I have my own Lisp interpreter in JavaScript that I work for some time now, and now I want to implement reader macros like in Common Lisp.
I've created Streams (almost working except for special symbols like ,# , ` ') but it freezes the browser for a few seconds when it's loading the page with included scripts (lisp files that have 400 lines of code). This is because my Streams are based on substring function. If I first split tokens and then use TokenStream that iterate over tokens, it works fine.
So my question is this, is string streams really something that is in Common Lisp? Can you add reader macros that create whole new syntax like Python inside CL, this simplify to question can I implement """ macro (not sure if you can have 3 characters as reader macro) or other character that will implement template literal inside lisp for instance:
(let ((foo 10) (bar 20))
{lorem ipsum ${baz} and ${foo}})
or
(let ((foo 10) (bar 20))
""lorem ipsum ${baz} and ${foo}"")
or
(let ((foo 10) (bar 20))
:"lorem ipsum ${baz} and ${foo}")
would yield string
"lorem ipsum 10 and 20"
is something like this possible in Common Lisp and how hard would be to implement #\{ or #\: as reader macro?
The only way I can think of to have template literals in Lisp is something like this:
(let ((foo 10) (bar 20))
(tag "lorem ipsum ${baz} and ${foo}")))
where tag is macro that return strings with ${} as free variable. Can reader macro also return lisp code that is evaluated?
And another question can you implement reader macros like this:
(list :foo:bar)
(list foo:bar)
where : is reader macro and if it's before symbols it convert symbol to
foo.bar
and if it's inside it throw error. I'm asking this because with token based macros :foo:bar and foo:bar will be symbols and will not be processed by my reader macros.
and one more question can reader macro be put in one line and second line use it? This will definitely be only possible with string streams and from what I've tested not possible with interpreter written in JavaScript.
There are some limitations in the sense that it is pretty hard to, for instance, intervene in the interpretation of tokens in any way short of 'implement your own token interpreter from scratch'. But, well, you could if you wanted to do just that: the problem is that your code would need to deal with numbers & things as well as the existing code does and things like floating-point parsing are pretty fiddly to get right.
But the macro functions associated with macro characters get the stream that is being read, and they are free to read as much or as little of the stream as they like, and return any kind of object (or no object, which is how comments are implemented).
I would strongly recommend reading chapters 2 & 23 of the hyperspec, and then playing with an implementation. When you play with the implementation be aware that it is just astonishingly easy to completely wedge things by mucking around with the reader. At the minimum I would suggest code like this:
(defparameter *my-readtable* (copy-readtable nil))
;;; Now muck around with *my-readtable*, *not* the default readtable
;;;
(defun experimentally-read ((&key (stream *standard-input*)
(readtable *my-raedtable*)))
(let ((*readtable* readtable))
(read stream)))
This gives you at least some chance to recover from catastrophe: if you can once abort experimentally-read then you are back in a position where *readtable* is something sensible.
Here is a fairly useless example which shows how much you can subvert the syntax with macro characters: a macro character definition which will cause ( ...) to be read as a string. This may not be fully debugged, and as I say I can see no use for it.
(defun mindless-parenthesized-string-reader (stream open-paren)
;; Cause parenthesized groups to be read as strings:
;; - (a b) -> "a b"
;; - (a (b c) d) -> "a (b c) d"
;; - (a \) b) -> "a ) b"
;; This serves no useful purpose that I can see. Escapes (with #\))
;; and nested parens are dealt with.
;;
;; Real Programmers would write this with LOOP, but that was too
;; hard for me. This may well not be completely right.
(declare (ignore open-paren))
(labels ((collect-it (escaping depth accum)
(let ((char (read-char stream t nil t)))
(if escaping
(collect-it nil depth (cons char accum))
(case char
((#\\)
(collect-it t depth accum))
((#\()
(collect-it nil (1+ depth) (cons char accum)))
((#\))
(if (zerop depth)
(coerce (nreverse accum) 'string)
(collect-it nil (1- depth) (cons char accum))))
(otherwise
(collect-it nil depth (cons char accum))))))))
(collect-it nil 0 '())))
(defvar *my-readtable* (copy-readtable nil))
(set-macro-character #\( #'mindless-parenthesized-string-reader
nil *my-readtable*)
(defun test-my-rt (&optional (stream *standard-input*))
(let ((*readtable* *my-readtable*))
(read stream)))
And now
> (test-my-rt)
12
12
> (test-my-rt)
x
x
> (test-my-rt)
(a string (with some parens) and \) and the end)
"a string (with some parens) and ) and the end"
Related
Short version:
I want to change the #+ and #- reader macros to apply to all immediately subsequent tokens starting with ##, in addition to the following token. Therefore, the following code...
