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
Related
I am trying to learn Common Lisp with the book Common Lisp: A gentle introduction to Symbolic Computation. In addition, I am using SBCL, Emacs, and Slime.
In the last chapter, on Macros, the author presents examples to re-write the built-in incf macro. He teaches the concept with two different approaches: using back-quote notation and without it. Such as:
(defmacro list-notation-my-incf (x)
(list 'setq x (list '+ x 1)))
(defmacro backquote-notation-my-incf (x)
`(setq ,x (+ ,x 1)))
Later, the author introduces another example:
In the example below, TWO-FROM-ONE is a macro that takes a function
name and another object as arguments; it expands into a call to the
function with two arguments, both of which are the quoted object.
He only uses back-quote character to do it:
(defmacro two-from-one (func object)
`(,func ',object ',object))
And it works as expected:
CL-USER> (two-from-one cons stack-overflow)
(STACK-OVERFLOW . STACK-OVERFLOW)
Using slime-macroexpad-1, I have:
(CONS 'STACK-OVERFLOW 'STACK-OVERFLOW)
As an exercise that I created for myself, I tried doing the same, but avoiding the back-quote notation. Unfortunately, I could not make it work:
(defmacro list-two-from-one (func object)
(list func (quote object) (quote object)))
Slime throws the error:
The variable OBJECT is unbound.
[Condition of type UNBOUND-VARIABLE]
Doing a macro expansion, I see:
(CONS OBJECT OBJECT)
If I try a different approach, it seems to be closer, but it does not work either:
(defmacro list-two-from-one (func object)
(list func object object))
Throws the error:
The variable STACK-OVERFLOW is unbound.
[Condition of type UNBOUND-VARIABLE]
And, finally, the macro expansion indicates:
(CONS STACK-OVERFLOW STACK-OVERFLOW)
I feel stuck. How do I successfully re-write the macro without using back-quote notation?
Thanks.
What you are looking for is something like
(defmacro list-two-from-one (func object)
(list func (list 'quote object) (list 'quote object)))
basically, the body of a macro should return the code, that, when evaluated, produces the desired result.
I.e., the macro body should produce (CONS 'STACK-OVERFLOW 'STACK-OVERFLOW).
Since 'a is the same as (quote a), you want your macro to produce
(CONS (QUOTE STACK-OVERFLOW) (QUOTE STACK-OVERFLOW))
which is what my defmacro above returns.
Your macro should expand to:
CL-USER 10 > (macroexpand '(two-from-one2 cons stack-overflow))
(CONS (QUOTE STACK-OVERFLOW) (QUOTE STACK-OVERFLOW))
So create lists with quote like this:
(defmacro two-from-one2 (func object)
(list func (list 'quote object) (list 'quote object)))
Test:
CL-USER 9 > (two-from-one2 cons stack-overflow)
(STACK-OVERFLOW . STACK-OVERFLOW)
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)
I'm trying to implement the Towers of Hanoi.I'm not printing out anything between my recursive calls yet, but I keep getting an error saying
'('(LIST) 'NIL 'NIL) should be a lambda expression
I've read that the reason this happens is because of a problem with the parenthesis, however I cannot seem to find what my problem is. I think it's happening in the pass-list function when I am trying to call the hanoi function. My code:
(defun pass-list(list)
(hanoi('('(list)'()'())))
)
(defun hanoi ('('(1) '(2) '(3)))
(hanoi '('(cdr 1) '(cons(car 1) 2) '(3)))
(hanoi '('(cons(car 3)1) '(2)'(cdr 3)))
)
This code has many syntax problems; there are erroneous quotes all over the place, and it looks like you're trying to use numbers as variables, which will not work. The source of the particular error message that you mentioned comes from
(hanoi('('(list)'()'())))
First, understand that the quotes in 'x and '(a b c) are shorthand for the forms (quote x) and (quote (a b c)), and that (quote anything) is the syntax for getting anything, without anything being evaluated. So '(1 2 3) gives you the list (1 2 3), and '1 gives you 1. quote is just a symbol though, and can be present in other lists, so '('(list)'()'()) is the same as (quote ((quote (list)) (quote ()) (quote ()))) which evaluates to the list ((quote (list)) (quote ()) (quote ())). Since () can also be written nil (or NIL), this last is the same as ('(list) 'NIL 'NIL). In Common Lisp, function calls look like
(function arg1 arg2 ...)
