(defun foo ()
(send-to-debug-log "Error. Function terminated." (get-current-function-name)))
I currently do this:
(defun foo ()
(send-to-debug-log "Error. Function terminated." 'foo)))
Hard coding the function name seems not to be a good practice.
Any suggestion to implement the get-current-function-name or get-function-name-that-call-me.
(defun foo ()
(send-to-debug-log "Error. Function terminated." (nth 1 (backtrace-frame 2))))
You could try something like:
(defconst my-defun-macros '(defun cl-defun defmacro cl-defmacro))
(defun my-add-defun-name (orig-fun name &rest args)
(let* ((my-current-defun-name (if (consp name) (cadr name) name))
(macroexpand-all-environment
(cons (cons 'my-get-current-function-name
(lambda () my-current-defun-name))
macroexpand-all-environment))
(code (apply orig-fun name args)))
(macroexpand-all code macroexpand-all-environment)))
(defmacro my-get-current-function-name ()
"Return current defun's name.
Only works within the advised macros in `my-defun-macros'."
(error "my-get-current-function-name used outside of a defun"))
(dolist (m my-defun-macros)
(advice-add m :around #'my-add-defun-name))
Here's something that seems to work pretty well:
(defun calling-function ()
(let ((n 6) ;; nestings in this function + 1 to get out of it
func
bt)
(while (and (setq bt (backtrace-frame n))
(not func))
(setq n (1+ n)
func (and bt
(nth 0 bt)
(nth 1 bt))))
func))
If you call it in a lambda, you get the whole lambda. It works with apply, funcall, and eval.
I think the most simple, robust way is to just explicitly write the function name, as you're doing now.
With the package which-function-mode (in melpa), you can programmatically call the function
(Which-function)
more info:
https://www.masteringemacs.org/article/which-function-mode
http://emacsredux.com/blog/2014/04/05/which-function-mode/
Related
I am trying to identify a safe way to check whether an object is a function (named or anonymous).
As functionp or fboundp do not work as I expected giving errors, I am trying with:
(defun function-check (x)
(and (boundp 'x)
(if (symbolp x) (fboundp x)
(functionp x))))
Apparently it works with several types of objects:
(setq lfun (lambda () "hello"))
(function-check lfun) ; -> t
(setq nfun 'buffer-name)
(function-check nfun) ; -> t
(setq slfun '(lambda () "hello"))
(function-check slfun) ; -> t
(function-check 'not-bound) ; -> safe nil
Anyway, looking at my code it seems too verbose and convoluted for such a simple task.
Is it possible to make it better?
Update:
As asked, I clarify what I mean with "functionp, fboundp do now work as I expected".
Say we want to detect a valid hook. This does not work:
(setq var 'buffer-name)
(functionp 'var) ;nil
(fboundp 'var) ;nil
We need to use:
(functionp var) ;t
(fboundp var) ;t
While this works, we need to make sure that var is not void, otherwise we get errors:
(functionp void-var) ;Lisp-error
(fboundp void-var) ;Lisp-error
Depending on the situations, this implies adding extra control code, compiling the code, etc.
A valid hook can be any callable object: macros, functions, lambdas are valid hooks. Anyway functionp does not work with macros:
(defmacro mac () "hello")
(functionp 'mac) ;nil
While fbound does not work with lambda expressions:
(functionp '(lambda () t)) ;t
(functionp (lambda () t)) ;t
(fboundp '(lambda () t)) ;Lisp error
(fboundp (lambda () t)) ;Lisp error
This happens also if assigning the expression to a variable:
(setq var '(lambda () t))
(functionp var) ;t
(fboundp var) ;Lisp error
which might require testing if var is a symbol.
As I understand, there is no straight way to test an object is callable, hence my attempt.
The (boundp 'x) check is a bit tautological here. If you're compiling your code with dynamic binding, (boundp 'x) will always be t, since x is bound when entering the function. If you're compiling your code with lexical binding, (boundp 'x) will probably be nil, unless you somehow create a "global" variable called x. In neither case will the result depend on the argument you pass to the function.
So I think you just need this:
(defun function-check (x)
(if (symbolp x)
(fboundp x)
(functionp x)))
That is, check that x is either a symbol that has a function binding, or a lambda function.
I found an interesting example of a function predicate used in the AUCTeX package.
(defun TeX-function-p (arg)
"Return non-nil if ARG is callable as a function."
(or (and (fboundp 'byte-code-function-p)
(byte-code-function-p arg))
(and (listp arg)
(eq (car arg) 'lambda))
(and (symbolp arg)
(fboundp arg))))
This test is used in AUCTeX before calling the (component of) TeX commands obtained by parsing the alist TeX-expand-list.