#+somefeature
##someattribute1
##someattribute2
(defun ...)
...would, in the absence of somefeature, result in no code.
Long version:
I have written my own readtable-macros which apply transformations to subsequent code. For example:
##traced
(defun ...)
This yields a function that writes its arguments and return values to a file, for debugging.
This fails, however, when used in conjunction with the #+ reader macro:
#+somefeature
##traced
(defun ...)
In the absence of somefeature, the function continues to be defined, albeit without the ##traced modification. This is obviously not the desired outcome.
One possible solution would be to use progn, as follows:
#+somefeature
(progn
##traced
(defun ...))
But that's kind of ugly.
I would like to modify the #+ and #- reader macros, such that they may consume more than one token. Something like this:
(defun conditional-syntax-reader (stream subchar arg)
; If the conditional fails, consume subsequent tokens while they
; start with ##, then consume the next token.
)
(setf *readtable* (copy-readtable))
(set-dispatch-macro-character #\# #\+ #'conditional-syntax-reader)
(set-dispatch-macro-character #\# #\- #'conditional-syntax-reader)
The problem is, I don't know how to "delegate" to the original reader macros; and I don't understand enough about how they were implemented to re-implement them myself in their entirety.
A naive approach would be:
(defun consume-tokens-recursively (stream)
(let ((token (read stream t nil t)))
(when (string= "##" (subseq (symbol-string token) 0 2))
(consume-tokens-recursively stream)))) ; recurse
(defun conditional-syntax-reader (stream subchar arg)
(unless (member (read stream t nil t) *features*)
(consume-tokens-recursively stream)))
However, I'm given to believe that this wouldn't be sufficient:
The #+ syntax operates by first reading the feature specification and then skipping over the form if the feature is false. This skipping of a form is a bit tricky because of the possibility of user-defined macro characters and side effects caused by the #. and #, constructions. It is accomplished by binding the variable read-suppress to a non-nil value and then calling the read function.
This seems to imply that I can just let ((*read-suppress* t)) when using read to solve the issue. Is that right?
EDIT 1
Upon further analysis, it seems the problem is caused by not knowing how many tokens to consume. Consider the following attributes:
##export expects one argument: the (defun ...) to export.
##traced expects two arguments: the debug level and the (defun ...) to trace.
Example:
#+somefeature
##export
##traced 3
(defun ...)
It turns out that #+ and #- are capable of suppressing all these tokens; but there is a huge problem!
When under a suppressing #+ or #-, (read) returns NIL!
Example:
(defun annotation-syntax-reader (stream subchar arg)
(case (read stream t nil t)
('export
(let ((defun-form (read stream t nil t)))))
; do something
('traced
(let* ((debug-level (read stream t nil t))
(defun-form (read stream t nil t)))))))
; do something
(setf *readtable* (copy-readtable))
(set-dispatch-macro-character #\# #\# #'annotation-syntax-reader)
#+(or) ##traced 3 (defun ...)
The ##traced token is being suppressed by the #+. In this situation, all the (read) calls in (annotation-syntax-reader) consume real tokens but return NIL!
Therefore, the traced token is consumed, but the case fails. No additional tokens are thus consumed; and control leaves the scope of the #+.
The (defun ...) clause is executed as normal, and the function comes into being. Clearly not the desired outcome.
The standard readtable
Changing the macros for #+ and #- is a bit excessive solution I think, but in any case remember to not actually change the standard readtable (as you did, but its important to repeat in the answer)
The consequences are undefined if an attempt is made to modify the standard readtable. To achieve the effect of altering or extending standard syntax, a copy of the standard readtable can be created; see the function copy-readtable.
§2.1.1.2 The Standard Readtable
Now, maybe I'm missing something (please give us a hint about how your reader macro is defined if so), but I think it is possible to avoid that and write your custom macros in a way that works for your use case.
