This question already has an answer here:
Lisp, instructions not working in defun [duplicate]
(1 answer)
Closed 8 months ago.
I have following code snippet:
(defun upsi()
(let ((firstpos (car *old-list*)) (secondpos (car(cdr *old-list*))))
(let ((found nil)(secondlocation (list (car (cdr firstpos)) (car secondpos))))
(when (= (found nil))
(setq *new-list* (list(firstlocation)))))))
This gives me following error "The function COMMON-LISP-USER::FOUND is undefined"
But when I try this code, it works perfectly.
(defun upsi()
(let ((firstpos (car *stenches*)) (secondpos (car(cdr *stenches*))))
(let ((found nil)(secondlocation (list (car (cdr firstpos)) (car secondpos))))
(print found)
(print secondlocation))))
What is the cause, that I cant use the variables found and secondlocation in my when condition. In fact I also cant use them in a loop. But when I print them I dont get any errors. It looks to me that they are out of scope but I dont understand why ?
I am just into common-lisp but this confuses me so much^^
Parentheses carry semantic meaning in Common Lisp, and you can't just place them anyplace you like as you can in many other languages. The first position of a list expression is interpreted as an identifier for a function, and that function is called on the arguments; this will fail when there is no such function.
In (when (= (found nil)) you are calling the function found, which doesn't exist. You probably meant (when (= found nil)), but this wouldn't work either because = is meant for comparing numbers in Common Lisp. You could use the null predicate, or it might be better to use an unless form instead of when.
The same problem also seems to occur in the body of the when form: (list (firstlocation)). Here the parentheses indicate that firstlocation is a function that you want to call. I don't think that this is the case given that secondlocation is a list, but I don't see firstlocation defined anyplace so I could be wrong.
A couple of other observations: there is a let* form that allows you to use identifiers which were previously bound in the same form. There is also cadr to use instead of calling car on the result of calling cdr, or you could use second. As presented here, your code is very difficult to read and reason about; you should really use an editor that properly indents your code, and look at some examples of lisp written by experienced programmers to get a feel for what it should look like.
Here are a couple of rewritten versions of your code; the first uses null and the second uses unless. I have assumed that firstlocation is a list, and just renamed secondlocation to firstlocation since secondlocation isn't used in the posted code for the sake of getting the code to function.
(defvar *old-list* '())
(defvar *new-list* '())
;; Using `let*`, `cadr`, and `when` with `null`:
(defun upsi()
(let* ((firstpos (car *old-list*))
(secondpos (car (cdr *old-list*))) ; could also use `cadr` or `second`
(found nil)
(firstlocation (list (cadr firstpos) (car secondpos))))
(when (null found)
(setq *new-list* (list firstlocation)))))
;; Using `let*`, `second`, and `unless`:
(defun upsi()
(let* ((firstpos (car *old-list*))
(secondpos (car (cdr *old-list*))) ; could also use `cadr` or `second`
(found nil)
(firstlocation (list (second firstpos) (car secondpos))))
(unless found
(setq *new-list* (list firstlocation)))))
Related
I am learning Common Lisp using Emacs, SBCL and Slime.
I would like to know exactly what is the code definition of the built-in functions.
I know how to use (documentation ...) and (describe ...). However, they provide only high level information. I would like to see the code details.
For instance, take the nth built-in function.
Documentation gives us:
CL-USER> (documentation 'nth 'function)
"Return the nth object in a list where the car is the zero-th element."
Describe gives me:
CL-USER> (describe 'nth)
COMMON-LISP:NTH
[symbol]
NTH names a compiled function:
Lambda-list: (SB-IMPL::N LIST)
Declared type: (FUNCTION (UNSIGNED-BYTE LIST) (VALUES T &OPTIONAL))
Derived type: (FUNCTION (T T) (VALUES T &OPTIONAL))
Documentation:
Return the nth object in a list where the car is the zero-th element.