If an expansion found in the alist is a function, then it is called with:
(apply expansion arguments)
TeX-function-p is thorough, as TeX-expand-list is huge and customisable, and expansions might also result in objects which are not symbols or symbols which are not bound.
This function answers my question in that it shows that functionp might not always be sufficient to test for a callable object.
I'm trying to pass one method to another in elisp, and then
have that method execute it. Here is an example:
(defun t1 ()
"t1")
(defun t2 ()
"t1")
(defun call-t (t)
; how do I execute "t"?
(t))
; How do I pass in method reference?
(call-t 't1)
First, I'm not sure that naming your function t is helping as 't' is used as the truth value in lisp.
That said, the following code works for me:
(defun test-func-1 () "test-func-1"
(interactive "*")
(insert-string "testing callers"))
(defun func-caller (callee)
"Execute callee"
(funcall callee))
(func-caller 'test-func-1)
Please note the use of 'funcall', which triggers the actual function call.
The note towards the end of "ยง13.7 Anonymous Functions" in the Emacs Lisp manual says that you can quote functions with #' instead of ' to signal to the byte compiler that the symbol always names a function.
Above answers are okey, but you can do something more interesting with defmacro, wich evaluates functions later for some reason:
(defun n1 ()
"n1")
(defmacro call-n (n)
(apply n))
(call-n (n1))
A practical example with a for loop that takes any amount of functions and their arguments:
(defmacro for (i &optional i++ &rest body)
"c-like for-loop"
(unless (numberp i++) (push i++ body) (setq i++ 1))
(while (/= i 0)
(let ((args 0))
(while (nth args body)
(apply (car (nth args body))
(cdr (nth args body)))
(setq args (1+ args))))
(setq i (- i i++))
)
)
I want to define a function with a parameter that defines another function with that parameter as the name.
example that dont work:
(DEFUN custom (name op const var)
(DEFUN name (var) (op const var)))
The problem is that name isnt evaluated and so the function that is defined is always called name.
I know that FUNCALL and APPLY can evaluate parameter dynamically, but i don't know how to call FUNCALL DEFUN... correctly.
I don't think you can use nested defuns.
You can either use a defun to return a lambda:
(defun custom (op const)
(lambda (arg) (funcall op const arg)))
and then use fdefinition:
(setf (fdefinition '+42) (custom '+ '42))
or use defmacro:
(defmacro custom (name op const)
(let ((arg (gensym)))
`(defun ,name (,arg)
(,op ,const ,arg))))
(custom +42 + 42)
PS. I think you need to explain why you are trying to do this, then we will be able to explain your options better.
It seems to me you might want to curry a function. Imagine you did this:
(defun curry (op arg1)
(lambda (&rest args) (apply op (cons arg1 args))))
(funcall (curry #'+ 10) 20) ; ==> 30
(mapcar (curry #'+ 10) '(1 2 3 4)) ; ==> (11 12 13 14)
Now defun makes a function in the global namespace always. It's not like in Scheme where it creates a closure. To do the same as defun we use symbol-function and setf:
(defun create-curried (name op arg1)
(setf (symbol-function name)
(lambda (&rest args) (apply op (cons arg1 args)))))
(create-curried '+x #'+ 10) ; ==> function
(+x 20) ; ==> 30
;; since it's a function, it even works with higher order functions
(mapcar create-curried '(+x -x /x *x) (list #'+ #'- #'/ #'*) '(10 10 10 10))
(/x 2) ; ==> 5
Last. With macros you can prettify it.
(defmacro defun-curried (newname oldname arg)
(if (and (symbolp newname) (symbolp oldname))
`(create-curried ',newname (function ,oldname) ,arg)
(error "Newname and Oldname need to be symbols")))
(defun-curried +xx + 20)
(+xx 10) ; ==> 30
oldname is taken from the lexical scope so you may use flet or labels with this but it ends up being global, just like with defun.
The main problem is that DEFUN isn't a function, it's a macro that treats its arguments specially (specifically, it doesn't evaluate "the function name", "the argument list", not "the function body").
You could probably make something work by careful use of (setf (symbol-function ...) ...), something like:
(defun define-custom-function (name op const)
(setf (symbol-function name) (lambda (var) (funcall op const var))))
This binds the function definition of the symbol to an anonymous function that calls your "operator" on your fed-in constant.
