car (+ - * /)) is evaluated, typing (car (+ - * /)) at the REPL prompt produces an error).
The expression (+ 2 3) is evaluated to 5.
I would like if I could get some other / in depth explanation.
Yes! However the procedures doesn't have literal representation so when you evaluate + you get some crazy textual representation like #<primitive-procedure-+> and copying it and pasting it into the repl won't give you the same object back. The same with lists. When you evaluate (list 1 2 3) you get (1 2 3) but you cannot just write (1 2 3) since it will think it should call 1 as a procedure with two arguments. (list 1 2 3) makes (cons 1 (cons 2 (cons 3 '()))) a chain of nested pairs. That the last cdr is () is what makes it proper. Thus the primitive that allows list to do it's thing is cons.
Wrong. You have an evaluated expression (list + - * /) to something like (#<primitive-procedure-+> #<...> #<...> #<...>). A list of evaluated variables and you now see their visual representations. Doing car on it gives you the first object #<primitive-procedure-+> which is the same you get when you evaluate the global variable +. There are no symbols involved in this step. The previous step didn't involve symbols either since bare symbols are variables. 'test becomes a symbol while test becomes whatever the variable pointed to. All procedures by name are just variables getting evaluated before application. It's the default behaviour in Scheme.
Since the object is the same as the value + is it will add the rest of the operands together after they have been evaluated. Since they all are numbers the arguments passed to apply stay the same, but if you have expressions there like (+ (* 3 3) (* 4 4)) then the operands like (* 3 3) need to get evaluated and it's result is what is applied.
You can apply substitution rules. It's not what the Scheme interpreter does, but any variable can be replaced by the value it has and every expression can be replaced by its result as long as you do not mutate anything:
((car (list + - * /)) (- 5 2) (/ 4 2)) ; == (car (list + - * /)) evaluates to the same value as the variable +
(+ (- 5 2) (/ 4 2)) ; == (- 5 2) evaluates to 3, (/4 2) evaluates to 2
(+ 3 2) ; == (+ 3 2) evaluates to 5
; ==> 5 (the result)
Notice I substitute (car (list + - * /)) with the variable + and not a symbol. They both are the same: (eq? (car (list + - * /)) +) ; ==> #t
The stepper in DrRacket is the perfect tool for exploring evaluation order.
The stepper allows you to see the effect of one evaluation step at a time.
You can even step backwards.
The stepper however only works in the Beginner and Intermediate languages.
The image is from the following program using the Intermediate language.
(car (list + - * /))
(+ 2 3)
The quote (') is used to introduce a pre-evaluated value, so (quote x) results in the symbol x and not what the symbol evalutes to.
Numbers, Booleans, characters and strings are self-evaluating in Scheme, so quoting them doesn't matter.
But why does (quote (1 2 3)) or (quote ()) answers #t to the predicate list?.
Should't the result be a "pre-evaluated" value? But in this case (1 2 3) has actually been evaluated to (list 1 2 3)?
Thank you.
pre-evaluated value
I'm not sure where you got that term from. I've never used it. It's not "pre-evaluated", it's unevaluated.
This is really all works from the fact Lisp (and Scheme) is Homoiconic: the structure of the program really uses lists and atoms underneath.
quote is the dual to eval: (eval (list '+ '1 '2 '3)) (and since a quoted number is just the number, (eval (list '+ 1 2 3)) does it as well) is the opposite of (quote '(+ 1 2 3)).
An evaluated list is a call, so an unevaluated call is a list.
Should't the result be a "pre-evaluated" value? But in this case (1 2 3) has actually been evaluated to (list? 1 2 3)?
You're missing some parentheses here! You get (list? '(1 2 3)) (or (list? (quote (1 2 3)). That is, (list? (list 1 2 3)). Which is true.
You can check the opposite with (eval (list '+ 1 2 3)): you get 6.
Note: Some values just evaluate to themselves (like numbers or functions. You can throw eval at it as many times as you want, and it won't change a thing: (eval (eval (eval 1))) is just 1.)
(quote (+ 1 2 3)) = '(+ 1 2 3) = (list '+ '1 '2 '3) = (list '+ 1 2 3) (numbers are self-evaluating, i.e. evaluating to self).
(eval '(+ 1 2 3)) = (+ 1 2 3) = 6. And (eval '(1 2 3)) = (1 2 3) = error.
The first identity is just syntactical. The central identity here is the second one, '(+ 1 2 3) = (list '+ '1 '2 '3). It holds because everything is evaluated in Lisp, but before that, it must be read. Which means, converted from textual source code to actual data structures.
