In one variant of Common Lisp (I think it was CMUCL, but I might be wrong—I can't find it any more) there was a function that was (I think) called function-lambda-expression. If it got a procedure, it would print out the lambda expression that had generated it. Example:
(let ((my-thunk (lambda () (+ 1 2))))
(write my-thunk)
(write (function-lambda-expression my-thunk)))
This would print out something like:
#<PROCEDURE>
(LAMBDA () (+ 1 2))
It was terribly useful for debugging and exploring the language.
I'm looking for a function like this in Racket. I've looked through the Racket Documentation but I can't find anything like it. (I wouldn't be surprised if I overlooked it, however.) Is there an equivalent in Racket?
No. Racket's lambda produces a closure that does not remember its S-expression (or syntax object) form. It does typically remember its name (or its abbreviated source location, if no name can be inferred), and that's often enough to help with debugging. (See object-name.)
You can build your own variant of lambda that has this feature, using Racket's applicable structures and a simple macro. Here's a basic example:
#lang racket
(struct exp-closure (f exp)
#:property prop:procedure (struct-field-index f))
(define-syntax-rule (exp-lambda formals . body)
(exp-closure (lambda formals . body)
(quote (exp-lambda formals . body))))
(let ([my-thunk (exp-lambda () (+ 1 2))])
(printf "fun is ~v\n" my-thunk)
(printf "exp is ~v\n" (exp-closure-exp my-thunk))
(printf "result is ~v\n" (my-thunk)))
This produces
fun is #<procedure:...tmp/lambda.rkt:11:19>
exp is '(exp-lambda () (+ 1 2))
result is 3
A better version of this macro might propagate the source location of the macro use to the lambda expression it creates, or the inferred name (see syntax-local-infer-name), or both.
Related
I'm in a process of implementing Hygienic macros in my Scheme implementation, I've just implemented syntax-rules, but I have this code:
(define odd?
(syntax-rules ()
((_ x) (not (even? x)))))
what should be the difference between that and this:
(define-syntax odd?
(syntax-rules ()
((_ x) (not (even? x)))))
from what I understand syntax-rules just return syntax transformer, why you can't just use define to assign that to symbol? Why I need to use define-syntax? What extra stuff that expression do?
Should first also work in scheme? Or only the second one?
Also what is the difference between let vs let-syntax and letrec vs letrec-syntax. Should (define|let|letrec)-syntax just typecheck if the value is syntax transformer?
EDIT:
I have this implementation, still using lisp macros:
;; -----------------------------------------------------------------------------
(define-macro (let-syntax vars . body)
`(let ,vars
,#(map (lambda (rule)
`(typecheck "let-syntax" ,(car rule) "syntax"))
vars)
,#body))
;; -----------------------------------------------------------------------------
(define-macro (letrec-syntax vars . body)
`(letrec ,vars
,#(map (lambda (rule)
`(typecheck "letrec-syntax" ,(car rule) "syntax"))
vars)
,#body))
;; -----------------------------------------------------------------------------
(define-macro (define-syntax name expr)
(let ((expr-name (gensym)))
`(define ,name
(let ((,expr-name ,expr))
(typecheck "define-syntax" ,expr-name "syntax")
,expr-name))))
This this code correct?
Should this code works?
(let ((let (lambda (x) x)))
(let-syntax ((odd? (syntax-rules ()
((_ x) (not (even? x))))))
(odd? 11)))
This question seems to imply some deep confusion about macros.
Let's imagine a language where syntax-rules returns some syntax transformer function (I am not sure this has to be true in RnRS Scheme, it is true in Racket I think), and where let and let-syntax were the same.
So let's write this function:
(define (f v)
(let ([g v])
(g e (i 10)
(if (= i 0)
i
(e (- i 1))))))
Which we can turn into this, of course:
(define (f v n)
(v e (i n)
(if (<= i 0)
i
(e (- i 1)))))
And I will tell you in addition that there is no binding for e or i in the environment.
What is the interpreter meant to do with this definition? Could it compile it? Could it safely infer that i can't possibly make any sense since it is used as a function and then as a number? Can it safely do anything at all?
The answer is that no, it can't. Until it knows what the argument to the function is it can't do anything. And this means that each time f is called it has to make that decision again. In particular, v might be:
(syntax-rules ()
[(_ name (var init) form ...)
(letrec ([name (λ (var)
form ...)])
(name init))]))
Under which the definition of f does make some kind of sense.
And things get worse: much worse. How about this?
(define (f v1 v2 n)
(let ([v v1])
(v e (i n)
...
(set! v (if (eq? v v1) v2 v1))
...)))
