Lately, I've been thinking a lot about the basis of Lisp; I've read several manuals and/or other materials on the Internet, including The Roots of Lisp by P. Graham:
In The Roots of Lisp, quote is described as a primitive that changes code into data, thereby quoting it, but there doesn't seem to be an equivalent inverse primitive, that is an unquote primitive. I thought it might have been eval's business, but eval often runs the data in a null lexical environment, which is not equivalent to changing data back into code.
Ergo, why isn't there an unquote Lisp primitive?
unquote is only useful in the context of quasiquote, and quasiquote can be implemented as a macro (that uses quote behind the scenes). So there's no need to have an unquote primitive; the quasiquote macro simply deals with unquote symbols as they are found.
(quasiquote is the Scheme name for the backtick quote. Thus:
`(foo bar ,baz)
is read in as
(quasiquote (foo bar (unquote baz)))
in Scheme.)
Here's a very simple Scheme quasiquote macro (it only handles lists, unlike standard quasiquote which also handles vectors and other data types):
(define-syntax quasiquote
(syntax-rules (unquote unquote-splicing)
((quasiquote (unquote datum))
datum)
((quasiquote ((unquote-splicing datum) . next))
(append datum (quasiquote next)))
((quasiquote (datum . next))
(cons (quasiquote datum) (quasiquote next)))
((quasiquote datum)
(quote datum))))
Equivalent version using all the standard reader abbreviations:
(define-syntax quasiquote
(syntax-rules (unquote unquote-splicing)
(`,datum
datum)
(`(,#datum . next)
(append datum `next))
(`(datum . next)
(cons `datum `next))
(`datum
'datum)))
I am also relatively new to Lisp, but I think that what you were thinking about is eval. evalis the way to change data back to code.
Namely, consider a simple function.
(defun foo (a b c) (list a b c))
Then, if you do something like this, you get a list of symbols:
CL-USER> (foo 'a 'b 'c)
(A B C)
If you add a quote in the front, the function call itself is treated as a piece of data (list):
CL-USER> '(foo 'a 'b 'c)
(FOO 'A 'B 'C)
Adding one more quote has an expected effect:
CL-USER> ''(foo 'a 'b 'c)
'(FOO 'A 'B 'C)
Let us now unwind it with eval, which in essence may be thought of as the inverse operation for the quote. It is the inverse. The x-axis is the data form. The y-axis is the code form. Hopefully this (somewhat stretched) analogy makes sense.
CL-USER> (eval ''(foo 'a 'b 'c))
(FOO 'A 'B 'C)
Can you guess what will happen if I chain two evals in a row?
Here it is:
CL-USER> (eval (eval ''(foo 'a 'b 'c)))
(A B C)
Matt Brown and Jens Palsberg actually provide a good definition of what unquote would be in Typed self-evaluation via intensional type functions. The eval procedure under that definition is a meta interpreter function which transforms a structured input into a different value with the same type. i.e., accept a Lisp program as an s-expression and return another s-expression as the result.
The quote form should be structure preserving such that if the program is run naturally and then the result is quoted, the resulting representation should be the same as quoting the program and then running eval.
For a more concrete example, suppose you represent JavaScript programs in JavaScript as strings (assume for simplicity that a program is a 0-argument function which returns some JavaScript object). Running this program naturally then taking the JS object output (which for example may be cyclic) and running quote on the output should return the same string as running eval on the string representation of the program.
function program() {
...
return obj;
}
// If we had a true quote operation in JS, we would be
// able to run const quotedProgram = quote(program);
const quotedProgram = `
function program() {
...
return obj;
}
`;
const result1 = program();
const result2 = quote(result1);
const result3 = eval(quotedProgram);
const result4 = unquote(result3);
The above example is a little weird because JS doesn't have a natural way to quote arbitrary functions as strings (toString sort of works in many cases). However, note that if quote/eval are correct, result2 and result3 should be the same; furthermore, if unquote is correct, result1 and result4 should be the same.
Related
I'm new to Racket and I was hoping to get more insights in the these two operators: , & ,#.
There's very little documentation of these new operators, however, to my understanding the former (,) unquotes everything if its is followed by a list. And the latter (,#) splices the values.
