In most of the examples on the Internet, symbol equality is straight-forward:
(eq 'sym 'sym)
t
In my program, I want to compare symbols in a custom package:
(defpackage #:my-package
(:use #:common-lisp)
(:export #:my-function))
(in-package #:my-package)
(defun my-function (value)
(cond
((eq value 'sym)
(format t "SYM: YES~%"))
(t
(format t "SYM: NO~%"))))
(in-package #:common-lisp)
(my-package:my-function 'sym)
But when this code is executed, it displays:
SYM: NO
It seems like the 2 symbols are different:
(eq 'sym 'my-package::sym)
nil
The reason seems easy to understand, a symbol interned in a given package is not equal to a symbol with the same name interned in another package. Fair enough! But what is the idiom to compare 2 symbols, regardless the package?
Should we convert it to a string in a first place and compare strings?
(defpackage #:my-package
(:use #:common-lisp)
(:export #:my-function))
(in-package #:my-package)
(defun my-function (value)
(cond
((string= (symbol-name value) "SYM")
(format t "SYM: YES~%"))
(t
(format t "SYM: NO~%"))))
(in-package #:common-lisp)
(my-package:my-function 'sym)
The result is better, but there is an obvious issue regarding the character case.
How to check that 2 symbols are the same regardless their package?
The usual idiom is string=, which compares the names of symbols without regard to identity.
For example:
(eq 'x:a 'y:a) => nil
(string= 'x:a 'y:a) => t
I think you are confused about what 'being the same symbol' means.
Two symbols are the same symbol iff they are eq, which means that if they are not eq they are not the same symbol. In particular if two symbols have home packages which are different (again: not eq) they are not the same symbol.
If you want to know if two symbols have the same name then simply compare their symbol-names, as strings. But two symbols which merely have the same name are not necessarily the same symbol (for instance (eq '#:foo '#:foo) is false).
There are cases where it is useful to know whether two symbols have the same name (for instance the loop macro must do this so that (loop for ...) and (loop :for ...) mean the same thing) but these cases are fairly rare.
If what you want to do is know if two symbols have the same names you quite likely should be using strings instead.
If what you actually want to know is whether two symbols accessed from different packages are really the same symbol (so (eq 'x:foo 'y:foo), say) then eq is the appropriate test, and understanding the package system also helps.
I'm a Lisp beginner and I'm struggling to understand why the following code gives me an error.
(dolist (elem '(mapcar
mapcon))
(when (fboundp `',elem) (print "hello")))
Thanks.
Edit:
A bit more context. I wrote the following in Elisp and I don't know how to fix it.
(dolist (ui-elem '(menu-bar-mode
tool-bar-mode
tooltip-mode
scroll-bar-mode
horizontal-scroll-bar-mode))
(when (fboundp `',ui-elem) (ui-elem -1)))
Note
In your question you mix common-lisp and elisp, but they are two different languages. The question however touches on concepts that are identical in both languages.
The need to quote symbols
The code you want to write checks if a symbol is bound to a function.
What you already know probably is that you can call fboundp on a symbol to determines this:
(fboundp 'menu-bar-mode)
=> t
When you evalute the above form, 'menu-bar-mode is the same as (quote menu-bar-mode), and is evaluated as the symbol object menu-bar-mode. This is the value that is given as an argument to fboundp.
In you example you want to iterate over a list of symbols, call fboundp on it and call the function if the symbol denotes a function. You can do this as follows:
(dolist (s '(menu-bar-mode and other symbols))
(when (fboundp s)
(funcall s -1)))
The list of symbols '(menu-bar-mode and other symbols) is quoted, which means that when dolist evaluates it, it sees a list of symbols. The value to which s is bound at each iteration of the loop is a symbol object, there is no need to quote them.
Quoting a symbol is something you have to do when writing them in your code so that they are not interpreted as variables. When you iterate over a list of symbols, you already manipulate symbols.
Note also that both Common Lisp and Emacs Lisp are "Lisp-2", meanings that you have to use (funcall ui-elem -1) instead of writing (ui-elem -1). When you write the latter form, that means calling the function literally named ui-elem because for function application, the first symbol in the list is not evaluated, it is taken literally.