What this means is that a system like this wouldn't know what the code it was meant to interpret meant until, the moment it was interpreting it, or even after that point, as you can see from the second function above.
So instead of this horror, Lisps do something sane: they divide the process of evaluating bits of code into phases where each phase happens, conceptually, before the next one.
Here's a sequence for some imagined Lisp (this is kind of close to what CL does, since most of my knowledge is of that, but it is not intended to represent any particular system):
there's a phase where the code is turned from some sequence of characters to some object, possibly with the assistance of user-defined code;
there's a phase where that object is rewritten into some other object by user- and system-defined code (macros) – the result of this phase is something which is expressed in terms of functions and some small number of primitive special things, traditionally called 'special forms' which are known to the processes of stage 3 and 4;
there may be a phase where the object from phase 2 is compiled, and that phase may involve another set of user-defined macros (compiler macros);
there is a phase where the resulting code is evaluated.
And for each unit of code these phases happen in order, each phase completes before the next one begins.
This means that each phase in which the user can intervene needs its own set of defining and binding forms: it needs to be possible to say that 'this thing controls what happens at phase 2' for instance.
That's what define-syntax, let-syntax &c do: they say that 'these bindings and definitions control what happens at phase 2'. You can't, for instance, use define or let to do that, because at phase 2, these operations don't yet have meaning: they gain meaning (possibly by themselves being macros which expand to some primitive thing) only at phase 3. At phase 2 they are just bits of syntax which the macro is ingesting and spitting out.
In this post, I ask tangentially why when I declare in SBCL
(defun a (&rest x)
x)
and then check what the function cell holds
(describe 'a)
COMMON-LISP-USER::A
[symbol]
A names a compiled function:
Lambda-list: (&REST X)
Derived type: (FUNCTION * (VALUES LIST &OPTIONAL))
Source form:
(LAMBDA (&REST X) (BLOCK A X))
I see this particular breakdown of the original function. Could someone explain what this output means? I'm especially confused by the last line
Source form:
(LAMBDA (&REST X) (BLOCK A X))
This is mysterious because for some reason not clear to me Lisp has transformed the original function into a lambda expression. It would also be nice to know the details of how a function broken down like this is then called. This example is SBCL. In Elisp
(symbol-function 'a)
gives
(lambda (&rest x) x)
again, bizarre. As I said in the other post, this is easier to understand in Scheme -- but that created confusion in the answers. So once more I ask, Why has Lisp taken a normal function declaration and seemingly stored it as a lambda expression?
I'm still a bit unclear what you are confused about, but here is an attempt to explain it. I will stick to CL (and mostly to ANSI CL), because elisp has a lot of historical oddities which just make things hard to understand (there is an appendix on elisp). Pre-ANSI CL was also a lot less clear on various things.
I'll try to explain things by writing a macro which is a simple version of defun: I'll call this defun/simple, and an example of its use will be
(defun/simple foo (x)
(+ x x))
So what I need to do is to work out what the expansion of this macro should be, so that it does something broadly equivalent (but simpler than) defun.
The function namespace & fdefinition
First of all I assume you are comfortable with the idea that, in CL (and elisp) the namespace of functions is different than the namespace of variable bindings: both languages are lisp-2s. So in a form like (f x), f is looked up in the namespace of function bindings, while x is looked up in the namespace of variable bindings. This means that forms like
(let ((sin 0.0))
(sin sin))
are fine in CL or elisp, while in Scheme they would be an error, as 0.0 is not a function, because Scheme is a lisp-1.
So we need some way of accessing that namespace, and in CL the most general way of doing that is fdefinition: (fdefinition <function name>) gets the function definition of <function name>, where <function name> is something which names a function, which for our purposes will be a symbol.
fdefinition is what CL calls an accessor: this means that the setf macro knows what to do with it, so that we can mutate the function binding of a symbol by (setf (fdefinition ...) ...). (This is not true: what we can access and mutate with fdefinition is the top-level function binding of a symbol, we can't access or mutate lexical function bindings, and CL provides no way to do this, but this does not matter here.)
So this tells us what our macro expansion needs to look like: we want to set the (top-level) definition of the name to some function object. The expansion of the macro should be like this:
(defun/simple foo (x)
x)
should expand to something involving
(setf (fdefinition 'foo) <form which makes a function>)
So we can write this bit of the macro now:
(defmacro defun/simple (name arglist &body forms)
`(progn
(setf (fdefinition ',name)
,(make-function-form name arglist forms))
',name))
This is the complete definition of this macro. It uses progn in its expansion so that the result of expanding it is the name of the function being defined, which is the same as defun: the expansion does all its real work by side-effect.