Reader macro
Let's define a simple macro as follows:
CL-USER> (defun my-reader (stream char)
(declare (ignore char))
(let ((name (read stream)
(form (read stream))
(unless *read-suppress*
`(with-decoration ,name ,form)))
MY-READER
[NB: This was edited to take into account *read-suppress*: the code always read two forms, but returns nil in case it is being ignored. In the comments you say that you may need to read an indefinite number of forms based on the name of the decoration, but with *read-suppress* the recursive calls to read return nil for symbols, so you don't know which decoration is being applied. In that case it might be better to wrap some arguments in a literal list, or parse the stream manually (read-char, etc.). Also, since you are using a dispatching macro, maybe you can add a numerical argument if you want the decoration to be applied to more than one form (#2#inline), but that could be a bad idea when later the decorated code is being modified.]
Here the reader does a minimal job, namely build a form that is intended to be macroexpanded later. I don't even need to define with-decoration for now, as I'm interested in the read step. The intent is to read the next token (presumably a symbol that indicates what decoration is being applied, and a form to decorate).
I'm binding this macro to a unused character:
CL-USER> (set-macro-character #\§ 'my-reader)
T
Here when I test the macro it wraps the following form:
CL-USER> (read-from-string "§test (defun)")
(WITH-DECORATION TEST (DEFUN))
13 (4 bits, #xD, #o15, #b1101)
And here it works with a preceding QUOTE too, the apostrophe reader grabs the next form, which recursively reads two forms:
CL-USER> '§test (defun)
(WITH-DECORATION TEST (DEFUN))
Likewise, a conditional reader macro will ignore all the next lines:
CL-USER> #+(or) t
; No values
CL-USER> #+(or) §test (defun)
; No values
CL-USER> #+(or) §one §two §three (defun)
; No values
Decoration macro
If you use this syntax, you'll have nested decorated forms:
CL-USER> '§one §two (defun test ())
(WITH-DECORATION ONE (WITH-DECORATION TWO (DEFUN TEST ())))
With respect to defun in toplevel positions, you can arrange for your macros to unwrap the nesting (not completely tested, there might be bugs):
(defun unwrap-decorations (form stack)
(etypecase form
(cons (destructuring-bind (head . tail) form
(case head
(with-decoration (destructuring-bind (token form) tail
(unwrap-decorations form (cons token stack))))
(t `(with-decorations ,(reverse stack) ,form)))))))
CL-USER> (unwrap-decorations ** nil)
(WITH-DECORATIONS (ONE TWO) (DEFUN TEST ()))
And in turn, with-decorations might know about DEFUN forms and how to annotate them as necessary.
For the moment, our original macro is only the following (it needs more error checking):
(defmacro with-decoration (&whole whole &rest args)
(unwrap-decorations whole nil))
For the sake of our example, let's define a generic annotation mechanism:
CL-USER> (defgeneric expand-decoration (type name rest))
#<STANDARD-GENERIC-FUNCTION COMMON-LISP-USER::EXPAND-DECORATION (0)>
It is used in with-decorations to dispatch on an appropriate expander for each decoration. Keep in mind that all the efforts here are to keep defun in a top-level positions (under a progn), a recursive annotation would let evaluation happens (in the case of defun, it would result in the name of the function being defined), and the annotation could be done on the result.