Too many levels of quoting
The actual error I have when I execute your code is:
(wrong-type-argument symbolp 'mapcar)
It may look like 'mapcar denotes a symbol, because when you want the interpreter to evaluate some code as a symbol, you need to quote it. However, Lisp printers write objects in a way that they can be read back to "similar" objects. The error message that is printed if I expect a symbol to be a number is the following, where symbol foo is printed unquoted:
(+ 'foo 3)
;; error: (wrong-type-argument number-or-marker-p foo)
In your error message, the form that you are trying to use as a symbol is (quote mapcar). Recall that when you directly call fboundp:
(fboundp 'mapcar)
It is the same as-if you wrote:
(fboundp (quote mapcar))
First, (quote mapcar) is evaluated, as the symbol mapcar. Then, fboundp is applied to that value.
But when you write the following, while ui-elem is bound to symbol mapcar:
(fboundp `',ui-elem)
This is equivalent to:
(fboundp `(quote ,ui-elem))
The argument to fboundp is evaluated as (quote mapcar). You have one extra level of quoting. You could write instead:
(fboundp `,ui-elem)
But then, you don't need to use backquote/comma, you can directly write:
(fboundp ui-elem)
In my function I am reading input from the user which is expected to be a lisp form given as a string e.g.:
(sym1 sym2 (sym3 sym4))
My goal is to substitute some of the symbols with other symbols e.g.:
(sublis '((sym1 . sym1%)
(sym2 . sym2%))
str)
Because I am getting the input as a string, I am first converting it to a lisp form. Here is how the final function looks like:
(defun sublis-when-string (str)
(sublis '((sym1 . sym1%)
(sym2 . sym2%))
(read-from-string str)))
When I compile the function and run it in the REPL with (sublis-when-string "(sym1 sym3 (sym2 sym4))") I correctly get:
(SYM1% SYM3 (SYM2% SYM4))
However when I run the whole program the substitutions do not work:
(SYM1 SYM3 (SYM2 SYM4))
This lead me to believe that the problems is with the package. When I changed the package in the REPL the substitutions were still not working.
My question is: How should I change my function so it works when called from other packages?
If you want to use your function independently from the package in which the symbols are read, you can change your definition by adding an explicit test: when the values to be checked are symbols, the string= operator is used instead of default eql. For instance:
(defun sublis-when-string (str)
(sublis '((sym1 . sym1%)
(sym2 . sym2%))
(read-from-string str)
:test (lambda (x y)
(if (and (symbolp x) (symbolp y))
(string= x y)
(eql x y)))))
See the definition of sublis.
You can define a package for your user:
(defpackage :my-user (:use) (:export #:sym1 #:sym2))
Of course, the exported symbols are the one you need to add in your substitution list. And then, you bind the *package* variable before reading:
(let ((*package* (find-package :my-user)))
(read-from-string string))
Notice that all the symbols will be read from the :my-user package.
Depending on how much you trust the source of your strings, you can also set *read-eval* to nil and define a minimalistic readtable too:
disable array notation that allocate very large arrays like #n() (bad user exhausting memory on purpose)
make : a terminating character to make qualified symbols throw an error (bad user interning symbols in other packages)
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.
How to print all the symbols in emacs using elisp.
It is possible to test wheather a lisp object is a symbol using symbolp
function. But how to collect all the symbols.
Is it possible to access symbol table of emacs?
Here's one way to do it:
(require 'cl)
(loop for x being the symbols
if (boundp x)
collect (symbol-name x))
loop is a Common Lisp macro, that has been ported to Emacs Lisp as well. It's part of the cl package (part of the standard Emacs distribution), that you'll have to require to use it.
Another option to consider is probably:
(apropos "." t)
The apropos invokation will take significantly more time to complete, but you'll get more information about the symbols that way.
Just for completeness, here's how you'd list all of the symbols without using the cl package:
Go to a newly-created buffer, and type M-:(mapatoms (lambda (s) (insert (symbol-name s) "\n")))RET. That will insert the names of all existing symbols in the buffer, one per line.