Inline proclamation: MAYBE-INLINE (inline expansion available)
Known attributes: foldable, flushable, unsafely-flushable
Source file: SYS:SRC;CODE;LIST.LISP
(SETF NTH) names a compiled function:
Lambda-list: (SB-KERNEL::NEWVAL SB-IMPL::N LIST)
Derived type: (FUNCTION (T UNSIGNED-BYTE LIST) (VALUES T &OPTIONAL))
Inline proclamation: INLINE (inline expansion available)
Source file: SYS:SRC;CODE;SETF-FUNS.LISP
(SETF NTH) has a complex setf-expansion:
Lambda-list: (SB-IMPL::N LIST)
(undocumented)
Source file: SYS:SRC;CODE;DEFSETFS.LISP
; No value
I would like to see something like:
(unknown-command 'nth)
Which would return something like:
(defun nth (x xs)
(if (equal x 0)
(car xs)
(my-nth (- x 1) (cdr xs))))
Lisp languages are fantastic and have a huge ecossystem built by awesome programmers. I hope there is some tool or command for that.
Thanks
First, some general clarifications
In your own code, hitting Meta-. should take you to the source of the code
This will also "just work" for libraries installed via Quicklisp.
Now for SBCL code itself:
If the code is in the "expected place", hitting Meta-. on built-in functions (like nth in your example above) will also take you to its source. I believe the default is /usr/share/sbcl-source/src/code/ but there's possibly a way to configure it.
However, there's another practical way to view this: if you look at the output of (describe ...) above, the line was:
Source file: SYS:SRC;CODE;LIST.LISP
Note: not the last line, that is for (setf nth), something slightly different
This tells you which file in the SBCL source code you can expect to find the function definition.
So, within [the repo](https:/ /github.com/sbcl/sbcl/tree/master/src), if you locate src/code/list.lisp, you should find the definition you're looking for; reproducing here:
(defun nth (n list)
"Return the nth object in a list where the car is the zero-th element."
(declare (explicit-check)
(optimize speed))
(typecase n
((and fixnum unsigned-byte)
(block nil
(let ((i n)
(result list))
(tagbody
loop
(the list result)
(if (plusp i)
(psetq i (1- i)
result (cdr result))
(return (car result)))
(go loop)))))
(t
(car (nthcdr n list)))))
When such information is available, it should be accessible via function-lambda-expression :
* (FUNCTION-LAMBDA-EXPRESSION #'nth)
(LAMBDA (SB-IMPL::N LIST)
(DECLARE (SB-INT:EXPLICIT-CHECK)
(OPTIMIZE SPEED))
(BLOCK NTH
(TYPECASE SB-IMPL::N
((AND FIXNUM UNSIGNED-BYTE)
(BLOCK NIL
(LET ((SB-IMPL::I SB-IMPL::N) (SB-IMPL::RESULT LIST))
(TAGBODY
LOOP
(THE LIST SB-IMPL::RESULT)
(IF (PLUSP SB-IMPL::I)
(PSETQ SB-IMPL::I (1- SB-IMPL::I)
SB-IMPL::RESULT (CDR SB-IMPL::RESULT))
(RETURN (CAR SB-IMPL::RESULT)))
(GO LOOP)))))
(T (CAR (NTHCDR SB-IMPL::N LIST))))))
NIL
NTH
However, it is not always available, in which case you would have to go to the SBCL source code repository.
I'm a beginner Scheme programmer and I want to enrich my knowledge in functional programming. I programm in DrRacket IDE. Recently I found some interesting piece of code:
(car ''(a b))
Output:
'quote
Can anyone explain me why is it evaluated in this way?
This is because ' is short for (quote ...). So,
(car ''(a b))
is actually
(car (quote (quote (a b))))
Which evaluates to:
'quote
As pointed out by Josh in the comments, The actual result is just quote And the REPL prints an expression that can evaluate to quote in this case 'quote or what is the same as seen above (quote quote).
Remember that 'x (for any expression x) is just shorthand for (quote x), so this code:
(car ''(a b))
Is equivalent to this one:
(car (quote (quote (a b))))
Now it's easy to see that you're evaluating the car of a list that looks like this:
'(quote (a b))
Which gets evaluated to this:
(list 'quote (list 'a 'b))
And if we take the car of the above line we'll get the symbol quote, which is precisely the result that you see printed.
