Consider this piece of code:
(defvar lst '(1 1))
(defmacro get-x (x lst)
`(nth ,x ,lst))
(defun get-y (y lst)
(nth y lst))
Now let us assume that I want to change the value of the elements of the list called lst, the car with get-x and the cdr with get-y.
As I try to change the value with get-x (with setf) everything goes fine but if I try it with get-y it signals an error (shortened):
; caught STYLE-WARNING:
; undefined function: (SETF GET-STUFF)
Why does this happen?
I myself suspect that this happens because the macro simply expands and the function nth simply returns a reference to the value of an element in the list and the function on the other hand evaluates the function-call to nth and returns the value of the referenced value (sounds confusing).
Am I correct in my suspicions?
If I am correct then how will one know what is simply a reference to a value and an actual value?
The error does not happen with the macro version, because, as you assumed, the expression (setf (get-x some-x some-list) some-value) will be expanded (at compile-time) into something like (setf (nth some-x some-list) some-value) (not really, but the details of setf-expansion are complex), and the compiler knows, how to deal with that (i.e., there is a suitable setf expander defined for function nth).
However, in the case of get-y, the compiler has no setf expander, unless you provide one. The easiest way to do so would be
(defun (setf get-y) (new-value x ls) ; Note the function's name: setf get-y
(setf (nth x ls) new-value))
Note, that there are a few conventions regarding setf-expanders:
The new value is always provided as the first argument to the setf function
All setf functions are supposed to return the new value as their result (as this is, what the entire setf form is supposed to return)
There is, BTW, no such concept as a "reference" in Common Lisp (at least not in the C++ sense), though there once were Lisp dialects which had locatives. Generalized place forms (ie., setf and its machinery) work very differently from plain C++ style references. See the CLHS, if you are curious about the details.
SETF is a macro.
The idea is that to set and read elements from data structures are two operations, but usually require two different names (or maybe even something more complex). SETF now enables you to use just one name for both:
(get-something x)
Above reads a datastructure. The inverse then simply is:
(setf (get-something x) :foobar)
Above sets the datastructure at X with :FOOBAR.
SETF does not treat (get-something x) as a reference or something like that. It just has a database of inverse operations for each operation. If you use GET-SOMETHING, it knows what the inverse operation is.
How does SETF know it? Simple: you have to tell it.
For The NTH operation, SETF knows how to set the nth element. That's builtin into Common Lisp.
For your own GET-Y operation SETF does not have that information. You have to tell it. See the Common Lisp HyperSpec for examples. One example is to use DEFUN and (SETF GET-Y) as a function name.
Also note following style problems with your example:
lst is not a good name for a DEFVAR variable. Use *list* as a name to make clear that it is a special variable declared by DEFVAR (or similar).
'(1 2) is a literal constant. If you write a Common Lisp program, the effects of changing it are undefined. If you want to change a list later, you should cons it with LIST or something like COPY-LIST.
Related
I am trying to do Recursive method to find max value in list.
Can anyone explain where I made the mistake on this code and how to approach it next time.
(defun f3 (i)
(setq x (cond (> (car (I)) (cdr (car (I))))
(f3 (cdr (I)))))
)
(f3 '(33 11 44 2) )
also I tried this following method and didn't work:
(defun f3 (i)
(cond ((null I )nil )
(setq x (car (i))
(f3(cdr (i)))
(return-from max x)
)
Thanks a lot for any help. I am coming from java if that helps.
If you're working in Common Lisp, then you do this:
(defun max-item (list)
(loop for item in list
maximizing item))
That's it. The maximizing item clause of loop determines the highest item value seen, and implicitly establishes that as the result value of loop when it terminates.
Note that if list is empty, then this returns nil. If you want some other behavior, you have to work that in:
(if list
(loop for item in list
maximizing item))
(... handle empty here ...))
If the number of elements in the list is known to be small, below your Lisp implementation's limit on the number of arguments that can be passed to a function, you can simply apply the list to the max function:
(defun max-item (list)
(apply #'max list))
If list is empty, then max is misused: it requires one or more arguments. An error condition will likely be signaled. If that doesn't work in your situation, you need to add code to supply the desired behavior.
