I have following code which confuse me now, I hope some can tell me the difference and how to fix this.
(defmacro tm(a)
`(concat ,(symbol-name a)))
(defun tf(a)
(list (quote concat) (symbol-name a)))
I just think they should be the same effect, but actually they seem not.
I try to following call:
CL-USER> (tf 'foo)
(CONCAT "FOO")
CL-USER> (tm 'foo)
value 'FOO is not of the expected type SYMBOL.
[Condition of type TYPE-ERROR]
So, what's the problem?
What i want is:
(tm 'foo) ==> (CONCAT "FOO")
The first problem is that 'foo is expanded by the reader to (quote foo), which is not a symbol, but a list. The macro tries to expand (tm (quote foo)). The list (quote foo) is passed as the parameter a to the macro expansion function, which tries to get its symbol-name. A list is not a valid argument for symbol-name. Therefore, your macro expansion fails.
The second problem is that while (tm foo) (note: no quote) does expand to (concat "FOO"), this form will then be executed by the REPL, so that this is also not the same as your tf function. This is not surprising, of course, because macros do different things than functions.
First, note that
`(concat ,(symbol-name a))
and
(list (quote concat) (symbol-name a))
do the exact same thing. They are equivalent pieces of code (backquote syntax isn't restricted to macro bodies!): Both construct a list whose first element is the symbol CONCAT and whose second element is the symbol name of whatever the variable A refers to.
Clearly, this only makes sense if A refers to a symbol, which, as Svante has pointed out, isn't the case in the macro call example.
You could, of course, extract the symbol from the list (QUOTE FOO), but that prevents you from calling the macro like this:
(let ((x 'foo))
(tm x))
which raises the question of why you would event want to force the user of the macro to explicitly quote the symbol where it needs to be a literal constant anyway.
Second, the way macros work is this: They take pieces of code (such as (QUOTE FOO)) as arguments and produce a new piece of code that, upon macroexpansion, (more or less) replaces the macro call in the source code. It is often useful to reuse macro arguments within the generated code by putting them where they are going to be evaluated later, such as in
(defmacro tm2 (a)
`(print (symbol-name ,a)))
Think about what this piece of code does and whether or not my let example above works now. That should get you on the right track.
Finally, a piece of advice: Avoid macros when a function will do. It will make life much easier for both the implementer and the user.
Related
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 Common Lisp, a macro definition must have been seen before the first use. This allows a macro to refer to itself, but does not allow two macros to refer to each other. The restriction is slightly awkward, but understandable; it makes the macro system quite a bit easier to implement, and to understand how the implementation works.
Is there any Lisp family language in which two macros can refer to each other?
What is a macro?
A macro is just a function which is called on code rather than data.
E.g., when you write
(defmacro report (x)
(let ((var (gensym "REPORT-")))
`(let ((,var ,x))
(format t "~&~S=<~S>~%" ',x ,var)
,var)))
you are actually defining a function which looks something like
(defun macro-report (system::<macro-form> system::<env-arg>)
(declare (cons system::<macro-form>))
(declare (ignore system::<env-arg>))
(if (not (system::list-length-in-bounds-p system::<macro-form> 2 2 nil))
(system::macro-call-error system::<macro-form>)
(let* ((x (cadr system::<macro-form>)))
(block report
(let ((var (gensym "REPORT-")))
`(let ((,var ,x)) (format t "~&~s=<~s>~%" ',x ,var) ,var))))))
I.e., when you write, say,
(report (! 12))
lisp actually passes the form (! 12) as the 1st argument to macro-report which transforms it into:
(LET ((#:REPORT-2836 (! 12)))
(FORMAT T "~&~S=<~S>~%" '(! 12) #:REPORT-2836)
#:REPORT-2836)
and only then evaluates it to print (! 12)=<479001600> and return 479001600.
Recursion in macros
There is a difference whether a macro calls itself in implementation or in expansion.