The main macro is then here, with a kind of fold (reduce) mechanism where the forms are decorated using the resulting expansion so far. This allows for expanders to place code before or after the main form (or do other fancy things):
(defmacro with-decorations ((&rest decorations) form)
(etypecase form
(cons (destructuring-bind (head . tail) form
(ecase head
(defun (destructuring-bind (name args . body) tail
`(progn
,#(loop
for b = `((defun ,name ,args ,#body)) then forms
for d in decorations
for forms = (expand-decoration d name b)
finally (return forms))))))))))
(nb. here above we only care about defun but the loop should probably be done outside of the dispatching thing, along with a way to indicate to expander methods that a function is being expanded; well, it could be better)
Say, for example, you want to declare a function as inline, then the declaration must happen before (so that the compiler can know the source code must be kept):
(defmethod expand-decoration ((_ (eql 'inline)) name rest)
`((declaim (inline ,name)) ,#rest))
Likewise, if you want to export the name of the function being defined, you can export it after the function is defined (order is not really important here):
(defmethod expand-decoration ((_ (eql 'export)) name rest)
`(,#rest (export ',name)))
The resulting code allows you to have a single (progn ...) form with a defun in toplevel position:
CL-USER> (macroexpand '§inline §export (defun my-test-fn () "hello"))
(PROGN
(DECLAIM (INLINE MY-TEST-FN))
(DEFUN MY-TEST-FN () "hello")
(EXPORT 'MY-TEST-FN))
I would like a more detailed explanation of how macro expansion works, at least in Emacs Lisp but an overview of other Lisps would be appreciated. The way I usually see it explained is that the arguments of the macro are passed unevaluated to the body, which is then executed and returns a new LISP form. However, if I do
(defun check-one (arg)
(eq arg 1))
(defmacro check-foo (checker foo)
(if (checker 1)
`(,foo "yes")
`(,foo "no")))
I would expect
(check-foo check-one print)
to first expand to
(if (check-one 1)
`(print "yes")
`(print "no))
and then finally to
(print "yes")
but instead I get a "checker" function is void error. On the other hand, if I had defined
(defmacro check-foo (checker foo)
(if (funcall checker 1)
`(,foo "yes")
`(,foo "no")))
then I would have the expected behavior. So the expressions do get replaced in the body unevaluated, but for some reason functions do not work? What is the step-by-step procedure the interpreter follows when macroexpanding? Is there a good text-book that explains this rigorously?
Macros are functions ...
A good way to think about macros is that they are simply functions, like any other function.
... which operate on source code
But they are functions whose arguments are source code, and whose value is also source code.
Looking at macro functions
Macros being functions is not quite explicit in elisp: some of the lower-level functionality is, I think, not exposed. But in Common Lisp this is quite literally how macros are implemented: a macro has an associated function, and this function gets called to expand the macro, with its value being the new source code. For instance, if you are so minded you could write macros in Common Lisp like this.
(defun expand-fn (form environment)
;; not talking about environment
(declare (ignore environment))
(let ((name (second form))
(arglist (third form))
(body (cdddr form)))
`(function (lambda ,arglist
(block ,name
,#body)))))
(setf (macro-function 'fn) #'expand-fn)
And now fn is a macro which will construct a function which 'knows its name', so you could write
(fn foo (x) ... (return-from foo x) ...)
which turns into
(function (lambda (x) (block foo ... (return-from foo x))))
In Common Lisp, defmacro is then itself a macro which arranges for a suitable macro function to be installed and also deals with making the macro available at compile time &c.
In elisp, it looks as if this lower layer is not specified by the language, but I think it's safe to assume that things work the same way.
So then the job of a macro is to take a bunch of source code and compute from it another bunch of source code which is the expansion of the macro. And of course the really neat trick is that, because source code (both arguments and values) is represented as s-expressions, Lisp is a superb language for manipulating s-expressions, you can write macros in Lisp itself.
Macroexpansion
There are a fair number of fiddly corner cases here such as local macros and so on. But here is, pretty much, how this works.
Start with some form <f>:
If <f> is (<a> ...) where <a> is a symbol, check for a macro function for <a>. If it has one, call it on the whole form, and call the value it returns <f'>: now simply recurse on <f'>.
If <f> is (<a> ...) where <a> is a symbol which names a special operator (something like if) then recurse on the subforms of the special operator which its rules say must be macroexpanded. As an example, in a form like (if <x> <y> <z>) all of <x>, <y>, & <z> need to be macroexpanded, while in (setq <a> <b>), only <b> would be subject to macroexpansion, and so on: these rules are hard-wired, which is why special operators are special.
If <f> is (<a> ...) where <a> is a symbol which is neither of the above cases, then it's a function call, and the forms in the body of the form are macroexpanded, and that's it.
If <f> is ((lambda (...) ...) ...) then the forms in the body of the lambda (but not its arguments!) are macroexpanded and then the case is the same as the last one.
Finally <f> might not be a compound form: nothing to do here.
I think that's all the cases. This is not a complete description of the process because there are complications like local macros and so on. But it's enough I think.
Order of macroexpansion
Note that macroexpansion happens 'outside in': a form like (a ...) is expanded until you get something which isn't a macro form, and only then is the body, perhaps, expanded. That's because, until the macro is completely expanded, you have no idea which, if any, of the subforms are even eligible for macroexpansion.