It doesn't!
(car ''(a b))
Which is
(car (quote (quote (a b))))
What happens is that car get evaluated to #<procedure:car> and since it'a procedure every argument gets evaluated once before applying..
(quote (quote (a b))) ; ==> (quote (a b))
Now #<procedure:car> of (quote (a b)) is quote. It's not 'quote. I can prove it. Try wrapping it in display like this: (display (car ''(a b))). The output is just quote
So why do you see 'quote in racket interactions window?
Racket has a feature that you can tweak how data is displayed when the expression results that are not explicitly printed by your program. The standard depends on the language selected and for #!racket it's to not to display the data as is but something that you can evaluate and becomes the data. In the REPL you can evaluate the output and get the same since that mode is in effect.
You can change the behaviour if you choose Choose language in the bottom left select and press Show details. The part output syntax lets you choose amongst 4 ways to "display results" in the interactions window. print is the language default and write is the same you get as when you display the data and what is expected from most Scheme books.
Can someone explain why I get different results for the following simple program with sbcl and clisp? Is what I am doing undefined by the language, or is one of the two lisp interpreters wrong?
; Modify the car of the passed-in list
(defun modify (a) (setf (car a) 123))
; Create a list and print car before and after calling modify
(defun testit () (let ((a '(0)))
(print (car a))
(modify a)
(print (car a))))
(testit)
SBCL (version 1.0.51) produces:
0
0
CLISP (version 2.49) produces (what I would expect):
0
123
I agree with Seth's and Vsevolod's comments in that this behavior is due to your modification of literal data. Try using (list 0) instead of '(0). Questions relating to this come up relatively frequently, so I'll quote the HyperSpec here.
3.7.1 Modification of Literal Objects:
The consequences are undefined if literal objects are destructively
modified.
The definition of "literal":
literal adj. (of an object) referenced directly in a program rather
than being computed by the program; that is, appearing as data in a
quote form, or, if the object is a self-evaluating object, appearing
as unquoted data. ``In the form (cons "one" '("two")), the expressions
"one", ("two"), and "two" are literal objects.''
Note that often (in many implementations), if you modify literal values, you'll really modify them in the code itself – writing self modifying code. Your example code will not work as you expect.
Your example code in CCL:
CL-USER> (defun modify (a) (setf (car a) 123))
MODIFY
CL-USER> (defun testit ()
(let ((a '(0)))
(print (car a))
(modify a)
(print (car a))))
TESTIT
CL-USER> (testit)
0
123
123
CL-USER> (testit)
123
123
123
Take a look at the second evaluation of testit, where the let itself really already contains the modified value, thus the first print also yields 123.
Also see: Lisp, cons and (number . number) difference, where I explained this in more detail, or the question linked in Vsevolod's comment above.
I have the following code inside a function that is reading in a file which is a map. I get an error that *numrows* is an illegal dimension. I think this is because lisp is processing these variables in parallel. How can I fix this?
(setq *numrows* (read map))
(setq *numcols* (read map))
(setq *map* (make-array '(*numrows* *numcols*) :initial-element nil))
You're misdiagnosing the problem. The first argument you're passing to MAKE-ARRAY is a list of two symbols, *NUMROWS* and *NUMCOLS*. However, the first argument to MAKE-ARRAY should be a list of non-negative integers. The easiest way to fix your example is to make a list with the values instead: (list *numrows* *numcols*). So the code would look like this instead:
(setq *numrows* (read map))
(setq *numcols* (read map))
(setq *map* (make-array (list *numrows* *numcols*) :initial-element nil))
You normally wouldn't use setq like this, though. It'd probably be better, depending on the context, to bind those variables with LET*:
(let* ((numrows (read map))
(numcols (read map))
(map-array (make-array (list numrows numcols) :initial-element nil))
; do something with map-array
)
Isn't it possible to treat functions in Scheme as any other list?