E.g., a possible implementation of the macro and is:
(defmacro my-and (&rest args)
(cond ((null args) T)
((null (cdr args)) (car args))
(t
`(if ,(car args)
(my-and ,#(cdr args))
nil))))
Note that it may expand into itself:
(macroexpand '(my-and x y z))
==> (IF X (MY-AND Y Z) NIL) ; T
As you can see, the macroexpansion contains the macro being defined.
This is not a problem, e.g., (my-and 1 2 3) correctly evaluates to 3.
However, if we try to implement a macro using itself, e.g.,
(defmacro bad-macro (code)
(1+ (bad-macro code)))
you will get an error (a stack overflow or undefined function or ...) when you try to use it, depending on the implementation.
Here's why mutually recursive macros can't work in any useful way.
Consider what a system which wants to evaluate (or compile) Lisp code for a slightly simpler Lisp than CL (so I'm avoiding some of the subtleties that happen in CL), such as the definition of a function, needs to do. It has a very small number of things it knows how to do:
it knows how to call functions;
it knows how to evaluate a few sorts of literal objects;
it has some special rules for a few sorts of forms – what CL calls 'special forms', which (again in CL-speak) are forms whose car is a special operator;
finally it knows how to look to see whether forms correspond to functions which it can call to transform the code it is trying to evaluate or compile – some of these functions are predefined but additional ones can be defined.
So the way the evaluator works is by walking over the thing it needs to evaluate looking for these source-code-transforming things, aka macros (the last case), calling their functions and then recursing on the results until it ends up with code which has none left. What's left should consist only of instances of the first three cases, which it then knows how to deal with.
So now think about what the evaluator has to do if it is evaluating the definition of the function corresponding to a macro, called a. In Cl-speak it is evaluating or compiling a's macro function (which you can get at via (macro-function 'a) in CL). Let's assume that at some point there is a form (b ...) in this code, and that b is known also to correspond to a macro.
So at some point it comes to (b ...), and it knows that in order to do this it needs to call b's macro function. It binds suitable arguments and now it needs to evaluate the definition of the body of that function ...
... and when it does this it comes across an expression like (a ...). What should it do? It needs to call a's macro function, but it can't, because it doesn't yet know what it is, because it's in the middle of working that out: it could start trying to work it out again, but this is just a loop: it's not going to get anywhere where it hasn't already been.
Well, there's a horrible trick you could do to avoid this. The infinite regress above happens because the evaluator is trying to expand all of the macros ahead of time, and so there's no base to the recursion. But let's assume that the definition of a's macro function has code which looks like this:
(if <something>
(b ...)
<something not involving b>)
Rather than doing the expand-all-the-macros-first trick, what you could do is to expand only the macros you need, just before you need their results. And if <something> turned out always to be false, then you never need to expand (b ...), so you never get into this vicious loop: the recursion bottoms out.
But this means you must always expand macros on demand: you can never do it ahead of time, and because macros expand to source code you can never compile. In other words a strategy like this is not compatible with compilation. It also means that if <something> ever turns out to be true then you'll end up in the infinite regress again.
Note that this is completely different to macros which expand to code which involves the same macro, or another macro which expands into code which uses it. Here's a definition of a macro called et which does that (it doesn't need to do this of course, this is just to see it happen):
(defmacro et (&rest forms)
(if (null forms)
't
`(et1 ,(first forms) ,(rest forms))))
(defmacro et1 (form more)
(let ((rn (make-symbol "R")))
`(let ((,rn ,form))
(if ,rn
,rn
(et ,#more)))))
Now (et a b c) expands to (et1 a (b c)) which expands to (let ((#:r a)) (if #:r #:r (et b c))) (where all the uninterned things are the same thing) and so on until you get
(let ((#:r a))
(if #:r
#:r
(let ((#:r b))
(if #:r
#:r
(let ((#:r c))
(if #:r
#:r
t))))))
Where now not all the uninterned symbols are the same
And with a plausible macro for let (let is in fact a special operator in CL) this can get turned even further into
((lambda (#:r)
(if #:r
#:r
((lambda (#:r)
(if #:r
#:r
((lambda (#:r)
(if #:r
#:r
t))
c)))
b)))
a)
And this is an example of 'things the system knows how to deal with': all that's left here is variables, lambda, a primitive conditional and function calls.