If the list is expected to be large, so that this approach is to be avoided, you can use reduce, treating max as a binary function:
(defun max-item (list)
(reduce #'max list))
Same remarks regarding empty list. These expressions are so small, many programmers will avoid writing a function and just use them directly.
Regarding recursion, you wouldn't use recursion to solve this problem in production code, only as a homework exercise for learning about recursion.
You are trying to compute the maximum value of a list, so please name your function maximum and your parameter list, not f3 or i. You can't name the function max without having to consider how to avoid shadowing the standard max function, so it is best for now to ignore package issues and use a different name.
There is a corner case to consider when the list is empty, as there is no meaningful value to return. You have to decide if you return nil or signal an error, for example.
The skeleton is thus:
(defun maximum (list)
(if (null list)
...
...))
Notice how closing parentheses are never preceded by spaces (or newlines), and opening parentheses are never followed by spaces (or newlines). Please note also that indentation increases with the current depth . This is the basic rules for Lisp formatting, please try following them for other developers.
(setq x <value>)
You are assigning an unknown place x, you should instead bind a fresh variable if you want to have a temporary variable, something like:
(let ((x <value>))
<body>)
With the above expression, x is bound to <value> inside <body> (one or more expressions), and only there.
(car (i))
Unlike in Java, parentheses are not used to group expressions for readability or to force some evaluation order, in Lisp they enclose compound forms. Here above, in a normal evaluation context (not a macro or binding), (i) means call function i, and this function is unrelated to your local variable i (just like in Java, where you can write int f = f(2) with f denoting both a variable and a method).
If you want to take the car of i, write (car i).
You seem to be using cond as some kind of if:
(cond (<test> <then>) <else>) ;; WRONG
You can have an if as follows:
(if <test> <then> <else>)
For example:
(if (> u v) u v) ;; evaluates to either `u` or `v`, whichever is greater
The cond syntax is a bit more complex but you don't need it yet.
You cannot return-from a block that was undeclared, you probably renamed the function to f3 without renaming that part, or copied that from somewhere else, but in any case return-from is only needed when you have a bigger function and probably a lot more side-effects. Here the computation can be written in a more functionnal way. There is an implicit return in Lisp-like languages, unlike Java, for example below the last (but also single) expression in add evaluates to the function's return value:
(defun add-3 (x)
(+ x 3))
Start with smaller examples and test often, fix any error the compiler or interpreter prints before trying to do more complex things. Also, have a look at the available online resources to learn more about the language: https://common-lisp.net/documentation
Although the other answers are right: you definitely need to learn more CL syntax and you probably would not solve this problem recursively in idiomatic CL (whatever 'idiomatic CL' is), here's how to actually do it, because thinking about how to solve these problems recursively is useful.
First of all let's write a function max/2 which returns the maximum of two numbers. This is pretty easy:
(defun max/2 (a b)
(if (> a b) a b))
Now the trick is this: assume you have some idea of what the maximum of a list of numbers is: call this guess m. Then:
if the list is empty, the maximum is m;
otherwise the list has a first element, so pick a new m which is the maximum of the first element of the list and the current m, and recurse on the rest of the list.
So, we can write this function, which I'll call max/carrying (because it 'carries' the m):
(defun max/carrying (m list)
(if (null list)
m
(max/carrying (max/2 (first list) m)
(rest list))))
And this is now almost all we need. The trick is then to write a little shim around max/carrying which bootstraps it:
to compute the maximum of a list:
if the list is empty it has no maximum, and this is an error;
otherwise the result is max/carrying of the first element of the list and the rest of the list.
I won't write that, but it's pretty easy (to signal an error, the function you want is error).
Reading this question got me thinking about what constitutes a valid car of an expression. Obviously, symbols and lambdas can be "called" using the usual syntax. According to the hyperspec,
function name n. 1. (in an environment) A symbol or a list (setf symbol) that is the name of a function in that environment. 2. A symbol or a list (setf symbol).
So, theoretically, (setf some-name) is a function name. I decided to give it a try.
(defun (setf try-this) ()
(format t "Don't name your functions like this, kids :)"))
((setf try-this))
(funcall '(setf try-this))
(setf (try-this))
GNU CLISP, SBCL, and ABCL will all let me define this function. However, SBCL and ABCL won't let me call it using any of the syntaxes shown in the snippet. CLISP, on the other hand, will run the first two but still errs on the third.