Your code
My guess is that what you want to happen is that (check-foo bog foo) should turn into (if (bog 1) (foo yes) (foo no)). So the way to get this is that this form is what the macro function needs to return. We could write this using the CL low-level facilities:
(defun check-foo-expander (form environment)
;; form is like (check-foo pred-name function-name)
(declare (ignore environment)) ;still not talking about environment
`(if (,(second form) 1)
(,(third form) "yes")
(,(third form) "no")))
And we can check:
> (check-foo-expander '(check-foo bog foo) nil)
(if (bog 1) (foo "yes") (foo "no"))
And then install it as a macro:
> (setf (macro-function 'check-foo) #'check-foo-expander)
And now
> (check-foo evenp print)
"no"
"no"
> (check-foo oddp print)
"yes"
"yes"
But it's easier to write it using defmacro:
(defmacro check-foo (predicate function)
`(if (,predicate 1)
(,function "yes")
(,function "no")))
This is the same thing (more-or-less), but easier to read.
A book [1] that I am reading says this:
One of the most interesting developments in programming languages has
been the creation of extensible languages—languages whose syntax and
semantics can be changed within a program. One of the earliest and
most commonly proposed schemes for language extension is the macro
definition.
Would you give an example (along with an explanation) of a Lisp macro that extends the syntax and semantics of the Lisp programming language, please?
[1] The Theory of Parsing, Translation, and Compiling, Volume 1 Parsing by Aho and Ullman, page 58.
Picture the scene: it is 1958, and FORTRAN has just been invented. Lit only by the afterglow of atomic tests, primitive Lisp programmers are writing loops in primitive Lisp the way primitive FORTRAN programmers always had:
(prog ((i 0)) ;i is 0
start ;label beginning of loop
(if (>= i 10)
(go end)) ;skip to end when finished
(do-hard-sums-on i) ;hard sums!
(setf i (+ i 1)) ;increment i
(go start) ;jump to start
end) ;end
(Except, of course this would all be in CAPITALS because lower-case had not been invented then, and the thing I wrote as setf would be something uglier, because setf (a macro!) also had not been invented then).
Enter, in clouds of only-slightly toxic smoke from their jetpack, another Lisp programmer who had escaped to 1958 from the future. 'Look', they said, 'we could write this strange future thing':
(defmacro sloop ((var init limit &optional (step 1)) &body forms)
(let ((<start> (make-symbol "START")) ;avoid hygiene problems ...
(<end> (make-symbol "END"))
(<limit> (make-symbol "LIMIT")) ;... and multiple evaluation problems
(<step> (make-symbol "STEP")))
`(prog ((,var ,init)
(,<limit> ,limit)
(,<step> ,step))
,<start>
(if (>= ,var ,<limit>)
(go ,<end>))
,#forms
(setf ,var (+ ,var ,<step>))
(go ,<start>)
,<end>)))
'And now', they say, 'you can write this':
(sloop (i 0 10)
(do-hard-sums i))
And thus were simple loops invented.
Back here in the future we can see what this loop expands into:
(sloop (i 0 10)
(format t "~&i = ~D~%" i))
->
(prog ((i 0) (#:limit 10) (#:step 1))
#:start (if (>= i #:limit) (go #:end))
(format t "~&i = ~D~%" i)
(setf i (+ i #:step))
(go #:start)
#:end)
Which is the code that the primitive Lisp programmers used to type in by hand. And we can run this:
> (sloop (i 0 10)
(format t "~&i = ~D~%" i))
i = 0
i = 1
i = 2
i = 3
i = 4
i = 5
i = 6
i = 7
i = 8
i = 9
nil
And in fact this is how loops work in Lisp today. If I try a simple do loop, one of Common Lisp's predefined macros, we can see what it expands into:
(do ((i 0 (+ i 1)))
((>= i 10))
(format t "~&i = ~D~%" i))
->
(block nil
(let ((i 0))
(declare (ignorable i))
(declare)
(tagbody
#:g1481 (if (>= i 10) (go #:g1480))
(tagbody (format t "~&i = ~D~%" i)
(setq i (+ i 1)))
(go #:g1481)
#:g1480)))
Well, this expansion is not the same and it uses constructs I haven't talked about, but you can see the important thing: this loop has been rewritten to use GO. And, although Common Lisp does not define the expansion of it's looping macros, it is almost certainly the case that all the standard ones expand into something like this (but more complicated in general).