But defun/simple relies on a helper function, called make-function-form, which I haven't defined yet, so you can't actually use it yet.
Function forms
So now we need to write make-function-form. This function is called at macroexpansion time: it's job is not to make a function: it's to return a bit of source code which will make a function, which I'm calling a 'function form'.
So, what do function forms look like in CL? Well, there's really only one such form in portable CL (this might be wrong, but I think it is true), which is a form constructed using the special operator function. So we're going to need to return some form which looks like (function ...). Well, what can ... be? There are two cases for function.
(function <name>) denotes the function named by <name> in the current lexical environment. So (function car) is the function we call when we say (car x).
(function (lambda ...)) denotes a function specified by (lambda ...): a lambda expression.
The second of these is the only (caveats as above) way we can construct a form which denotes a new function. So make-function-form is going to need to return this second variety of function form.
So we can write an initial version of make-function-form:
(defun make-function-form (name arglist forms)
(declare (ignore name))
`(function (lambda ,arglist ,#forms)))
And this is enough for defun/simple to work:
> (defun/simple plus/2 (a b)
(+ a b))
plus/2
> (plus/2 1 2)
3
But it's not quite right yet: one of the things that functions defined by defun can do is return from themselves: they know their own name and can use return-from to return from it:
> (defun silly (x)
(return-from silly 3)
(explode-the-world x))
silly
> (silly 'yes)
3
defun/simple can't do this, yet. To do this, make-function-form needs to insert a suitable block around the body of the function:
(defun make-function-form (name arglist forms)
`(function (lambda ,arglist
(block ,name
,#forms))))
And now:
> (defun/simple silly (x)
(return-from silly 3)
(explode-the-world x))
silly
> (silly 'yes)
3
And all is well.
This is the final definition of defun/simple and its auxiliary function.
Looking at the expansion of defun/simple
We can do this with macroexpand in the usual way:
> (macroexpand '(defun/simple foo (x) x))
(progn
(setf (fdefinition 'foo)
#'(lambda (x)
(block foo
x)))
'foo)
t
The only thing that's confusing here is that, because (function ...) is common in source code, there's syntactic sugar for it which is #'...: this is the same reason that quote has special syntax.
It's worth looking at the macroexpansion of real defun forms: they usually have a bunch of implementation-specific stuff in them, but you can find the same thing there. Here's an example from LW:
> (macroexpand '(defun foo (x) x))
(compiler-let ((dspec::*location* '(:inside (defun foo) :listener)))
(compiler::top-level-form-name (defun foo)
(dspec:install-defun 'foo
(dspec:location)
#'(lambda (x)
(declare (system::source-level
#<eq Hash Table{0} 42101FCD5B>))
(declare (lambda-name foo))
x))))
t
Well, there's a lot of extra stuff in here, and LW obviously has some trick around this (declare (lambda-name ...)) form which lets return-from work without an explicit block. But you can see that basically the same thing is going on.
Conclusion: how you make functions
In conclusion: a macro like defun, or any other function-defining form, needs to expand to a form which, when evaluated, will construct a function. CL offers exactly one such form: (function (lambda ...)): that's how you make functions in CL. So something like defun necessarily has to expand to something like this. (To be precise: any portable version of defun: implementations are somewhat free to do implementation-magic & may do so. However they are not free to add a new special operator.)
What you are seeing when you call describe is that, after SBCL has compiled your function, it's remembered what the source form was, and the source form was exactly the one you would have got from the defun/simple macro given here.
Notes
lambda as a macro
In ANSI CL, lambda is defined as a macro whose expansion is a suitable (function (lambda ...)) form:
> (macroexpand '(lambda (x) x))
#'(lambda (x) x)
t
> (car (macroexpand '(lambda (x) x)))
function
This means that you don't have to write (function (lambda ...)) yourself: you can rely on the macro definition of lambda doing it for you. Historically, lambda wasn't always a macro in CL: I can't find my copy of CLtL1, but I'm pretty certain it was not defined as one there. I'm reasonably sure that the macro definition of lambda arrived so that it was possible to write ISLisp-compatible programs on top of CL. It has to be in the language because lambda is in the CL package and so users can't portably define macros for it (although quite often they did define such a macro, or at least I did). I have not relied on this macro definition above.
defun/simple does not purport to be a proper clone of defun: its only purpose is to show how such a macro can be written. In particular it doesn't deal with declarations properly, I think: they need to be lifted out of the block & are not.