* (defun define-custom-function (name op const)
(setf (symbol-function name) (lambda (var) (funcall op const var))))
DEFINE-CUSTOM-FUNCTION
* (define-custom-function 'add3 #'+ 3)
#<CLOSURE (LAMBDA (VAR) :IN DEFINE-CUSTOM-FUNCTION) {1002A4760B}>
* (add3 5)
8
However, unless you absolutely need to define (or re-define) these custom functions dynamically, you are probably better off defining a custom DEFUN-like macro.
I'm working on a function that takes in a list of structures and then using that list of structures produces a function that processes a list of symbols into a number. Each structure is made up of a symbol, that will be in the second list consumed, and a number. This function produced has to turn the list of symbols into a number by assigning each symbol a value based on the previous structures. Using abstract list functions btw.
Example: ((function (list (make-value 'value1 10) (make-value 'value2 20)))
(list 'value1 'value2 'nothing 'value1)) would produced 40.
Heres my code but it only works for specific cases.
(define (function lst)
(lambda (x) (foldr + 0 (map (lambda (x)
(cond
[(equal? x (value-name(first lst)))(value-value (first lst))]
[else (value-value (second lst))]))
(filter (lambda (x) (member? x (map value-name lst)))x)))))
Looks like a homework. Basic shape of your solution is ok. I think the reason you have a problem here is that there is no decomposition in your code so it's easy to get lost in parentheses.
Let's start with your idea of fold-ing with + over list of integers as a last step of computation.
For this subtask you have:
1) a list of (name, value) pairs
2) a list of names
and you need to get a list of values. Write a separate function which does exactly that and use it. Like this
(define (function lst)
(lambda (x) (foldr +
0
(to-values x lst)))
(define (to-values names names-to-values)
(map (lambda (name)
(to-value name names-to-values))))
(define (to-value n ns-to-vs)
...)
Here we map over the names with another little function. It will lookup the n value in ns-to-vs and return it or 0 if there is no one.
There are two approaches for solving the problem with foldr, it'd be interesting to study and understand both of them. The first one, attempted in the question, is to first produce a list with all the values and let foldr take care of adding them. It can be implemented in a simpler way like this:
(define (function lst)
(lambda (x)
(foldr +
0
(map (lambda (e)
(cond ((assoc e lst) => value-value)
(else 0)))
x))))
Alternatively: maybe using foldr is overkill, applying + is simpler:
(define (function lst)
(lambda (x)
(apply +
(map (lambda (e)
(cond ((assoc e lst) => value-value)
(else 0)))
x))))
In the second approach we take the input list "as is" and let foldr's lambda perform the addition logic. This is more efficient than the first approach using foldr, because there's no need to create an intermediate list - the one generated by map in the first version:
(define (function lst)
(lambda (x)
(foldr (lambda (e a)
(cond ((assoc e lst) => (lambda (p) (+ a (value-value p))))
(else a)))
0
x)))
In both approaches I'm using assoc for finding the element in the list; it's easy to implement as a helper function if you're not allowed to use it or if it doesn't work for the values created with make-value: assoc takes a list of name-value pairs and returns the first pair with the given name. The => syntax of cond passes the pair returned by assoc to a lambda's parameter and executes it.
And because you're using Racket, there's a bit of syntactic sugar that can be used for returning a function from another function, try this equivalent code, for simplicity's sake:
(define ((function lst) x)
(foldr +
0
(map (lambda (e)
(cond ((assoc e lst) => value-value)
(else 0)))
x)))
Or this:
(define ((function lst) x)
(foldr (lambda (e a)
(cond ((assoc e lst) => (lambda (p) (+ a (value-value p))))
(else a)))
0
x))
Anyway, the result is as expected:
((function (list (make-value 'value1 10) (make-value 'value2 20)))
(list 'value1 'value2 'nothing 'value1))
=> 40
Say I have a function that takes a list and does something:
(defun foo(aList)
(loop for element in aList ...))
But if the list is nested I want to flatten it first before the loop does stuff, so I want to use another function (defun flatten(aList)) that flattens any list:
(defun foo(flatten(aList))
(loop for element in aList ...))
Lisp doesn't like this. Is there another direct way around this?
Here's one way:
(defun foo (alist)
(loop for element in (flatten alist) ...)
You can pass the function as an &optional argument.
(defun foo (alist &optional fn)
(if (not (null fn))
(setf alist (funcall fn alist)))
(dostuff alist))
A sample run where dostuff just print its argument:
(foo '(1 2 (3)))
=> (1 2 (3))
(foo '(1 2 (3)) #'flatten)
=> (1 2 3)
This approach is more flexible as you are not tied to just one 'pre-processor' function.