I'm curious about which compiler is behaving correctly. Since SBCL and ABCL agree, I would hazard a guess that a correct implementation should reject that code. As a second question, how would I call my incredibly-contrived not-useful function from the code snippet, since the things I tried above aren't working portably. Or, perhaps more usefully,
A SETF function has to take at least one argument, which is the new value to be stored in the place. It can take additional arguments as well, these will be filled in from arguments in the place expression in the call to SETF.
When you use SETF, it has to have an even number of arguments: every place you're assigning to needs a value to be assigned.
So it should be:
(defun (setf try-this) (new-value)
(format t "You tried to store ~S~%" new-value))
(setf (try-this) 3)
(funcall #'(setf try-this) 'foo)
You can't use
((setf try-this) 'bar)
because the car of a form does not contain a function name. It can only be a symbol or a lambda expression (although implementations may allow other formats as extensions).
This question already has answers here:
setq and defvar in Lisp
(4 answers)
Closed 6 years ago.
I was following a tutorial on lisp and they did the following code
(set 'x 11)
(incf x 10)
and interpreter gave the following error:
; in: INCF X
; (SETQ X #:NEW671)
;
; caught WARNING:
; undefined variable: X
;
; compilation unit finished
; Undefined variable:
; X
; caught 1 WARNING condition
21
what is the proper way to increment x ?
This is indeed how you are meant to increment x, or at least one way of doing so. However it is not how you are meant to bind x. In CL you need to establish a binding for a name before you use it, and you don't do that by just assigning to it. So, for instance, this code (in a fresh CL image) is not legal CL:
(defun bad ()
(setf y 2))
Typically this will cause a compile-time warning and a run-time error, although it may do something else: its behaviour is not defined.
What you have done, in particular, is actually worse than this: you have rammed a value into the symbol-value of x (with set, which does this), and then assumed that something like (incf x) will work, which it is extremely unlikely to do. For instance consider something like this:
(defun worse ()
(let ((x 2))
(set 'x 4)
(incf x)
(values x (symbol-value 'x))))
This is (unlike bad) legal code, but it probably does not do what you want it to do.
Many CL implementations do allow assignment to previously unbound variables at the top-level, because in a conversational environment it is convenient. But the exact meaning of such assignments is outwith the language standard.
CMUCL and its derivatives, including SBCL, have historically been rather more serious about this than other implementations were at the time. I think the reason for this is that the interpreter was a bunch more serious than most others and/or they secretly compiled everything anyway and the compiler picked things up.
A further problem is that CL has slightly awkward semantics for top-level variables: if you go to the effort to establish a toplevel binding in the normal way, with defvar & friends, then you also cause the variable to be special -- dynamically scoped -- and this is a pervasive effect: it makes all bindings of that name special. That is often a quite undesirable consequence. CL, as a language, has no notion of a top-level lexical variable.
What many implementations did, therefore, was to have some kind of informal notion of a top-level binding of something which did not imply a special declaration: if you just said (setf x 3) at the toplevel then this would not contage the entire environment. But then there were all sorts of awkward questions: after doing that, what is the result of (symbol-value 'x) for instance?
Fortunately CL is a powerful language, and it is quite possible to define top-level lexical variables within the language. Here is a very hacky implementation called deflexical. Note that there are better implementations out there (including at least one by me, which I can't find right now): this is not meant to be a bullet-proof solution.
(defmacro deflexical (var &optional value)
;; Define a cheap-and-nasty global lexical variable. In this
;; implementation, global lexicals are not boundp and the global
;; lexical value is not stored in the symbol-value of the symbol.
;;
;; This implementation is *not* properly thought-through and is
;; without question problematic
`(progn
(define-symbol-macro ,var (get ',var 'lexical-value))
(let ((flag (cons nil nil)))
;; assign a value only if there is not one already, like DEFVAR
(when (eq (get ',var 'lexical-value flag) flag)
(setf (get ',var 'lexical-value) ,value))
;; Return the symbol
',var)))
I've written an ad hoc parser generator that creates code to convert an old and little known 7-bit character set into unicode. The call to the parser generator expands into a bunch of defuns enclosed in a progn, which then get compiled. I only want to expose one of the generated defuns--the top-level one--to the rest of the system; all the others are internal to the parser and only get called from within the dynamic scope of the top-level one. Therefore, the other defuns generated have uninterned names (created with gensym). This strategy works fine with SBCL, but I recently tested it for the first time with CLISP, and I get errors like:
*** - FUNCALL: undefined function #:G16985
It seems that CLISP can't handle functions with uninterned names. (Interestingly enough, the system compiled without a problem.) EDIT: It seems that it can handle functions with uninterned names in most cases. See the answer by Rörd below.