where each argi is a form, and function is either a symbol (e.g., list, hanoi, car) or a list, in which case it must be a lambda expression, e.g., (lambda (x) (+ x x)). So, in your line
(hanoi('('(list)'()'())))
we have a function call. function is hanoi, and arg1 is ('('(list)'()'())). But how will this arg1 be evaluated? Well, it's a list, which means it's a function application. What's the function part? It's
'('(list)'()'())
which is the same as
'('(list 'NIL 'NIL))
But as I just said, the only kind of list that can be function is a lambda expression. This clearly isn't a lambda expression, so you get the error that you're seeing.
I can't be sure, but it looks like you were aiming for something like the following. The line marked with ** is sort of problematic, because you're calling hanoi with some arguments, and when it returns (if it ever returns; it seems to me like you'd recurse forever in this case), you don't do anything with the result. It's ignored, and then you go onto the third line.
(defun pass-list(list)
(hanoi (list list) '() '()))
(defun hanoi (a b c)
(hanoi (rest a) (cons (first a) b) c) ; **
(hanoi (cons (first c) a) b (rest c)))
If hanoi is supposed to take a single list as an argument, and that list is supposed to contain three lists (I'm not sure why you'd do it that way instead of having hanoi take just three arguments, but that's a different question, I suppose), it's easy enough to modify; just take an argument abc and extract the first, second, and third lists from it, and pass a single list to hanoi on the recursive call:
(defun hanoi (abc)
(let ((a (first abc))
(b (second abc))
(c (third abc)))
(hanoi (list (rest a) (cons (first a) b) c))
(hanoi (list (cons (first c) a) b (rest c)))))
I'd actually probably use destructuring-bind here to simplify getting a, b, and c out of abc:
(defun hanoi (abc)
(destructuring-bind (a b c) abc
(hanoi (list (rest a) (cons (first a) b) c))
(hanoi (list (cons (first c) a) b (rest c)))))
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.
I'm trying to write a macro in Lisp that re-implements let using itself. This is a trivial exercise which has no practical purpose; however after giving a response to a related question, I realized I should probably learn more about macros. They're touted as one of the great things about Lisp, but I rarely use them.
Anyway, here's what I tried first:
(defmacro mylet (args &rest exp) `(let ,args (dolist (x ,exp) x)))
but when I try something like:
(mylet ((a 5) (b 2)) (print (+ a b)))
this throws up an error:
#1=(PRINT (+ A B)) is not a symbol or lambda expression in the form (#1#) .
args (a and b) are set properly, but the print statement doesn't work. I think it's because I'm using two levels of indirection-- referring to a variable that I've created within the macro. But I can't seem to figure out how to fix it! Any ideas?
Your macro expands to:
(LET ((A 5) (B 2))
(DOLIST (X ((PRINT (+ A B)))) X))
which is invalid because ((PRINT (+ A B))) is not a valid expression. There is also an issue that using an interned symbol in macro expansion can lead to variable capture, but that is not directly relevant (read more in PCL).
Using DOLIST here is unnecessary, and compilcated to get right (you would have to convert all subforms to anonymous function in order to stick them in a list, funcall them in sequence and then store the final result in order to conform to PROGN behaviour). You can just use PROGN, or, since LET includes an implicit PROGN, just splice the body using the ,# feature of backquote mechanism:
(defmacro mylet (args &body exp) `(let ,args ,(cons 'progn exp)))
(defmacro mylet (args &body exp) `(let ,args ,#exp))