Listing currently bound global variables in lisp - lisp

I am quite new to Lisp and was wondering:
Is there a way to list all (user-defined) global variables?

One possibility would be to check which of the symbols of a package are boundp:
(defun user-defined-variables (&optional (package :cl-user))
(loop with package = (find-package package)
for symbol being the symbols of package
when (and (eq (symbol-package symbol) package)
(boundp symbol))
collect symbol))
It returns a list of bound symbols that are not inherited from another package.
CL-USER> (user-defined-variables) ; fresh session
NIL
CL-USER> 'foo ; intern a symbol
FOO
CL-USER> (defun bar () 'bar) ; define a function
BAR
CL-USER> (defparameter *baz* 'baz) ; define a global variable
*BAZ*
CL-USER> (user-defined-variables)
(*BAZ*) ; only returns the global variable

Related

Assigning values to global variables in Lisp

I am new to Lisp and I am trying to understand how variables and assigning variables work. I am trying to check if a given number is 5 or not. If it is, it is supposed to change it to 10 and print it. If not it just tells you that it is not 5 but my code below is not working.
(print "Give me a number")
(setq *x* (read))
(defun check(*x*)
(if (= *x* 5) (setq *x* 10) (format t "~d is not 5" *x*))
)
check(*x*)
print(*x*)
check(*x*) and print(*x*) isn't Lisp syntax. Replace that with (check *x*) and (print *x*)
Your function operates with some *x*, but it isn't your global *x*, but local variable*x* from argument list. Don't name function arguments same as global variables. Check this example:
(defparameter a 8)
(defun add (a)
(setf a (+ a 2))
(print a))
(add a)
(print a)
Returned values are 10 and then 8. Compare it with:
(defparameter *a* 8)
(defun add (a)
(setf *a* (+ *a* 2))
(print *a*))
(add *a*)
(print *a*)
Returned values are 10 and 10.
Don't use setf or setq for creating variable, these functions are only for setting new value. Use defparameter or defvar instead.
This code assigns value to global variable:
(print "Give me a number")
(defparameter *x* (read *query-io*))
(defun check ()
(if (= *x* 5)
(setf *x* 10)
(format t "~d is not 5" *x*)))
(check)
(print *x*)
You can also do this task without global variables.
(print "Give me a number")
(defun check (n)
(if (= n 5)
(setf n 10)
(format t "~d is not 5" n))
(print n))
(check (read *query-io*))
If I compile a file with this definition in SBCL:
(print "Give me a number")
(setq *x* (read))
(defun check(*x*)
(if (= *x* 5) (setq *x* 10) (format t "~d is not 5" *x*))
)
There are two warnings:
warning: undefined variable: *X*
This is located at the global setq, because setq does not declare global variables. So the compiler does not know any global variable named *x*.
Then, inside the function:
style-warning:
using the lexical binding of the symbol (*X*), not the
dynamic binding, even though the name follows
the usual naming convention (names like *FOO*) for special variables
--> PROGN SB-IMPL::%DEFUN SB-IMPL::%DEFUN SB-INT:NAMED-LAMBDA
==>
#'(SB-INT:NAMED-LAMBDA AAAB::CHECK
(*X*)
(DECLARE (SB-C::TOP-LEVEL-FORM))
(BLOCK CHECK
(IF (= *X* 5)
(SETQ *X* 10)
(FORMAT T "~d is not 5" *X*))))
This is pretty self-explanatory, but here the binding is lexical.
This is however not a property of defun, this is because there is no declared special variable named *x*.
If instead, you use defvar or defparameter, there is no warning (e.g. (defvar *x* 0)). The same goes if you only declare the variable as special: (declaim (special *x*)), before you define the function that binds it.
Inside the function, assuming *x* is declared as a special variable, the bindings works as a let, which means you can use special variables as function parameters. In your case this does not change much because you only set the current binding with the inner setq (the parameter shadows the global binding), so in effect this is a lot like a lexical variable.
But you can use a special variable to rebind dynamically a parameter:
(defun foo (*a*) ...)
This is the same as:
(defun foo (a)
(let ((*a* a))
...))

Is defun or setf preferred for creating function definitions in common lisp and why?