In other words: Lisp doesn't have any primitive looping constructs, but all such constructs are added to the language by macros. These macros as well as other macros to extend the language in other ways, can be written by users: they do not have to be provided by the language itself.
Lisp is a programmable programming language.
Well, perhaps the explanations will be terse, but you could look at the macros used in the lisp language itself, defun, for example.
http://clhs.lisp.se/Body/m_defun.htm
In lisp, macros are a big part of the language itself, basically allowing you to rewrite code before it's compiled.
There's more than just macros defined by defmacro. There are also Reader Macros ! as Paul Graham said in On Lisp :
The three big moments in a Lisp expression’s life are read-time,
compile-time, and runtime. Functions are in control at runtime. Macros
give us a chance to perform transformations on programs at
compile-time. …read-macros… do their work at read-time.
Macros and read-macros see your program at different stages. Macros
get hold of the program when it has already been parsed into Lisp
objects by the reader, and read-macros operate on a program while it
is still text. However, by invoking read on this text, a read-macro
can, if it chooses, get parsed Lisp objects as well. Thus read-macros
are at least as powerful as ordinary macros.
With Reader Macros, you can define new semantics way beyond ordinary macros, like for example :
adds support for string interpolation ( cl-interpol )
adds support for JSON directly into the language : See this article to know more.
I am trying to emulate the single namespace of scheme within common lisp, with a macro (based on Doug Hoyte's) that expands to a lambda, where every use of an f! symbol (similar to Doug Hoyte's o! and g! symbols) in the function position expands to the same expression, but with funcall added in the function position of each invocation. For example:
(fplambda (f!z x) (f!z x x))
would expand to:
(LAMBDA (F!Z X) (FUNCALL F!Z X X))
The macro currently looks like this:
(defmacro fplambda (parms &body body)
(let ((syms (remove-duplicates
(remove-if-not #'f!-symbol-p
(flatten body)))))
`(lambda ,parms
(macrolet ,(mapcar
(lambda (f)
`(,f (&rest parmlist) `(funcall ,',f ',#parmlist)))
syms))
,#body)))
but given the above input, it expands (as far as I can see) to this:
(LAMBDA (F!F X)
(MACROLET ((F!F (&REST PARMLIST) `(FUNCALL ,'F!F ',#PARMLIST))))
(F!F X X))
In the macrolet definition, F!F should not be quoted or unquoted, and parmlist should just be unquoted. What is going on?
Thanks in advance!
Your definition is mostly right. You just made two pretty simple mistakes. The first one being a mismatched paren. The macrolet does not include the body (in the output the macrolet and the body are at the same level of indentation).
As for the nested backquote, the only mistake is the quote before parmlist. Other than that everything else is correct. The comma and quote before F!F is actually correct. From the hyperspec:
"An implementation is free to interpret a backquoted form F1 as any form F2 that, when evaluated, will produce a result that is the same under equal as the result implied by the above definition". Since the inner backquote has not been expanded yet, it does not have to be free of quotes and unquotes. The expression `(,'x) is actually the same as `(x).
Nested backquotes are notoriously complicated. What is probably the easiest way to understand them is to read Steele's explanation of them.
Edit:
The answer to your question about whether it is possible to use a fplambda expression in the function position is no. From the part of the hyperspec that deals with the evaluation of code: "If the car of the compound form is not a symbol, then that car must be a lambda expression, in which case the compound form is a lambda form.". Since the car of the form, (fplambda ...), is not a lambda expression, your code is no longer valid Common Lisp code.
There is a workaround to this that I figured out, but it's kind of ugly. You can define a reader macro that will allow you to write something like ([fplambda ...] ...) and have it read as
((LAMBDA (&REST #:G1030) (APPLY (FPLAMBDA ...) #:G1030)) ...)
which would do what you want. Here is code that will allow you to do that:
(set-macro-character #\[ 'bracket-reader)
(set-macro-character #\] (get-macro-character #\)))
(defun bracket-reader (stream char)
"Read in a bracket."
(declare (ignore char))
(let ((gargs (gensym)))
`(lambda (&rest ,gargs)
(apply ,(read-delimited-list #\] stream t)
,gargs))))
The only other solution I can think of would be to use some sort of code walker (I can't help you there).