My questions is: Is this a problem with CLISP, or is it a limitation of Common Lisp that certain implementations (e.g. SBCL) happen to overcome?
EDIT:
For example, the macro expansion of the top-level generated function (called parse) has an expression like this:
(PRINC (#:G75735 #:G75731 #:G75733 #:G75734) #:G75732)
Evaluating this expression (by calling parse) causes an error like the one above, even though the function is definitely defined within the very same macro expansion:
(DEFUN #:G75735 (#:G75742 #:G75743 #:G75744) (DECLARE (OPTIMIZE (DEBUG 2)))
(DECLARE (LEXER #:G75742) (CONS #:G75743 #:G75744))
(MULTIPLE-VALUE-BIND (#:G75745 #:G75746) (POP-TOKEN #:G75742)
...
The two instances of #:G75735 are definitely the same symbol--not two different symbols with the same name. As I said, this works with SBCL, but not with CLISP.
EDIT:
SO user Joshua Taylor has pointed out that this is due to a long standing CLISP bug.
You don't show one of the lines that give you the error, so I can only guess, but the only thing that could cause this problem as far as I can see is that you are referring to the name of the symbol instead of the symbol itself when trying to call it.
If you were referring to the symbol itself, all your lisp implementation would have to do is lookup that symbol's symbol-function. Whether it's interned or not couldn't possibly matter.
May I ask why you haven't considered another way to hide the functions, i.e. a labels statement or defining the functions within a new package that exports only the one external function?
EDIT: The following example is copied literally from an interaction with the CLISP prompt.
As you can see, calling the function named by a gensym is working as expected.
[1]> (defmacro test ()
(let ((name (gensym)))
`(progn
(defun ,name () (format t "Hello!"))
(,name))))
TEST
[2]> (test)
Hello!
NIL
Maybe your code that's trying to call the function gets evaluated before the defun? If there's any code in the macro expansion besides the various defuns, it may be implementation-dependent what gets evaluated first, and so the behaviour of SBCL and CLISP may differ without any of them violating the standard.
EDIT 2: Some further investigation shows that CLISP's behaviour varies depending upon whether the code is interpreted directly or whether it's first compiled and then interpreted. You can see the difference by either directly loading a Lisp file in CLISP or by first calling compile-file on it and then loading the FASL.
You can see what's going on by looking at the first restart that CLISP offers. It says something like "Input a value to be used instead of (FDEFINITION '#:G3219)." So for compiled code, CLISP quotes the symbol and refers to it by name.
It seems though that this behaviour is standard-conforming. The following definition can be found in the HyperSpec:
function designator n. a designator for a function; that is, an object that denotes a function and that is one of: a symbol (denoting the function named by that symbol in the global environment), or a function (denoting itself). The consequences are undefined if a symbol is used as a function designator but it does not have a global definition as a function, or it has a global definition as a macro or a special form. See also extended function designator.
I think an uninterned symbol matches the "a symbol is used as a function designator but it does not have a global definition as a function" case for unspecified consequences.
EDIT 3: (I can agree that I'm not sure whether CLISP's behaviour is a bug or not. Someone more experienced with details of the standard's terminology should judge this. It comes down to whether the function cell of an uninterned symbol - i.e. a symbol that cannot be referred to by name, only by having a direct hold on the symbol object - would be considered a "global definition" or not)
Anyway, here's an example solution that solves the problem in CLISP by interning the symbols in a throwaway package, avoiding the matter of uninterned symbols:
(defmacro test ()
(let* ((pkg (make-package (gensym)))
(name (intern (symbol-name (gensym)) pkg)))
`(progn
(defun ,name () (format t "Hello!"))
(,name))))
(test)
EDIT 4: As Joshua Taylor notes in a comment to the question, this seems to be a case of the (10 year old) CLISP bug #180.