One of the nice things about CL is that, although there is a lot of useful sugar, you can still poke around in the guts of things if you like. And in particular, you still see that macros are just functions that transform source code. The following does exactly what the defmacro versions do (not quite: defmacro does the necessary cleverness to make sure the macros are available early enough: I'd need to use eval-when to do that with the below):
(setf (macro-function 'et)
(lambda (expression environment)
(declare (ignore environment))
(let ((forms (rest expression)))
(if (null forms)
't
`(et1 ,(first forms) ,(rest forms))))))
(setf (macro-function 'et1)
(lambda (expression environment)
(declare (ignore environment))
(destructuring-bind (_ form more) expression
(declare (ignore _))
(let ((rn (make-symbol "R")))
`(let ((,rn ,form))
(if ,rn
,rn
(et ,#more)))))))
There have been historic Lisp systems that allow this, at least in interpreted code.
We can allow a macro to use itself for its own definition, or two or more macros to mutually use each other, if we follow an extremely late expansion strategy.
That is to say, our macro system expands a macro call just before it is evaluated (and does that each time that same expression is evaluated).
(Such a macro expansion strategy is good for interactive development with macros. If you fix a buggy macro, then all code depending on it automatically benefits from the change, without having to be re-processed in any way.)
Under such a macro system, suppose we have a conditional like this:
(if (condition)
(macro1 ...)
(macro2 ...))
When (condition) is evaluated, then if it yields true, (macro1 ...) is evaluated, otherwise (macro2 ...). But evaluation also means expansion. Thus only one of these two macros is expanded.
This is the key to why mutual references among macros can work: we are able rely on the conditional logic to give us not only conditional evaluation, but conditional expansion also, which then allows the recursion to have ways of terminating.
For example, suppose macro A's body of code is defined with the help of macro B, and vice versa. And when a particular invocation of A is executed, it happens to hit the particular case that requires B, and so that B call is expanded by invocation of macro B. B also hits the code case that depends on A, and so it recurses into A to obtain the needed expansion. But, this time, A is called in a way that avoids requiring, again, an expansion of B; it avoids evaluating any sub-expression containing the B macro. Thus, it calculates the expansion, and returns it to B, which then calculates its expansion returns to the outermost A. A finally expands and the recursion terminates; all is well.
What blocks macros from using each other is the unconditional expansion strategy: the strategy of fully expanding entire top-level forms after they are read, so that the definitions of functions and macros contain only expanded code. In that situation there is no possibility of conditional expansion that would allow for the recursion to terminate.
Note, by the way, that a macro system which expands late doesn't recursively expand macros in a macro expansion. Suppose (mac1 x y) expands into (if x (mac2 y) (mac3 y)). Well, that's all the expansion that is done for now: the if that pops out is not a macro, so expansion stops, and evaluation proceeds. If x yields true, then mac2 is expanded, and mac3 is not.
I've written a macro that works as intended. The problem is that it contains an eval. I'd like to get rid of it but try as I might, I can't find the correct combination of backquotes and commas to do so.
(defmacro mymacro (x &body body)
`(myothermacro ,(fun1 (eval x))
,#body))
Here myothermacro is a macro and fun1 is a function.
Here is the desired behaviour:
(defvar v 88)
(defun fun1 (z) (1+ z))
(defmacro mymacro (x &body body)
`(myothermacro ,(fun1 (eval x))
,#body))
(macroexpand-1 '(mymacro v 42 43 44))
=> (MYOTHERMACRO 89 42 43 44)
There is no amount of backquotes that can help you here. In a scenario you have more than one backquote there are equally as many quasiquote and thus you can get different layers of quoted data but not data evaluated more than once.