Elisp
Elisp is much more horrible than CL. In particular, in CL there is a well-defined function type, which is disjoint from lists:
> (typep '(lambda ()) 'function)
nil
> (typep '(lambda ()) 'list)
t
> (typep (function (lambda ())) 'function)
t
> (typep (function (lambda ())) 'list)
nil
(Note in particular that (function (lambda ())) is a function, not a list: function is doing its job of making a function.)
In elisp, however, an interpreted function is just a list whose car is lambda (caveat: if lexical binding is on this is not the case: it's then a list whose car is closure). So in elisp (without lexical binding):
ELISP> (function (lambda (x) x))
(lambda (x)
x)
And
ELISP> (defun foo (x) x)
foo
ELISP> (symbol-function 'foo)
(lambda (x)
x)
The elisp intepreter then just interprets this list, in just the way you could yourself. function in elisp is almost the same thing as quote.
But function isn't quite the same as quote in elisp: the byte-compiler knows that, when it comes across a form like (function (lambda ...)) that this is a function form, and it should byte-compile the body. So, we can look at the expansion of defun in elisp:
ELISP> (macroexpand '(defun foo (x) x))
(defalias 'foo
#'(lambda (x)
x))
(It turns out that defalias is the primitive thing now.)
But if I put this definition in a file, which I byte compile and load, then:
ELISP> (symbol-function 'foo)
#[(x)
"\207"
[x]
1]
And you can explore this a bit further: if you put this in a file:
(fset 'foo '(lambda (x) x))
and then byte compile and load that, then
ELISP> (symbol-function 'foo)
(lambda (x)
x)
So the byte compiler didn't do anything with foo because it didn't get the hint that it should. But foo is still a fine function:
ELISP> (foo 1)
1 (#o1, #x1, ?\C-a)
It just isn't compiled. This is also why, if writing elisp code with anonymous functions in it, you should use function (or equivalently #'). (And finally, of course, (function ...) does the right thing if lexical scoping is on.)
Other ways of making functions in CL
Finally, I've said above that function & specifically (function (lambda ...)) is the only primitive way to make new functions in CL. I'm not completely sure that's true, especially given CLOS (almost any CLOS will have some kind of class instances of which are functions but which can be subclassed). But it does not matter: it is a way and that's sufficient.
DEFUN is a defining macro. Macros transform code.
In Common Lisp:
(defun foo (a)
(+ a 42))
Above is a definition form, but it will be transformed by DEFUN into some other code.
The effect is similar to
(setf (symbol-function 'foo)
(lambda (a)
(block foo
(+ a 42))))
Above sets the function cell of the symbol FOO to a function. The BLOCK construct is added by SBCL, since in Common Lisp named functions defined by DEFUN create a BLOCK with the same name as the function name. This block name can then be used by RETURN-FROM to enable a non-local return from a specific function.
Additionally DEFUN does implementation specific things. Implementations also record development information: the source code, the location of the definition, etc.
Scheme has DEFINE:
(define (foo a)
(+ a 10))
This will set FOO to a function object.
I am trying to make a list of callback functions, which could look like this:
(("command1" . 'callback1)
("command2" . 'callback2)
etc)
I'd like it if I could could do something like:
(define-callback callback1 "command1" args
(whatever the function does))
Rather than
(defun callback1 (args)
(whatever the function does))
(add-to-list 'callback-info ("command1" . 'callback1))
Is there a convenient way of doing this, e.g., with macros?
This is a good example of a place where it's nice to use a two-layered approach, with an explicit function-based layer, and then a prettier macro layer on top of that.
Note the following assumes Common Lisp: it looks just possible from your question that you are asking about elisp, in which case something like this can be made to work but it's all much more painful.
First of all, we'll keep callbacks in an alist called *callbacks*:
(defvar *callbacks* '())
Here's a function which clears the alist of callbacks
(defun initialize-callbacks ()
(setf *callbacks* '())
(values)
Here is the function that installs a callback. It does this by searching the list to see if there is a callback with the given name, and if there is then replacing it, and otherwise installing a new one. Like all the functions in the functional layer lets us specify the test function which will let us know if two callback names are the same: by default this is #'eql which will work for symbols and numbers, but not for strings. Symbols are probably a better choice for the names of callbacks than strings, but we'll cope with that below.
(defun install-callback (name function &key (test #'eql))
(let ((found (assoc name *callbacks* :test test)))
(if found
(setf (cdr found) function)
(push (cons name function) *callbacks*)))
name)
Here is a function to find a callback, returning the function object, or nil if there is no callback with that name.
(defun find-callback (name &key (test #'eql))
(cdr (assoc name *callbacks* :test test)))
And a function to remove a named callback. This doesn't tell you if it did anything: perhaps it should.