I've tested both workarounds suggested in that bug report and found that replacing the progn with locally actually doesn't help, but replacing it with let () does.
You can most certainly define functions whose names are uninterned symbols. For instance:
CL-USER> (defun #:foo (x)
(list x))
#:FOO
CL-USER> (defparameter *name-of-function* *)
*NAME-OF-FUNCTION*
CL-USER> *name-of-function*
#:FOO
CL-USER> (funcall *name-of-function* 3)
(3)
However, the sharpsign colon syntax introduces a new symbol each time such a form is read read:
#: introduces an uninterned symbol whose name is symbol-name. Every time this syntax is encountered, a distinct uninterned symbol is created. The symbol-name must have the syntax of a symbol with no package prefix.
This means that even though something like
CL-USER> (list '#:foo '#:foo)
;=> (#:FOO #:FOO)
shows the same printed representation, you actually have two different symbols, as the following demonstrates:
CL-USER> (eq '#:foo '#:foo)
NIL
This means that if you try to call such a function by typing #: and then the name of the symbol naming the function, you're going to have trouble:
CL-USER> (#:foo 3)
; undefined function #:foo error
So, while you can call the function using something like the first example I gave, you can't do this last one. This can be kind of confusing, because the printed representation makes it look like this is what's happening. For instance, you could write such a factorial function like this:
(defun #1=#:fact (n &optional (acc 1))
(if (zerop n) acc
(#1# (1- n) (* acc n))))
using the special reader notation #1=#:fact and #1# to later refer to the same symbol. However, look what happens when you print that same form:
CL-USER> (pprint '(defun #1=#:fact (n &optional (acc 1))
(if (zerop n) acc
(#1# (1- n) (* acc n)))))
(DEFUN #:FACT (N &OPTIONAL (ACC 1))
(IF (ZEROP N)
ACC
(#:FACT (1- N) (* ACC N))))
If you take that printed output, and try to copy and paste it as a definition, the reader creates two symbols named "FACT" when it comes to the two occurrences of #:FACT, and the function won't work (and you might even get undefined function warnings):
CL-USER> (DEFUN #:FACT (N &OPTIONAL (ACC 1))
(IF (ZEROP N)
ACC
(#:FACT (1- N) (* ACC N))))
; in: DEFUN #:FACT
; (#:FACT (1- N) (* ACC N))
;
; caught STYLE-WARNING:
; undefined function: #:FACT
;
; compilation unit finished
; Undefined function:
; #:FACT
; caught 1 STYLE-WARNING condition
I hope I get the issue right. For me it works in CLISP.
I tried it like this: using a macro for creating a function with a GENSYM-ed name.
(defmacro test ()
(let ((name (gensym)))
`(progn
(defun ,name (x) (* x x))
',name)))
Now I can get the name (setf x (test)) and call it (funcall x 2).
Yes, it is perfectly fine defining functions that have names that are unintenred symbols. The problem is that you cannot then call them "by name", since you can't fetch the uninterned symbol by name (that is what "uninterned" means, essentially).
You would need to store the uninterned symbol in some sort of data structure, to then be able to fetch the symbol. Alternatively, store the defined function in some sort of data structure.
Surprisingly, CLISP bug 180 isn't actually an ANSI CL conformance bug. Not only that, but evidently, ANSI Common Lisp is itself so broken in this regard that even the progn based workaround is a courtesy of the implementation.
Common Lisp is a language intended for compilation, and compilation produces issues regarding the identity of objects which are placed into compiled files and later loaded ("externalized" objects). ANSI Common Lisp requires that literal objects reproduced from compiled files are only similar to the original objects. (CLHS 3.2.4 Literal Objects in Compiled Files).
Firstly, according to the definition similarity (3.2.4.2.2 Definition of Similarity), the rules for uninterned symbols is that similarity is name based. If we compile code with a literal that contains an uninterned symbol, then when we load the compiled file, we get a symbol which is similar and not (necessarily) the same object: a symbol which has the same name.
What if the same uninterned symbol is inserted into two different top-level forms which are then compiled as a file? When the file is loaded, are those two similar to each other at least? No, there is no such requirement.
