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))
Related
I'm learning common lisp. I have written a version of the once-only macro, which suffers from an unusual variable capture problem.
My macro is this:
(defmacro my-once-only (names &body body)
(let ((syms (mapcar #'(lambda (x) (gensym))
names)))
``(let (,,#(mapcar #'(lambda (sym name) ``(,',sym ,,name))
syms names))
,(let (,#(mapcar #'(lambda (name sym) `(,name ',sym))
names syms))
,#body))))
The canonical version of only-once is this:
(defmacro once-only ((&rest names) &body body)
(let ((gensyms (loop for n in names collect (gensym))))
`(let (,#(loop for g in gensyms collect `(,g (gensym))))
`(let (,,#(loop for g in gensyms for n in names collect ``(,,g ,,n)))
,(let (,#(loop for n in names for g in gensyms collect `(,n ,g)))
,#body)))))
The difference, as far as I can tell, is that the canonical version generates new symbols for every expansion of the macro using only-once. For example:
CL-USER> (macroexpand-1 '(once-only (foo) foo))
(LET ((#:G824 (GENSYM)))
`(LET (,`(,#:G824 ,FOO))
,(LET ((FOO #:G824))
FOO)))
T
CL-USER> (macroexpand-1 '(my-once-only (foo) foo))
`(LET (,`(,'#:G825 ,FOO))
,(LET ((FOO '#:G825))
FOO))
T
The variable my macro uses to store the value of foo is the same for every expansion of this form, in this case it would be #:G825. This is akin to defining a macro like the following:
(defmacro identity-except-for-bar (foo)
`(let ((bar 2))
,foo))
This macro captures bar, and this capture manifests when bar is passed to it, like so:
CL-USER> (let ((bar 1))
(identity-except-for-bar bar))
2
However, I cannot think of any way to pass #:G825 to a macro that uses my-only-once so that it breaks like this, because the symbols gensym returns are unique, and I cannot create a second copy of it outside of the macro. I assume that capturing it is unwanted, otherwise the canonical version wouldn't bother adding the additional layer of gensym. How could capturing a symbol like #:G826 be a problem? Please provide an example where this capture manifests.
We can demonstrate a behavioral difference between my-once-only and once-only:
Let's store our test form in a variable.
(defvar *form* '(lexalias a 0 (lexalias b (1+ a) (list a b))))
This test form exercises a macro called lexalias, which we will define in two ways. First with once-only:
(defmacro lexalias (var value &body body)
(once-only (value)
`(symbol-macrolet ((,var ,value))
,#body)))
(eval *form*) -> (0 1)
Then with my-once-only:
(defmacro lexalias (var value &body body)
(my-once-only (value)
`(symbol-macrolet ((,var ,value))
,#body)))
(eval *form*) -> (1 1)
Oops! The problem is that under my-once-only, both a and b end up being symbol-macrolet aliases for exactly the same gensym; the returned expression (list a b) ends up being something like (list #:g0025 #:g0025).
If you're writing a macro-writing helper that implements once-only evaluation, you have no idea how the symbol is going to be used by the code which calls the macro, whose author uses your once-only tool. There are two big unknowns: the nature of the macro and of its use.
As you can see, if you don't make fresh gensyms, it will not work correctly in all conceivable scenarios.
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))
Assume that the macro would take the boolean types a and b . If a is nil, then the macro should return nil (without ever evaluating b), otherwise it returns b. How do you do this?
This really depends on what you can use.
E.g., is or available? if? cond?
Here is one example:
(defmacro and (a b)
`(if ,a ,b nil)
EDIT. In response to a comment, or is more complicated because we have to avoid double evaluation:
(defmacro or (a b)
(let ((v (gensym "OR")))
`(let ((,v ,a))
(if ,v ,v ,b))))
sds's answer is nice and concise, but it has two limitations:
It only works with two arguments, whereas the built in and and or take any number of arguments. It's not too hard to update the solution to take any number of arguments, but it would be a bit more complicated.
More importantly, it's based very directly in terms of delayed operations that are already present in the language. I.e., it takes advantage of the fact that if doesn't evaluate the then or else parts until it has first evaluated the condition.