(defun remove-callback (name &key (test #'eql))
(setf *callbacks* (delete name *callbacks* :key #'car :test test))
name)
Now comes the macro layer. The syntax of this is going to be (define-callback name arguments ...), so it looks a bit like a function definition.
There are three things to know about this macro.
It is a bit clever: because you can know at macro-expansion time what sort of thing the name of the callback is, you can decide then and there what test to use when installing the callback, and it does this. If the name is a symbol it also wraps a block named by the symbol around the body of the function definition, so it smells a bit more like a function defined by defun: in particular you can use return-from in the body. It does not do this if the name is not a symbol.
It is not quite clever enough: in particular it does not deal with docstrings in any useful way (it ought to pull them out of the block I think). I am not sure this matters.
The switch to decide the test uses expressions like '#'eql which reads as (quote (function eql)): that is to avoid wiring in functions into the expansion because functions are not externalisable objects in CL. However I am not sure I have got this right: I think what is there is safe but it may not be needed.
So, here it is
(defmacro define-callback (name arguments &body body)
`(install-callback ',name
,(if (symbolp name)
`(lambda ,arguments
(block ,name
,#body))
`(lambda ,arguments
,#body))
:test ,(typecase name
(string '#'string=)
(symbol '#'eql)
(number '#'=)
(t '#'equal))))
And finally here are two different callbacks being defined:
(define-callback "foo" (x)
(+ x 3))
(define-callback foo (x)
(return-from foo (+ x 1)))
These lists are called assoc lists in Lisp.
CL-USER 120 > (defvar *foo* '(("c1" . c1) ("c2" . c2)))
*FOO*
CL-USER 121 > (setf *foo* (acons "c0" `c1 *foo*))
(("c0" . C1) ("c1" . C1) ("c2" . C2))
CL-USER 122 > (assoc "c1" *foo* :test #'equal)
("c1" . C1)
You can write macros for that, but why? Macros are advanced Lisp and you might want to get the basics right, first.
Some issues with you example you might want to check out:
what are assoc lists?
what are useful key types in assoc lists?
why you don't need to quote symbols in data lists
variables are not quoted
data lists need to be quoted
You can just as easy create such lists for callbacks without macros. We can imagine a function create-callback, which would be used like this:
(create-callback 'callback1 "command1"
(lambda (arg)
(whatever the function does)))
Now, why would you use a macro instead of a plain function?
In the end, assisted by the responders above, I got it down to something like:
(defmacro mk-make-command (name &rest body)
(let ((func-sym (intern (format "mk-cmd-%s" name))))
(mk-register-command name func-sym)
`(defun ,func-sym (args &rest rest)
(progn
,#body))))
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...
I am having a problem with a lisp macro. I would like to create a macro
which generate a switch case according to an array.
Here is the code to generate the switch-case:
(defun split-elem(val)
`(,(car val) ',(cdr val)))
(defmacro generate-switch-case (var opts)
`(case ,var
,(mapcar #'split-elem opts)))
I can use it with a code like this:
(generate-switch-case onevar ((a . A) (b . B)))
But when I try to do something like this:
(defparameter *operators* '((+ . OPERATOR-PLUS)
(- . OPERATOR-MINUS)
(/ . OPERATOR-DIVIDE)
(= . OPERATOR-EQUAL)
(* . OPERATOR-MULT)))
(defmacro tokenize (data ops)
(let ((sym (string->list data)))
(mapcan (lambda (x) (generate-switch-case x ops)) sym)))
(tokenize data *operators*)
I got this error: *** - MAPCAR: A proper list must not end with OPS, but I don't understand why.
When I print the type of ops I get SYMBOL I was expecting CONS, is it related?
Also, for my function tokenize, how many times is the lambda evaluated (or the macro expanded)?
Thanks.
This makes no sense. You trying to use macros, where functions are sufficient.
What you want is similar to this:
(defun tokenize (data ops)
(mapcar (lambda (d)
(cdr (assoc d ops)))
(string->list data)))
CASE is a macro that expects a bunch of fixed clauses. It does not take clauses that are computed at runtime. If list data should drive computation, then use functions like ASSOC.
GENERATE-SWITCH-CASE is also an odd name, since the macro IS a switch case.
GENERATE-SWITCH-CASE also does expect a list as a second argument. But in TOKENIZE you call it with a symbol OPS. Remember, macros compute with Lisp source code.
Next, there are also no ARRAYs involved. Lisp has arrays, but in your example is none.
Typical advice:
if you want to write a MACRO, think again. Write it as a function.
if you still want to write a macro, Go to 1.