Since ( ... ) parentheses denote lists, reading ( ... ) forms creates lists. Then, evaluating the quoted form just returns that form as is (i.e. non-evaluated). The token + gets read as a symbol +; the literals 1, 2, 3 get read as numbers 1, 2, 3. So the end result is the same as the result of evaluating the form (list '+ '1 '2 '3).
And of course all this still applies without the + inside, as well.
The quote introduces an unevaluated value, not pre-evaluated, whatever that means.
When the expression (quote X) is treated as a form to be evaluated, it simply evaluates to X itself. X is not treated as a form to be evaluated to a value, but rather the resulting value is the syntax X itself.
Quote is a way of the program expressing, "I want to use a piece of my own syntax as a value". And that's precisely what a "literal" is in computer science: a piece of program text reflected back into the program as a run-time value.
In Lisp, atoms other than symbols denote themselves when evaluated. The syntax 3 evaluates to the integer 3. They are the same thing: Lisp syntax is a data structure, and in that data structure, an integer literal 3 is already represented by the object that it denotes.
Thus, under evaluation, there is no difference between (quote 3) and just 3. "Give me the syntax 3 itself" and "give me the value of the syntax 3" are the same: just 3.
Under (quote (1 2 3)), the syntax being quoted is (1 2 3). That syntax is the list object that it looks like; quote just regurgitates it. If were to evaluate the (1 2 3) form, it would be an error: it looks like 1 is being used as an operator or function, with arguments 2 and 3. When quote is itself evaluated, it suppresses this evaluation and just yields the (1 2 3) as-is.
Because the (1 2 3) is a piece of the program syntax, however, there is a restriction in the language that this list may not be modified. An operation like (inc (car (quote (1 2 3)))) which tries to change the list to (2 2 3) invokes undefined behavior. Essentially, the program is trying to modify its own syntax; if for a moment we disregard the additional complexity that Lisp is a compiled language, this is de facto self-modifying code.
Why is '(1 2 3) written instead of (1 2 3) ?
> (list 1 2 3)
'(1 2 3)
Racket's default printer prints a value as an expression that would evaluate to an equivalent value (when possible). It uses quote (abbreviated ') when it can; if a value contains an unquotable data structure, it uses constructor functions instead. For example:
> (list 1 2 3)
'(1 2 3)
> (list 1 2 (set 3)) ;; sets are not quotable
(list 1 2 (set 3))
Most Lisps and Schemes print values using the write function instead. You can change Racket's printer to write mode using the print-as-expression parameter, like this:
> (print-as-expression #f)
> (list 1 2 3)
(1 2 3)
See the docs on the Racket printer for more information.
Are these 2 expressions using quotation and list the same when evaluating the expression such as a derivative?
The intrepreter outputs the same values for both:
(define a '(+ 3 4))
(define b (list '+ 3 4))
a
b
(car a)
(car b)
Output:
=> (+ 3 4)
=> (+ 3 4)
+
+
For these expressions:
(define a '(+ 3 4))
(define b (list '+ 3 4))
Do they have the same value? yes:
(equal? a b)
=> #t
Do they refer to the same object? no:
(eq? a b)
=> #f
I guess that you want to somehow process the lists (say, for calculating a derivative). It doesn't matter how you created the lists as long as you're only interested in treating them as lists of symbols, and you can safely switch back and forth between either representation. It'd be different if you were to treat the operators as actual procedures, these are different things:
(define a '(+ 3 4))
(define b (list + 3 4)) ; quote was removed
(equal? a b)
=> #f
Consider this example:
(define a '(+ 3 4))
(define b (list '+ 3 4))
(define c '(+ 3 4))
Here a will point to an immutable* list (+ 3 4). b will point to a list whose elements are the same as a but it's produced when that statement is evaluated (a can exist before the program runs since it's a constant and for a compiled program the order and how it gets produced is not important). In b you are guaranteed that all the cons cells are newly created and unique. Thus (equal? a b) ; ==>#t while (eq? a b) ; ==> #f.
The output of (eq? a c) is undefined. Some Scheme implementations will create a new immutable list while others will point to the same list created earlier because they are both immutable according to the spec.
*Immutable by the specifications in the Scheme standard but this is seldom enforced so many implementations lets you alter immutable data with undefined behavior as consequence.
Given
#;> (cons (cons 1 2) 3)
((1 . 2) . 3)
When we try
#;> (cons 3 (cons 1 2))
(3 1 . 2)
What governs where the . is used? What would the memory representation of these constructs be?
Scheme implementations usually print things that look like lists in list form:
-> (cons 1 (cons 2 '()))
'(1 2)
In your example, (cons 3 (cons 1 2)) would be a list if it weren't for the last 2. So your implementation makes a best effort to print it as a list until the 2. The other example does not contain any part that looks like a list, so it just prints as nested pairs.