It's important to understand that a macro does not do anything runtime. Thus if you are to use a macro eg (mymacro variable (my-function x)) the macro function mymacro is fed variable and (my-function x) right away and the result in put in place. variable might not exist yet so evaluating it would be premature. When you define a function that uses the macro it will most likely expand the macros before storing the function. When in runtime there are no macros because they are all expanded, but this is the very first time it's possible to make conclusions if the arguments passed to the macro and its expansion actually makes sense according to the lexical environment and global bindings.
Perhaps if you added more information would there be a way to help you solve your actual problem since I get a feeling this is a XY problem.
You need to use use read-time evaluation if I got your idea right.
Something like this:
(eval-when (:compile-toplevel :load-toplevel :execute)
(defparameter *foo* "foo"))
(defmacro bar (arg)
`(list #.*foo* ,arg))
CL-USER> (macroexpand '(bar "bar"))
(LIST "foo" "bar")
You may be better with defconstant instead of defparameter because it's clearer about intentions.
The Lisp forum thread Define macro alias? has an example of creating function alias using a form such as
(setf (symbol-function 'zero?) #'zerop)
This works fine, making zero? a valid predicate. Is it possible to parametrize this form without resorting to macros? I'd like to be able to call the following and have it create function?:
(define-predicate-alias 'functionp)`
My take was approximately:
(defun defalias (old new)
(setf (symbol-function (make-symbol new))
(symbol-function old)))
(defun define-predicate-alias (predicate-function-name)
(let ((alias (format nil "~A?" (string-right-trim "-pP" predicate-function-name))))
(defalias predicate-function-name alias)))
(define-predicate-alias 'zerop)
(zero? '())
This fails when trying to call zero? saying
The function COMMON-LISP-USER::ZERO? is undefined.
make-symbol creates an uninterned symbol. That's why zero? is undefined.
Replace your (make-symbol new) with e.g. (intern new *package*). (Or you may want to think more carefully in which package to intern your new symbol.)
Your code makes a symbol, via MAKE-SYMBOL, but you don't put it into a package.
Use the function INTERN to add a symbol to a package.
To expand on Lars' answer, choose the right package. In this case the default might be to use the same package from the aliased function:
About style:
Anything that begins with DEF should actually be a macro. If you have a function, don't use a name beginning with "DEF". If you look at the Common Lisp language, all those are macro. For example: With those defining forms, one would typically expect that they have a side-effect during compilation of files: the compiler gets informed about them. A function can't.
If I put something like this in a file
(define-predicate-alias zerop)
(zero? '())
and then compile the file, I would expect to not see any warnings about an undefined ZERO?. Thus a macro needs to expand (define-predicate-alias 'zerop) into something which makes the new ZERO? known into the compile-time environment.
I would also make the new name the first argument.
Thus use something like MAKE-PREDICATE-ALIAS instead of DEFINE-PREDICATE-ALIAS, for the function.
There are already some answers that explain how you can do this, but I'd point out:
Naming conventions, P, and -P
Common Lisp has a naming convention that is mostly adhered to (there are exceptions, even in the standard library), that if a type name is multiple words (contains a -), then its predicate is named with -P suffix, whereas if it doesn't, the suffix is just P. So we'd have keyboardp and lcd-monitor-p. It's good then, that you're using (string-right-trim "-pP" predicate-function-name)), but since the …P and …-P names in the standard, and those generated by, e.g., defstruct, will be using P, not p, you might just use (string-right-trim "-P" predicate-function-name)). Of course, even this has the possible issues with some names (e.g., pop), but I guess that just comes with the territory.