What is the fundamental difference in the functions defined using defun and setf as below and is one method preferred over another outside of style considerations?
Using defun:
* (defun myfirst (l)
(car l) )
MYFIRST
* (myfirst '(A B C))
A
Using setf:
* (setf (fdefinition 'myfirst) #'(lambda (l) (car l)))
#<FUNCTION (LAMBDA (L)) {10021B477B}>
* (myfirst '(A B C))
A
If, as according to Wikipedia:
named functions are created by storing a lambda expression in a symbol using the defun macro
Using setf to create a variable in a different way requires the use of funcall:
* (defvar myfirst)
MYFIRST
* (setf myfirst (lambda (l) (car l)))
#<Interpreted Function (LAMBDA (X) (+ X X)) {48035001}>
* (funcall myfirst '(A B C))
A
My understanding is that this type of variable is different than the previous in that this variable is not found in the same namespace as the defun bound symbol as described in Why multiple namespaces?.
First of all, one should never underestimate the importance of style.
We write code not just for computers to run, but, much more importantly, for people to read.
Making code readable and understandable for people is a very important aspect of software development.
Second, yes, there is a big difference between (setf fdefinition) and defun.
The "small" differences are that defun can also set the doc string of the function name (actually, depending on how your imeplementation works, it might do that with lambda also), and creates a named block (seen in the macroexpansions below) which you would otherwise have to create yourself if you want to.
The big difference is that the compiler "knows" about defun and will process it appropriately.
E.g., if your file is
(defun foo (x)
(+ (* x x) x 1))
(defun bar (x)
(+ (foo 1 2 x) x))
then the compiler will probably warn you that you call foo in bar with the wrong number of arguments:
WARNING: in BAR in lines 3..4 : FOO was called with 3 arguments, but it requires 1
argument.
[FOO was defined in lines 1..2 ]
If you replace the defun foo with (setf (fdefinition 'foo) (lambda ...)), the compiler is unlikely to handle it as carefully. Moreover, you will probably get a warning along the lines of
The following functions were used but not defined:
FOO
You might want to examine what defun does in your implementation by macroexpanding it:
(macroexpand-1 '(defun foo (x) "doc" (print x)))
CLISP expands it to
(LET NIL (SYSTEM::REMOVE-OLD-DEFINITIONS 'FOO)
(SYSTEM::EVAL-WHEN-COMPILE
(SYSTEM::C-DEFUN 'FOO (SYSTEM::LAMBDA-LIST-TO-SIGNATURE '(X))))
(SYSTEM::%PUTD 'FOO
(FUNCTION FOO
(LAMBDA (X) "doc" (DECLARE (SYSTEM::IN-DEFUN FOO)) (BLOCK FOO (PRINT X)))))
(EVAL-WHEN (EVAL)
(SYSTEM::%PUT 'FOO 'SYSTEM::DEFINITION
(CONS '(DEFUN FOO (X) "doc" (PRINT X)) (THE-ENVIRONMENT))))
'FOO)
SBCL does:
(PROGN
(EVAL-WHEN (:COMPILE-TOPLEVEL) (SB-C:%COMPILER-DEFUN 'FOO NIL T))
(SB-IMPL::%DEFUN 'FOO
(SB-INT:NAMED-LAMBDA FOO
(X)
"doc"
(BLOCK FOO (PRINT X)))
(SB-C:SOURCE-LOCATION)))
The point here is that defun has a lot "under the hood", and for a reason. setf fdefinition is, on the other hand, more of "what you see is what you get", i.e., no magic involved.
This does not mean that setf fdefinition has no place in a modern lisp codebase. You can use it, e.g., to implement a "poor man's trace" (UNTESTED):
(defun trace (symbol)
(setf (get symbol 'old-def) (fdefinition symbol)
(fdefinition symbol)
(lambda (&rest args)
(print (cons symbol args))
(apply (get symbol 'old-def) args))))
(defun untrace (symbol)
(setf (fdefinition symbol) (get symbol 'old-def))
(remprop symbol 'odd-def))

Can't call functions defined in macro with names generated by make-symbol

I'm trying to write an ELisp macro to generate a multiple functions based on some common data. For example, when I want to compute the fn names I write something like (I'm ignoring hygiene for the moment, I'm passing a symbol literal into the macro so evaluation shouldn't matter):
(cl-defmacro def-fns (sym)
"SYM."
(let ((s1 (make-symbol (concat (symbol-name sym) "-1")))
(s2 (make-symbol (concat (symbol-name sym) "-2"))))
`(progn (defun ,s1 () (+ 1 2 3))
(defun ,s2 () "six"))))
which I expect to generate 2 fns when invoked, called foo-1 and foo-2.
I should then be able to invoke the macro and fns like so:
(def-fns foo)
(foo-1)
;; => 6
(foo-2)
;; -> "six
Even the macroexpansion of (def-fns foo) in Emacs suggests that this should be the case:
(progn
(defun foo-1 nil (+ 1 2 3))
(defun foo-2 nil "six"))
However, when I evaluate the def-fns definition and invoke it it does not generate those functions. Why is this the case? This technique works in Common Lisp and in Clojure (which have very similar macro systems), so why not in ELisp?
Your code would not work in CL either.
The problem is with make-symbol - it creates a new symbol, so that
(eq (make-symbol "A") (make-symbol "A"))
==> nil
This means that your macro creates the functions but binds them to symbols which you no longer have a handle on.
When you evaluate (foo-1), Emacs Lisp reader tries to find the function binding of the interned symbol foo-1, not the fresh uninterned symbol your macro created.
You need to use intern instead: it makes the symbol "generally available", so to speak:
(eq (intern "a") (intern "a))
==> t
So, the corrected code looks like this:
(defmacro def-fns (sym)
"SYM."
(let ((s1 (intern (concat (symbol-name sym) "-1")))
(s2 (intern (concat (symbol-name sym) "-2"))))
`(progn (defun ,s1 () (+ 1 2 3))
(defun ,s2 () "six"))))
(def-fns foo)
(foo-1)
==> 6
(foo-2)
==> "six"
Notes:
If you were using CL, the uninterned symbols would have been printed as #:foo-1 and the source of your problem would have been obvious to you.
It is exceedingly rare that you really need to use make-symbol. Usually, you want to use either intern or gensym.