For example if the following is typed in the Dr. Racket interpreter:
(define scores '(1 3 2))
(define pets '(dog cat))
and then the following query is made:
`(,scores ,#pets)
this would yield : '((1 3 2) dog cat)
It would be appreciated if I could get more details, definitions and more examples about these operators.
Thanks in advance.
A single quote followed by the written representation of a value
will produce that value:
Example:
'(1 x "foo")
will produce a value that prints as (1 x "foo").
Suppose now that I don't want a literal symbol x in the list.
I have a variable x in my program, and I want to insert
the value to which x is bound.
To mark that I want the value of x rather than the symbol x,
I insert a comma before x:
'(1 ,x "foo")
It won't work as-is though - I now get a value that has a literal comma as well as a symbol x. The problem is that quote does not know about the comma convention.
Backtick or backquote knows about the comma-convention, so that will give the correct result:
> `(1 ,x "foo")
(1 3 "foo") ; if the value of x is 3
Now let's say x is the list (a b).
> `(1 ,x "foo")
(1 (a b) "foo") ; if the value of x is (a b)
This looks as expected. But what if I wanted (1 a b "foo")
as the result? We need a way so show "insert the elements of a list".
That's where ,# comes into the picture.
> `(1 ,#x "foo")
(1 a b "foo") ; if the value of x is (a b)
They are "reader abbreviations" or "reader macros". They are introduced in the section of the Racket guide on quasiquotation. To summarize:
`e reads as (quasiquote e)
,e reads as (unquote e)
,#e reads as (unquote-splicing e)
Because Racket's printer uses the same abbreviations by default, it can be confusing to test this yourself. Here are a few examples that should help:
> (equal? (list 'unquote 'abc) (read (open-input-string ",abc")))
#t
> (writeln (read (open-input-string ",abc")))
(unquote abc)
A more exhaustive description of the Racket reader is in the section on The Reader in the Racket Reference. A list of reader abbreviations is in the Reading Quotes subsection.
In Common Lisp, given that "a" is simply a character, what is the difference between #\a, 'a #'a?
My question comes from the tutorialspoint.com tutorial on Lisp. At one point the tutorial introduces:
; a character array with all initial elements set to a
; is a string actually
(write(make-array 10 :element-type 'character :initial-element #\a))
(terpri)
; a two dimensional array with initial values a
(setq myarray (make-array '(2 2) :initial-element 'a :adjustable t))
(write myarray)
(terpri)
With the output:
"aaaaaaaaaa"
#2A((A A) (A A))
#' is not included in this example but I'm including it in the question because it can be confusing as well. 🙂
Thank you very much! 😊
To start, a is not "simply a character." The Lisp reader parses #\a as the character literal a, which is an object in Common Lisp. Note that #\a and #\A are different character objects.
When the Lisp reader encounters a single quote, the expression following the single quote is not evaluated. Specifically, 'a is treated as (quote a), where quote returns its argument unevaluated. Now, a is a symbol, so 'a evaluates to that symbol. But the Lisp reader upcases most characters it reads by default, so 'a really evaluates to the symbol A. The good news is that whether you type a or A, the Lisp reader will read A (unless you mess with the readtable), and both 'a and 'A evaluate to the symbol A.
When the Lisp reader encounters #'a, the entire expression is treated as (function a), which when evaluated returns the function associated with the name a. But, note that it is an error to use function, and by extension #', on an identifier that does not denote a function.
To clarify this last part a bit, consider the following REPL interaction:
CL-USER> (defvar a 1)
A
CL-USER> a
1
CL-USER> #'a
The function COMMON-LISP-USER::A is undefined.
[Condition of type UNDEFINED-FUNCTION]
Here the variable a is defined and given the value 1, but when we try to access the function denoted by a we get an error message because there is no such function. Continuing:
; Evaluation aborted on #<UNDEFINED-FUNCTION A {1002DDC303}>.