It might be a good exercise, then, to note that when a macro needs to delay evaluation of some forms, it's often the simplest strategy (in terms of implementation, but not necessarily the most efficient) to use a macro that expands to a function call that takes a function. For instance, a naive implementation of with-open-file might be:
(defun %call-with-open-file (pathname function)
(funcall function (open pathname)))
(defmacro my-with-open-file ((var pathname) &body body)
`(%call-with-open-file
,pathname
(lambda (,var)
,#body)))
Using a technique like this, you can easily get a binary and (and or):
(defun %and (a b)
(if (funcall a)
(funcall b)
nil))
(defmacro my-and (a b)
`(%and (lambda () ,a)
(lambda () ,b)))
CL-USER> (my-and t (print "hello"))
"hello" ; printed output
"hello" ; return value
CL-USER> (my-and nil (print "hello"))
NIL
or is similar:
(defun %or (a b)
(let ((aa (funcall a)))
(if aa
aa
(funcall b))))
(defmacro my-or (a b)
`(%or (lambda () ,a)
(lambda () ,b)))
To handle the n-ary case (since and and or actually take any number of arguments), you could write a function that takes a list of lambda functions and calls each of them until you get to one that would short circuit (or else reaches the end). Common Lisp actually already has functions like that: every and some. With this approach, you could implement and in terms of every by wrapping all the arguments in lambda functions:
(defmacro my-and (&rest args)
`(every #'funcall
(list ,#(mapcar #'(lambda (form)
`(lambda () ,form))
args))))
For instance, with this implementation,
(my-and (listp '()) (evenp 3) (null 'x))
expands to:
(EVERY #'FUNCALL
(LIST (LAMBDA () (LISTP 'NIL))
(LAMBDA () (EVENP 3))
(LAMBDA () (NULL 'X))))
Since all the forms are now wrapped in lambda functions, they won't get called until every gets that far.
The only difference is that and is specially defined to return the value of the last argument if all the preceding ones are true (e.g., (and t t 3) returns 3, not t, whereas the specific return value of every is not specified (except that it would be a true value).
With this approach, implementing or (using some) is no more complicated than implementing and:
(defmacro my-or (&rest args)
`(some #'funcall ,#(mapcar #'(lambda (form)
`(lambda () ,form))
args)))
I store some macros in quoted form (because in fact they produce lambdas with tricky lexical environment and I prefer store and serialize them as lists). So now I'm trying:
(defun play (s)
(funcall (macroexpand s)))
Macroexpand evaluates quoted lambda, so funcall can't run it. How to unquote result of macroexpand without eval? Because in my case it would cause indefensible security hole.
MORE INFO:
What I get look like this (in simplest case):
FUNCALL: #1=#'(LAMBDA (#:G6008) (SYMBOL-MACROLET NIL T)) is not a function name; try using a symbol instead
and symbol-macrolet is what actually builds up "tricky lexical environment" inside lambda.
Macroexpand evaluates quoted lambda, so funcall can't run it. How to
unquote result of macroexpand without eval? Because in my case it
would cause indefensible security hole.
I think that Sylwester's comment about the XY problem is probably right here; it sounds like you're trying to do something that might be done better in a different way. That said, if you have a list that's a lambda expression, you can use coerce to get a function object instead of using eval. That is, you can do this:
CL-USER> (funcall '(lambda () 42))
; Error, like you've been having
CL-USER> (funcall (coerce '(lambda () 42) 'function))
42 ; turned the list (lambda () 42) into a function and called it
This is described in the documentation for coerce; when the "output" type is function, this is what happens with the object argument:
If the result-type is function, and object is any function name that
is fbound but that is globally defined neither as a macro name nor as
a special operator, then the result is the functional value of object.
If the result-type is function, and object is a lambda expression,
then the result is a closure of object in the null lexical
environment.
Thus, if you have a function that returns list of the form (lambda ...), you can use coerce and funcall with its result. This includes macroexpansions, although you may want to use macroexpand-1 rather than macroexpand, because lambda is already a macro, so if you expand too far, (lambda () ...) turns into (function (lambda () ...)).