How is sharp quote (#') different from symbol-function?

To me these operators seem to do the same thing. Both take a symbol and return the function associated with it. Is there any difference?
elisp evaluation returns the following:
(defun foo (x) (+ 1 x))
foo
(foo 3)
4
#'foo
Which I don't understand either.
Furthermore is there a difference between common lisp and elisp? I'm learning from resources on either.
Common Lisp:
SYMBOL-FUNCTION can't retrieve functions from lexically bound functions. FUNCTION references the lexically bound function by default. #'foo is just a shorter notation for (FUNCTION foo).
CL-USER 1 > (defun foo () 'foo)
FOO
CL-USER 2 > (flet ((foo () 'bar))
(list (funcall (symbol-function 'foo))
(funcall #'foo)
(funcall (function foo))
(eq (function foo) (symbol-function 'foo))))
(FOO BAR BAR NIL)
CL-USER 3 > (eq (function foo) (symbol-function 'foo))
T
Rainer's answer discusses the things that you can do with function that you can't do with symbol-function, namely retrieve the value of lexically scoped functions, but there are a few other differences too.
The special operator FUNCTION
The special operator FUNCTION provides a way to find the functional value of a name in a lexical environment. Its argument is either a function name or a lambda expression. That function can take a lambda expression means that you can write: (function (lambda (x) (list x x))). The term function name includes more than just symbols. It also includes lists of the form (setf name), which means that you can do things like (function (setf car)).
The accessor SYMBOL-FUNCTION
The accessor symbol-function, on the other hand, lets you retrieve and set the functional value of a symbol. Since it requires a symbol, you can't do (symbol-function (lambda …)) or (function (setf name)). Symbol-function also can't see lexical environments; it only works with global definitions. E.g.,
(flet ((foo () 'result))
(symbol-function 'foo))
;=> NIL
Since symbol-function is an accessor, you can change the value of a function's symbol with it. E.g.:
CL-USER> (setf (symbol-function 'foo) (lambda () 42))
#<FUNCTION (LAMBDA ()) {1005D29AAB}>
CL-USER> (foo)
42
CL-USER> (setf (symbol-function 'foo) (lambda () 26))
#<FUNCTION (LAMBDA ()) {1005D75B9B}>
CL-USER> (foo)
26
The accessor FDEFINITION
There's also the accessor fdefinition which is kind of like symbol-function in that is can only access globally defined functions, but kind of like function in that it can access of symbols and (setf name) lists. However, it does not retrieve the value of lambda functions.

define variable with defparameter with name determined at runtime

I would like to initiate dynamically a hash table with defmethod or defun using one of the arguments to create the name. For instance:
(defun foo (arg)
(let ((NAME (read-from-string (format nil "\*~S\*" arg))))
(defparameter NAME (make-hash-table))))
Of course, foo create hash table with the symbol NAME, instead of the value of NAME in let. What can I do to get the value of NAME to create this hash table?
General Remarks
It is almost always wrong to create global variables in functions.
It is also almost always wrong to create new symbols using read-from-string instead of intern.
Use a Macro
What you probably want is
(defmacro def-ht (name)
(let ((var (intern (concatenate 'string "*" (symbol-name name) "*")
(symbol-package name))))
`(defparameter ,var (make-hash-table))))
(def-ht foo)
Use a Function
You might be able to do it in a function too - by inspecting the macroexpansion of a defparameter form and placing the needed stuff in the function:
(defun make-ht-var (name)
(let ((var (intern (concatenate 'string "*" (symbol-name name) "*")
(symbol-package name))))
(setf (symbol-value var) (make-hash-table))
(proclaim (list var 'special))))
(make-ht-var 'foo)
Note that the argument to the function is quoted, but the argument to the macro is not.
You need to use a macro instead of a function. DEFPARAMETER will bind value of MAKE-HASH-TABLE to the symbol NAME because it evaluates at macro-expansion time which occurs earlier than run-time, which is when the function FOO binds the lexical value of NAME.
Look up the CL evaluation model for a deeper understanding.
(defmacro foo (arg)
(let ((name (read-from-string (format nil "*~S*" arg))))
`(defparameter ,name (make-hash-table))))
(foo "my-hash")
=> <hash-table 0x121>
*my-hash*
=> <hash-table 0x121>