CL-USER> (defun a (x) x)
A
CL-USER> (a 'b)
B
CL-USER> a
1
CL-USER> #'a
#<FUNCTION A>
Now we have defined a function named a that simply returns its argument. You can see that when we call a with an argument 'b we get the expected result: (a 'b) --> b. But, then when we evaluate a alone we still get 1. Symbols in Common Lisp are objects that have, among other cells, value cells and function cells. After the above interaction, the symbol a now has 1 in its value cell, and it has the function we have defined in its function cell. When the symbol a is evaluated the value cell is accessed, but when (function a) or #'a is evaluated, the function cell is accessed. You can see above that when #'a is evaluated, the function we defined is returned, and the REPL prints #<FUNCTION A> to show this.
As an aside, I wouldn't recommend using Tutorialspoint to learn Common Lisp. Glancing over the site, right away I see this:
LISP expressions are case-insensitive, cos 45 or COS 45 are same.
This is just wrong. And, Lisp is not written in all-caps. None of this inspires faith. Instead, find a good book. There are some recommendations on the common-lisp tag-info page.
#\
This is to introduce a character.
CL-USER> #\a
#\a
CL-USER> (character 'a)
#\A
CL-USER> (character "a")
#\a
'
This is quote, to quote and not evaluate things and construct object literals.
CL-USER> a
=> error: the variable a is unbound.
CL-USER> 'a
A
CL-USER> (inspect 'a)
The object is a SYMBOL.
0. Name: "A"
1. Package: #<PACKAGE "COMMON-LISP-USER">
2. Value: "unbound"
3. Function: "unbound"
4. Plist: NIL
> q
CL-USER> (equal (list 1 2) (quote (1 2))) ;; aka '(1 2)
T ;; but watch out with object literals constructed with quote, prefer constructor functions.
and #'
This is sharpsign-quote to reference a function.
CL-USER> #'a
=> error: The function COMMON-LISP-USER::A is undefined.
CL-USER> (defun a () (print "hello A"))
A
CL-USER> (a)
"hello A"
"hello A"
CL-USER> #'a
#<FUNCTION A>
CL-USER> (function a)
#<FUNCTION A>
One can ask Lisp to describe the data objects you've mentioned.
If we look at the expressions:
CL-USER 13 > (dolist (object (list '#\a ''a '#'a))
(terpri)
(describe object)
(terpri))
#\a is a CHARACTER
Name "Latin-Small-Letter-A"
Code 97
(QUOTE A) is a LIST
0 QUOTE
1 A
(FUNCTION A) is a LIST
0 FUNCTION
1 A
NIL
If we look at the evaluated expressions:
CL-USER 5 > (dolist (object (list #\a 'a #'a))
(terpri)
(describe object)
(terpri))
#\a is a CHARACTER
Name "Latin-Small-Letter-A"
Code 97
A is a SYMBOL
NAME "A"
VALUE #<unbound value>
FUNCTION #<interpreted function A 422005BD54>
PLIST NIL
PACKAGE #<The COMMON-LISP-USER package, 73/256 internal, 0/4 external>
#<interpreted function A 422005BD54> is a TYPE::INTERPRETED-FUNCTION
CODE (LAMBDA (B)
A)
How does lisp quote work internally?
For example:
(quote (+ 1 (* 1 2)) )
seems to be equivalent to
(list '+ 1 (list '* 1 2))
which means it is some how symbolizing the Head values recursively. Is this function a built in?
Run (equal (quote (+ 1 (* 1 2))) (list '+ 1 (list '* 1 2))) if you don't believe me.
How does it work?
quote is really really simple to implement. It does mostly nothing. The quote special operator just returns the enclosed object like it is. Nothing more. No evaluation. The object is not changed in any way.
Evaluation of quoted forms
Probably a good time to read McCarthy, from 1960:
Recursive Functions of Symbolic Expressions and Their Computation by Machine, Part I
Pages 16/17 explain evaluation with eval. Here:
eq [car [e]; QUOTE] → cadr [e];
or in s-expression notation:
(cond
...
((eq (car e) 'quote)
(cadr e))
...)
Above code implements the evaluation rule for QUOTE: If the expression is a list and the first element of the list is the symbol QUOTE, then return the second element of the list.