CL-USER> (defmacro my-constantly (value)
`(lambda () ,value))
MY-CONSTANTLY
CL-USER> (macroexpand-1 '(my-constantly 36))
(LAMBDA () 36)
T
CL-USER> (funcall (coerce (macroexpand-1 '(my-constantly 36)) 'function))
36
If you try that with the plain macroexpand, though, there's a problem. Consider yourself warned:
CL-USER> (macroexpand '(my-constantly 36))
#'(LAMBDA () 36) ; not a list, but a function
T
CL-USER> (funcall (coerce (macroexpand '(my-constantly 36)) 'function))
; Error. :(
I think this is a case where you will find that the REPL is your friend. To get you started:
cl-user> (defmacro foo () #'(lambda () :hi))
foo
cl-user> (foo)
#<Compiled-function (:internal foo) (Non-Global) #x3020014F82FF>
cl-user> (funcall *)
:hi
cl-user> (macroexpand '(foo))
#<Compiled-function (:internal foo) (Non-Global) #x3020014F82FF>
t
cl-user> (funcall *)
:hi
cl-user>
I'll note in passing that the lambda form that appears in your example takes an argument, while your funcall doesn't provide one.
There are several issues with this. First I think this is a XY problem so if you show more of your problem I guess we might find a solution for it.
It has nothing with unquoting since when evaluation a quoted expression it's not longer quoted, but it's turn into data representation of the original quoted expression. You must eval data if you want to run it.
With eval you won't get the current lexical scope. So since you mention eval and lexical environment in the same post makes me thing you won't get what you want.
Now it's no problem making a list of functions, like this:
(list (lambda (x) (+ x x))
This is not quoted since I use list and teh lambda is evaluated to a closure. If it was assigned to variable x you could call it with (funcall (car x) 10)) ; ==> 20
EDIT
I actually made the same error message with this code:
(defmacro test () '#'(lambda (x) x))
(macroexpand '(test)) ; ==> #'(lambda (x) x) ; t
(funcall (macroexpand '(test)) 5) ; ==>
*** - funcall: #'(lambda (x) x) is not a function name; try using a symbol
instead
It doesn't work since you cannot call a lambda unevaluated. You need to call the closure (function) which is the result of the evaluation of the lambda form. If you would instead not quote it you'll have that evaluation in the macro:
(defmacro test () #'(lambda (x) x))
(macroexpand '(test)) ; ==> #<function :lambda (x) x> ;
(funcall (macroexpand '(test)) 5) ; ==> 5
Actually I cannot see why you would need to make this a macro. Lets make it a function instead.
(defun test () #'(lambda (x) x))
(funcall (test) 5 ) ; ==> 5
Even if this was more comples in most cases you would do with a closure.
One of the exercises in Paul Grahams ANSI Common Lisp book is this: Define a macro that takes a list of variables and a body of code, and ensures that the variables revert to their original values after the body of code is evaluated.
The problem I'm having with this exercise is how to save the symbol-names of the input variables. Below I have a start where I only save the values that the symbols are bound to.
(defmacro save-run (varlist &body body)
`(let ((valuelist (list ,#varlist)))
(format t "valuelist: ~A" valuelist)))
(let ((a 5)(b 6))
(values '(a b))
(save-run (a b)
(setf a 7)
(setf b 8)))
[507]> valuelist: (5 6)
Edit: Here's a solution where the variables are saved and then restored (using tips from finnw below). But shadowing the variables as in Vatine's answer is probably more elegant.
(defmacro save-run (varlist &body body)
`(let ((valuelist (list ,#varlist)))
,#body
(multiple-value-setq ,varlist (values-list valuelist))))
To get the list of symbols, all you need to do is quote the contents of VARLIST:
(defmacro save-run (varlist &body body)
`(let ((namelist ',varlist)
(valuelist (list ,#varlist)))
(format t "namelist: ~A~%" namelist)
(format t "valuelist: ~A~%" valuelist)))
I suspect this will not be useful in the final defnition however. There is not much you can do with the list of symbols at run-time. Instead, look for a good place to insert the list in the macro expansion.
Also you might want to use a GENSYM instead of the hard-coded variable name VALUELIST:
(defmacro save-run (varlist &body body)
(let ((valuelist (gensym)))
`(let ((,valuelist (list ,#varlist)))
,#body
(setf (values ,#varlist) (values-list ,valuelist)))))
Personally, I'd introduce another binding layer for the variables we want to save.
You have the list of variables in varlist so something like this may work:
(defmacro save-run (varlist &body body)
`(let ,(loop for var in varlist
collect (list var var))
,#body))
I really liked Vatine's approach. Here is an implementation that expands to the same code, but uses mapcar instead of the loop macro:
(defmacro save-run (varlist &body body)
`(let ,(mapcar #'list varlist varlist)
,#body))