Symbol names, format, and *print-case*
More importantly, using format to create symbol names for subsequent interning is dangerous, because format doesn't always print a symbol's name with the characters in the same case that they actually appear in its name. E.g.,
(let ((*print-case* :downcase))
(list (intern (symbol-name 'foo))
(intern (format nil "~A" 'foo))))
;=> (FOO |foo|) ; first symbol has name "FOO", second has name "foo"
You may be better off using string concatenation and extracting symbol names directly. This means you could write code like (this is slightly different use case, since the other questions already explain how you can do what you're trying to do):
(defmacro defpredicate (symbol)
(flet ((predicate-name (symbol)
(let* ((name (symbol-name symbol))
(suffix (if (find #\- name) "-P" "P")))
(intern (concatenate 'string name suffix)))))
`(defun ,(predicate-name symbol) (x)
(typep x ',symbol)))) ; however you're checking the type
(macroexpand-1 '(defpredicate zero))
;=> (DEFUN ZEROP (X) (TYPEP X 'ZERO))
(macroexpand-1 '(defpredicate lcd-monitor))
;=> (DEFUN LCD-MONITOR-P (X) (TYPEP X 'LCD-MONITOR))
What is the best practice for selectively passing evaluated arguments to a macro form?
To elaborate: The usefulness of macros lies in its ability to receives unevaluated parameter, unlike the default evaluation rule for function forms. However, there is a legitimate use cases for evaluating macro arguments.
Consider a contrived example:
(defparameter *func-body* '((print i) (+ i 1)))
Suppose it would be nice that *func-body* could serve as the body of a macro our-defun that is defined as:
(defmacro our-defun (fun args &body body)
`(defun ,fun ,args ,#body))
So after (our-defun foo (i) (+ 1 i)), we could say (foo 1) to get 2. However, if we use (our-defun foo (i) *func-body*), the result of (foo 1) will be ((PRINT I) (+ I 1)) (i.e., the value of *func-body*). It would be nice if we can force the evaluation of *func-body* as an argument to the macro our-defun.
Currently, I can think of a technique of using compile and funcall to do this, as in
(funcall (compile nil `(lambda () (our-defun foo (i) ,#*func-body*))))
after which (our-defun 1) will print out 1 and return 2, as intended. I can think of case of making this work with eval, but I would rather stay away from eval due to its peculiarity in scoping.
This leads to my question at the begining, is there a more straightforward or native way to do this?
P.S.,
A not-so-contrived example is in the function (UPDATE-HOOK), which uses two library macros (ADD-HOOK) and (REMOVE-HOOK) and needs to evaluate its parameters. The (funcall (compile nil `(lambda () ...))) technique above is used here.
(defun update-hook (hook hook-name &optional code)
(funcall (compile nil `(lambda () (remove-hook ,hook ',hook-name))))
(unless (null code)
(compile hook-name `(lambda () ,#code))
(funcall (compile nil `(lambda () (add-hook ,hook ',hook-name))))))
That's slightly confused. A macro does not receive unevaluated parameters.
A macro gets source code and creates source code from that. Remember also that source code in Lisp is actually provided as data. The macro creates code, which evaluates some forms and some not.
Macros need to work in a compiling system. Before runtime. During compile time. All the macro sees is source code and then it creates source code from that. Think of macros as code transformations, not about evaluating arguments or not.
It would be nice if we can force the evaluation of *func-body* as an argument to the macro our-defun
That is not very clean. In a compiled system, you would need to make sure that *func-body* actually has a useful binding and that it can be resolved at COMPILE TIME.
If you have a macro like DEFUN, it makes sense to have the source code static. If you want to insert some source code into a form, then it could make sense to do that at read time:
(defun foo (i) #.`(,#*foo*))
But that's code I usually would want to avoid.
two library macros (ADD-HOOK) and (REMOVE-HOOK) and needs to evaluate its parameters.
Why should ADD-HOOK and REMOVE-HOOK be macros? If you don't have a real reason, they simply should be functions. Already since they make reuse difficult.
If you want to make ADD-HOOK and REMOVE-HOOK macros for some reason, then UPDATE-HOOK usually should be a macro, too.
The list you are giving to your macro has the form
(Quote (...))
So the list you actually want is the CADR of the list you get.