Equivalence of a quoted list with a list created by LIST
(equal (quote (+ 1 (* 1 2)))
(list '+ 1 (list '* 1 2)))
The result is T. This means that both result lists are structurally equivalent.
(eq (quote (+ 1 (* 1 2)))
(list '+ 1 (list '* 1 2)))
The result is NIL. This means that the first cons cell of the linked lists are not the same objects. EQ tests whether we really have the same cons cell object.
QUOTE returns a literal data object. The consequences of modifying this object is undefined. So, don't do it.
LIST returns a new freshly consed list each time it is called. The fresh new list will not share any cons cells with any earlier allocated list.
So the main difference is that QUOTE is a built-in operator, which returns literal and unevaluated data. Whereas LIST is a function which creates a new,fresh list with its arguments as contents.
See the effects with respect to EQ and EQUAL:
CL-USER 6 >
(flet ((foo () (quote (+ 1 (* 1 2))))
(bar () (list '+ 1 (list '* 1 2))))
(list (list :eq-foo-foo (eq (foo) (foo)))
(list :eq-foo-bar (eq (foo) (bar)))
(list :eq-bar-bar (eq (foo) (bar)))
(list :equal-foo-foo (equal (foo) (foo)))
(list :equal-foo-bar (equal (foo) (bar)))
(list :equal-bar-bar (equal (foo) (bar)))))
((:EQ-FOO-FOO T)
(:EQ-FOO-BAR NIL)
(:EQ-BAR-BAR NIL)
(:EQUAL-FOO-FOO T)
(:EQUAL-FOO-BAR T)
(:EQUAL-BAR-BAR T))
is quote a function?
quote can't be a function, since it returns its enclosed data unevaluated. Thus it is a special evaluation rule.
If quote were a function, it's arguments were evaluated. But that's exactly what is NOT what quote is supposed to do.
why does Lisp need QUOTE?
Lisp usually uses s-expressions to write Lisp code. So s-expressions have a both purpose to denote data and we use it to write programs. In a Lisp program lists are used for function calls, macro forms and special forms. symbols are used as variables:
(+ n 42)
Here (+ n 42) is a list and n is a symbol. But we also want to use lists as data in our programs and we want to use symbols as data. Thus we have to quote them, so that Lisp will not see them as programs, but as data:
(append '(+ n) '(42)) evaluates to (+ n 42)
Thus in a Lisp program, lists and variables are by default part of the language elements, for example as function calls and variables. If we want to use lists and symbols as literal data, we have to quote them, to prevent the evaluator treating them as Lisp code to evaluate.
quote does nothing more than return its argument unevaluated. But what is an unevaluated argument?
When a Lisp program is defined, it is either read from textual source into s-expression form or constructed directly in terms of s-expressions. A macro would be an example of generating s-expressions. Either way there is a data structure comprising (mostly) symbols and conses that represents the program.
Most Lisp expressions will call upon evaluation and compilation machinery to interpret this data structure as terms in a program. quote is treated specially and passed these uninterpreted symbols and conses as its argument. In short, quote does almost nothing - the value it returns already exists and is simply passed through.
You can observe the difference between passing through and fresh construction by using eq to test the identity of the return value of quote:
(defun f () '(1 2))
(defun g () (list 1 2))
(eq (f) (f)) => T
(eq (g) (g)) => NIL
As you can see, quote returns the same conses each time through.
actually i am trying to perfectly understand clojure and particularly symbols
(def a 1)
(type a)
;;=>java.lang.Long
(type 'a)
;;=>clojure.lang.Symbol
I know that type is a function so its arguments get evaluated first so i perfectly understand why the code above work this way .In the flowing code i decided to delay the evaluation using macro
(defmacro m-type [x] (type x))
(m-type a)
;;==>clojure.lang.Symbol
and i am fine with that but what i fail to uderstand is this:
(m-type 'a)
;;=>clojure.lang.Cons
why the type of 'a is a cons
the character ' is interpreted by the clojure reader as a reader-macro which expands to a list containing the symbol quote followed by whatever follows the ', so in your call to (m-type 'a) the 'a is expanding to:
user> (macroexpand-1 ''a)
(quote a)
then calling type on the list (quote a) which is a Cons.