I've been reading an article by Olin Shivers titled Stylish Lisp programming techniques and found the second example there (labeled "Technique n-1") a bit puzzling. It describes a self-modifying macro that looks like this:
(defun gen-counter macro (x)
(let ((ans (cadr x)))
(rplaca (cdr x)
(+ 1 ans))
ans))
It's supposed to get its calling form as argument x (i.e. (gen-counter <some-number>)). The purpose of this is to be able to do something like this:
> ;this prints out the numbers from 0 to 9.
(do ((n 0 (gen-counter 1)))
((= n 10) t)
(princ n))
0.1.2.3.4.5.6.7.8.9.T
>
The problem is that this syntax with the macro symbol after the function name is not valid in Common Lisp. I've been unsuccessfully trying to obtain similar behavior in Common Lisp. Can someone please provide a working example of analogous macro in CL?
Why the code works
First, it's useful to consider the first example in the paper:
> (defun element-generator ()
(let ((state '(() . (list of elements to be generated)))) ;() sentinel.
(let ((ans (cadr state))) ;pick off the first element
(rplacd state (cddr state)) ;smash the cons
ans)))
ELEMENT-GENERATOR
> (element-generator)
LIST
> (element-generator)
OF
> (element-generator)
This works because there's one literal list
(() . (list of elements to be generated)
and it's being modified. Note that this is actually undefined behavior in Common Lisp, but you'll get the same behavior in some Common Lisp implementations. See Unexpected persistence of data and some of the other linked questions for a discussion of what's happening here.
Approximating it in Common Lisp
Now, the paper and code you're citing actually has some useful comments about what this code is doing:
(defun gen-counter macro (x) ;X is the entire form (GEN-COUNTER n)
(let ((ans (cadr x))) ;pick the ans out of (gen-counter ans)
(rplaca (cdr x) ;increment the (gen-counter ans) form
(+ 1 ans))
ans)) ;return the answer
The way that this is working is not quite like an &rest argument, as in Rainer Joswig's answer, but actually a &whole argument, where the the entire form can be bound to a variable. This is using the source of the program as the literal value that gets destructively modified! Now, in the paper, this is used in this example:
> ;this prints out the numbers from 0 to 9.
(do ((n 0 (gen-counter 1)))
((= n 10) t)
(princ n))
0.1.2.3.4.5.6.7.8.9.T
However, in Common Lisp, we'd expect the macro to be expanded just once. That is, we expect (gen-counter 1) to be replaced by some piece of code. We can still generate a piece of code like this, though:
(defmacro make-counter (&whole form initial-value)
(declare (ignore initial-value))
(let ((text (gensym (string 'text-))))
`(let ((,text ',form))
(incf (second ,text)))))
CL-USER> (macroexpand '(make-counter 3))
(LET ((#:TEXT-1002 '(MAKE-COUNTER 3)))
(INCF (SECOND #:TEXT-1002)))
Then we can recreate the example with do
CL-USER> (do ((n 0 (make-counter 1)))
((= n 10) t)
(princ n))
023456789
Of course, this is undefined behavior, since it's modifying literal data. It won't work in all Lisps (the run above is from CCL; it didn't work in SBCL).
But don't miss the point
The whole article is sort of interesting, but recognize that it's sort of a joke, too. It's pointing out that you can do some funny things in an evaluator that doesn't compile code. It's mostly satire that's pointing out the inconsistencies of Lisp systems that have different behaviors under evaluation and compilation. Note the last paragraph:
Some short-sighted individuals will point out that these programming
techniques, while certainly laudable for their increased clarity and
efficiency, would fail on compiled code. Sadly, this is true. At least
two of the above techniques will send most compilers into an infinite
loop. But it is already known that most lisp compilers do not
implement full lisp semantics -- dynamic scoping, for instance. This
is but another case of the compiler failing to preserve semantic
correctness. It remains the task of the compiler implementor to
adjust his system to correctly implement the source language, rather
than the user to resort to ugly, dangerous, non-portable, non-robust
``hacks'' in order to program around a buggy compiler.
I hope this provides some insight into the nature of clean, elegant
Lisp programming techniques.
—Olin Shivers
Common Lisp:
(defmacro gen-counter (&rest x)
(let ((ans (car x)))
(rplaca x (+ 1 ans))
ans))
But above only works in the Interpreter, not with a compiler.
With compiled code, the macro call is gone - it is expanded away - and there is nothing to modify.
Note to unsuspecting readers: you might want to read the paper by Olin Shivers very careful and try to find out what he actually means...