Basically, what I want do to is something like this:
(define (foo) "hello")
(cdr foo) ; or similar, should return the list ((foo) "hello")
I've found a similar discussion about this, and I feel a bit disappointed if this is not possible with Scheme. If so, why is this impossible? Is it possible in other lisps?
EDIT: Changed (cdr 'foo) to (cdr foo) -- it was misleading. I'm asking, why can't I access a function as a list?
I have often wanted to be able to do the same thing csl.
Below is a quick example of how you could go about doing this in mzscheme.
DrScheme 4.2
(module qdefine mzscheme
(provide ;(all-from-except mzscheme let)
(rename define olddefine)
(rename quote-define define)
(rename quote-cdr cdr)
(rename quote-car car))
(define define-list '())
(define define-list-add
(lambda (key value)
(set! define-list (cons `(,key ,value) define-list))))
(define-syntax quote-define
(syntax-rules ()
((_ (pro-name args ...) body ...)
(begin
(define (pro-name args ...) body ...)
(define-list-add pro-name '((pro-name args ...) body ...))))
((_ pro-name pro) (begin
(define pro-name pro)
(define-list-add pro-name 'pro)))
))
(define quote-cdr (lambda (lst)
(if (procedure? lst)
(cdr (cadr (assoc lst define-list)))
(cdr lst))))
(define quote-car (lambda (lst)
(if (procedure? lst)
(car (cadr (assoc lst define-list)))
(car lst))))
)
(require 'qdefine)
(define testfunc (lambda (args) args))
(cdr testfunc)
(car testfunc)
(define (testfunc2 test) 'blah)
(cdr testfunc2)
(car testfunc2)
(define testlist '(1 2 3 4 5 6 'a))
(cdr testlist)
(car testlist)
Outputs:
((args) args)
lambda
('blah)
(testfunc2 test)
(2 3 4 5 6 'a)
1
>
Your define form is not a function but a function definition. In fact, it is a shorthand for
(define foo
(lambda ()
"hello"))
Lambda can be thought of as a "compiler invocation". In this case, it produces a function which returns that string. Define then binds this function to the symbol 'foo.
Compare this to
(define foo "hello")
which binds just the string to the symbol 'foo. What would (cdr foo) return?
Now, it is imaginable that some Scheme implementation actually saves or has the option to save the lambda form when binding a function to a symbol. You will have to check the documentation, but the kind of pure interpretation this implies surely would have an impact on performance.
If you manage to get this, though, it will return the lambda form, not the define form.
MIT Scheme has the ability to do this. (If you really want, comment on this and I'll give you the code. I had to find some undocumented functions to make it happen.)
However, it's not in the Scheme language definition, so implementations don't have to allow it. The reason for this is that in order to make functions faster, a good Scheme implementation will modify the functions. This means both rewriting them in a different language (either machine code or something fairly low-level) and taking out any bits you don't need - for instance, the + function must in general check whether its arguments are numbers, and if so, what sort of numbers, but if your function is a loop which calls +, you can just check once at the beginning, and make the function a lot faster.
Of course, you could still keep the lists around without too much trouble, even with all of these things. But if you tried to modify the lists, how would it work?
(Again, you could make it work. It would just be more work for implementers, and since it's not usually used in programs, most people probably just don't want to bother.)
In guile,
guile> (define (foo bar) 'baz)
guile> (procedure-source foo)
(lambda (bar) (quote baz))
guile> (cdr (procedure-source foo))
((bar) (quote baz))
guile>
'foo evaluates to a symbol, you can't take the CDR of a symbol.
What you may want to do is (cdr foo), but this does not work. The value of a FOO is a procedure, not a list.
You might be able to access the function as a list using pp or pretty-print. That said, you might also need to run your code in debug mode. This is very implementation dependent however. I know it can work in Gambit-C.
(define (foo) ...) produces a compiled object, and it's a value - a procedure.
You cannot iterate over it because it is not an s-expression.
Like what other suggested, you should check out your programming environment and see
if it has any facilities for such tasks.