(defmacro foo (x)
`(defun ,xt ()
(format t "hullo")))
(foo bar)
will not define a function bart, since ,xt is read as the variable xt rather than the variable x plus a t. But is there a way to get a function bart by supplying the argument bar?
You need to create the function name (which is a string then) and then convert it to a symbol, for example:
(defmacro foo (x)
`(defun ,(intern (format nil "~aT" x)) ()
(format t "hullo")))
then
? (foo bar)
BART
? (bart)
hullo
NIL
Related
I'm trying to create a macro (bar) that should be used like this:
(let ((my-var "foo"))
(bar ("some")
:buzz (lambda () (format t "~a~%" my-var))))
The macro should basically just FUNCALL the lambda with taking MY-VAR into account.
What I've come up with is this:
(defmacro bar ((thing) &body body)
`(funcall (coerce (getf (list ,#body) :buzz) 'function)))
This works, it prints "foo".
But I'd like to know if this is how this is done or if there is a better way of doing this.
Well, for a start if you don't need the extra, unused argument, then this is just funcall, since (lambda ...) denotes a function.
(let ((my-var "foo"))
(funcall (lambda () (format t "~a~%" my-var))))
So even if you didn't want to call this funcall you could write it as a function, not a macro: it's not doing any code transformation. But presumably you actually do need this extra argument, and you are intending to use it for some purpose in the expansion, so what you need then is a macro which takes keyword arguments and just expands into a suitable funcall form:
(defmacro bar ((thing) &key (buzz '(lambda ())))
(declare (ignore thing))
`(funcall ,buzz))
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.
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>
I'm tring to apply closure in emacs lisp.And I find a post here:
How do I do closures in Emacs Lisp?
with some code like:
(defun foo (x) `(lambda () ,x)) (message (string (funcall (foo
66))))
But following the emacs documentation lambda should be formated like
'(lambda () x) ==> using this format ,I got an ERROR :Symbol's value as variable is void: x
When " , " is add betwenn "()" and "x" ,everything goes right .
Why?
This happens because Emacs Lisp is dynamically scoped thus foo returns a lambda where x is free. That's what the error tells you.
To do a closure in Emacs Lisp you have to use lexical-let which simulates a lexical binding and therefore allows you to make a real closure.
(defun foo (x)
(lexical-let ((x x))
(lambda () x)))
(message (string (funcall (foo 66))))
Here are some links from the Emacs Wiki:
Dynamic Binding Vs Lexical Binding
Fake Closures
Note that you could have defined x with a let like this:
(defun foo (x)
(lambda () x))
(message (string (let ((x 66)) (funcall
(foo 'i-dont-care)))))
This answer gives a bit of detail behind the first part of #Daimrod's correct answer.
Your question is why this works:
(defun foo (x) `(lambda () ,x)) (message (string (funcall (foo 66))))
and this doesn't work:
(defun foo (x) '(lambda () x)) (message (string (funcall (foo 66))))
First, the quote (') is unnecessary in the second, because in Emacs Lisp lambda forms are self-evaluating. That is, '(lambda (...) ...) generally acts the same as (lambda (...) ...).
But what does the backquote (`), instead of quote ('), do?
Inside a backquoted expression, a comma (,) means replace the next expression by its value, that is, evaluate it. So this:
`(lambda () ,x)
means create and return a list whose first element is the symbol lambda (that's not evaluated), whose second element is () (that's not evaluated), and whose third element is the value of variable x. It's equivalent to evaluating this code, which uses function list:
(list 'lambda '() x)
That's what you want: replace x by its current value (in this case, its value inside the function foo, that is, the value of foo's argument).
But this:
'(lambda () x)
(and likewise, because a lambda form is self-evaluating, (lambda () x)) returns this list: (lambda () x). And when that's evaluated using dynamic scoping (the default scoping regime in Emacs Lisp), x is unbound - it has no value. So a void-variable error is raised.
Suppose I have this wonderful function foo
[92]> (defun foo () (lambda() 42))
FOO
[93]> (foo)
#<FUNCTION :LAMBDA NIL 42>
[94]>
Now, suppose I want to actually use foo and return 42.
How do I do that? I've been scrounging around google and I can't seem to come up with the correct syntax.
You want the FUNCALL function:
* (defun foo () (lambda () 42))
FOO
* (funcall (foo))
42
The relevant terms here are "Lisp-1" and "Lisp-2".
Your attempt at calling would work in a Lisp-1 like e.g. Scheme or Clojure.
Common Lisp is a Lisp-2 however which roughly means that variable names and function names are separate.
So, in order to call the function bound to a variable you either need to use the special forms funcall or apply as others have pointed out or set the function value of the symbol foo rather than the variable value.
The former basically takes the variable value of the symbol, assumes/checks that value is a function and then calls the function (with whatever arguments you passed to funcall/apply.
You don't really want to do the latter as that is quite silly in all but very specialised cases, but for completness sake this is roughly how you'd do it:
CL-USER> (setf (symbol-function 'foo) (lambda () 42))
#<FUNCTION (LAMBDA ()) {C43DCFD}>
CL-USER> (foo)
42
You should also want to look into the labels and flet special forms (which are commonly used) - there you actually do use the latter methods (these forms create a temporary function binding for symbols).
So your problem would there look like this:
(flet ((foo ()
42))
(foo))
i.e. here you temporarily bind the function value of the symbol foo to the function returning 42. Within that temporary context you can then call (foo) like regular global functions.
Your function foo returns a function. Use the funcall function to apply a function to arguments, even if the argument set is empty.
Here you can see that foo returns a value of type function:
CL-USER> (let ((f (foo)))
(type-of f))
FUNCTION
CL-USER> (let ((f (foo)))
(funcall f))
42
CL-USER> (type-of (foo))
FUNCTION
CL-USER> (funcall (foo))
42
Another option besides FUNCALL is APPLY:
(apply (foo) nil)
FUNCALL is the idiomatic way here, but you'll need APPLY when you have a list of parameters.