This may be a bit more clear if we make the m-type macro print the arguments as it sees them while it is evaluating:
user> (defmacro m-type [x] (println "x is " x) (type x))
#'user/m-type
user> (m-type 'a)
x is (quote a)
clojure.lang.Cons
I made this IF-THEN-ELSE Lambda Calculus code
(defvar IF-THEN-ELSE
#'(lambda(con)
#'(lambda(x)
#'(lambda(y)
#'(lambda(acc1)
#'(lambda (acc2)
(funcall (funcall (funcall (funcall con x) y) acc1) acc2))))))
)
(defun IF-THEN-ELSEOP(c x y a1 a2)
(funcall (funcall (funcall (funcall (funcall IF-THEN-ELSE c) x) y) a1) a2)
)
And this Greater or Equal operator
(defvar GEQ
#'(lambda(p)
#'(lambda(q)
(funcall #'LEQOP q p)))
)
LEQOP is a function for "Less or Equal" and it works OK. So when I call IF-THEN-ELSE like this ("six" and "two" are church numbers)
(if-then-elseop GEQ six two (print "THIS") (print "THAT"))
as output I've got
"THIS"
"THAT"
"THIS"
Both functions that I'm passing are being called. How can I avoid it in order to get only as output "THIS"?
This happens with every function I use, and this is a trouble because I want to use IF-THEN-ELSE in a recursive call, so just one function must be called dependign on the IF-THEN-ELSE eval.
Any help would be appreciated
Thanks.
Passing your print statements by wrapping them in lambdas should work, but maybe it's worth an explanation as to why this is necessary.
You're implementing a lambda calculus. By definition, all 'things' in the calculus are higher order functions. Your six and two and any other church numerals you may have defined are also higher order functions.
IF-THEN-ELSE is a lambda abstraction (also a higher-order function because it's 'arguments' are also functions). So this would have been valid:
(if-then-elseop GEQ six two one two)
Where one and two are church numbers. By doing that, you're expressing in lambda calculus what you would in plain lisp as:
(if (>= 6 2)
1
2)
But I'm guessing what you were aiming for was:
(if (>= 6 2)
(print "this")
(print "that"))
(more later about why messing with print might be a distraction to your exercise)
So the 'real' 1 has a church encoding one, which I'me assuming you've defined. That way, it can be applied to the lambda abstraction IF-THEN-ELSE - In the same way that
(>= 6 2)
evaluates to TRUE in the lisp world, your lambda calculus implementation of the same,
((GEQ six) two)
will evaluate to the lambda encoding of TRUE, which is again, encoded as a higher-order function.
(defvar TRUE #'(lambda (x) #'(lambda (y) x)))
(defvar FALSE #'(lambda (x) #'(lambda (y) y)))
So the rule to remember is that everything you are passing around and getting back in the lambda calculus are functions:
0 := λf.λx.x
1 := λf.λx.f x
2 := λf.λx.f (f x)
3 := λf.λx.f (f (f x))
... and so on
Which is why, if you did:
(if-then-elseop GEQ six two
#'(lambda () (print "THIS"))
#'(lambda () (print "THAT")))
should work. (sort of, read ahead)
(I'd stick to the faithful interpretation of IF-THEN-ELSE though:
(defvar IFTHENELSE
#'(lambda (p)
#'(lambda (a)
#'(lambda (b) (funcall (funcall p a) b)))))
Where p is your condition... )
As a side note, it's worth pointing out that it might not be too helpful to bring in print and other code that 'does stuff' within lambda calculus - the calculus does not define IO, and is restricted to evaluation of lambda expressions. The church encodings are a way of encoding numbers as lambda terms; There's no simple and meaningful way to represent
a statement with side-effects such a (print "hello") as a lambda term; #'(lambda () (print "THIS")) works but as an academic exercise it's best to stick to only evaluating things and getting back results.
What about lisp itself? if in lisp is not a function so (if cond then-expr else-expr) works the way you expect (that is, only one of then-expr or else-expr will actually be evaluated) because it is a special form. If you were to define your own, you would need a macro (as #wvxvw rightly suggests). But that's another topic.