But it gets worse: there is also no requirement that two occurrences of the same uninterned symbol in the same form will be externalized in such a way that their relative identity is preserved: that the re-loaded version of that object will have the same symbol object in all the places where the original was. In fact, the definition of similarity contains no provision for preserving the circular structure and substructure sharing. If we have a literal like '#1=(a b . #1#), as a literal in a compiled file, there appears to be no requirement that this be reproduced as a circular object with the same graph structure as the original (a graph isomorphism). The similarity rule for conses is given as naive recursion: two conses are similar if their respective cars and cdrs are similar. (The rule can't even be evaluated for circular objects; it doesn't terminate).
That the above works is because of implementations going beyond what is required in the spec; they are providing an extension consistent with (3.2.4.3 Extensions to Similarity Rules).
Thus, purely according to ANSI CL, we cannot expect to use macros with gensyms in compiled files, at least in some ways. The expectation expressed in code like the following runs afoul of the spec:
(defmacro foo (arg)
(let ((g (gensym))
(literal '(blah ,g ,g ,arg)))
...))
(defun bar ()
(foo 42))
The bar function contains a literal with two insertions of a gensym, which according to the similarity rules for conses and symbols need not reproduce as a list containing two occurrences of the same object in the second and third positions.
If the above works as expected, it's due to "extensions to the similarity rules".
So the answer to the "Why can't CLISP ..." question is that although CLISP does provide an extension for similarity which preserves the graph structure of literal forms, it doesn't do it across the entire compiled file, only within individual top level items within that file. (It uses *print-circle* to emit the individual items.) The bug is that CLISP doesn't conform to the best possible behavior users can imagine, or at least to a better behavior exhibited by other implementations.
Does emacs lisp have a function that provides a unique object identifier, such as e.g. a memory address? Python has id(), which returns an integer guaranteed to be unique among presently existing objects. What about elisp?
The only reason I know for wanting a function like id() is to compare objects, and ensure that they only compare equal if they are the same (as in, in the same memory location). In Lisps, this is done a bit differently from in Python:
In most lisps, including elisp, there are several different notions of equality. The most expensive, and weakest equivalence is equal. This is not what you want, since two lists (say) are equal if they have the same elements (tested recursively with equal). As such
(equal (list 1 2) (list 1 2)) => T
is true. At the other end of the spectrum is eq, which tests "identity" rather than equality:
(eq (list 1 2) (list 1 2)) => NIL
This is what you want, I think.
So, it seems that Python works by providing one equality test, and then a function that gives you a memory location for each object, which then can be compared as integers. In Elisp (and at least Common Lisp too), on the other hand, there is more than one meaning of "equality".
Note, there is also "eql", which lies somewhere between the two.
(EDIT: My original answer probably wasn't clear enough about why the distinction between eq and equal probably solves the problem the original poster was having)
There is no such feature in Emacs Lisp, as far as I know. If you only need equality, use eq, which performs a pointer comparison behind the scenes.
If you need a printable unique identifier, use gensym from the cl package.
If you need a unique identifier to serve as an index in a data structure, use gensym (or maintain your own unique id — gensym is simpler and less error-prone).
Some languages bake a unique id into every object, but this has a cost: either every object needs extra memory to store the id, or the id is derived from the address of the object, which precludes modifying the address. Python chooses to pay the cost, Emacs chooses not to.
My whole point in asking the question was that I was looking for a way to distinguish between the printed representations of different symbols that have the same name. Thanks to the elisp manual, I've discovered the variable print-gensym, which, when non-nil, causes #: to be prepended to uninterned symbols printed. Moreover, if the same call to print prints the same uninterned symbol more than once, it will mark the first one with #N= and subsequent ones with `#N#. This is exactly the kind of functionality I was looking for. For example:
(setq print-gensym t)
==> t
(make-symbol "foo")
==> #:foo
(setq a (make-symbol "foo"))
==> #:foo
(cons a a)
==> (#1=#:foo . #1#)
(setq b (make-symbol "foo"))
==> #:foo
(cons a b)
==> (#:foo . #:foo)
The #: notation works for read as well:
(setq a '#:foo)
==> #:foo
(symbol-name a)
==> "foo"
Note the ' on '#:foo--the #: notation is a symbol-literal. Without the ', the uninterned symbol is evaluated:
(symbol-name '#:foo)
==> "foo"
(symbol-name #:foo)
==> (void-variable #:foo)