I understand that both are used to temporarily change the value of a function. Beyond the fact that cl-flet is a function and cl-letf is a macro, when do you use them?
The bindings can be recursive
If the function definition calls itself by name, which function will be called? (compare cl-flet vs cl-labels behaviour).
Scoping is lexical ... capturing them in closures ...
Read about lexical binding/scope vs dynamic binding/scope.
cl-letf can be used to set dynamically-bound function values, by using a PLACE of (symbol-function 'FUNC) for some FUNC. This is analogous to the deprecated flet.
Any supported PLACE can be specified, though -- cl-letf isn't only for function bindings.
when do you use them?
When you want to temporarily define (or override) a function. The scoping rules you require for any given use-case will determine which option you would use.
(cl-flet ((FUNC ARGLIST BODY...) ...) FORM...)
FUNC is visible only to the code in FORM.
(cl-labels ((FUNC ARGLIST BODY...) ...) FORM...)
FUNC is visible both to the code in FORM and also to the code in FUNC's own BODY.
(cl-letf (((symbol-function 'FUNC) VALUE) ...) BODY...)
FUNC is visible to absolutely everything until BODY has finished being evaluated.
Some (fairly contrived) examples...
In the first example, the temporary function we have defined is recursive -- it calls itself -- and therefore we use cl-labels:
(n.b. this isn't a robust factorial implementation; it's just for demonstration purposes.)
(defun my-factorial (number)
"Show the factorial of the argument."
(interactive "nFactorial of: ")
(cl-labels ((factorial (n) (if (eq n 1)
1
(* n (factorial (1- n))))))
(message "Factorial of %d is %d" number (factorial number))))
If you change cl-labels to cl-flet you will get an error as soon as the inner (factorial (1- n)) is evaluated, because within our temporary function, no function factorial is known.
If you were to then define a global factorial function which unconditionally returns the value 1:
(defun factorial (n) 1)
Then the factorial function defined by cl-flet will see that when it calls factorial, and my-factorial would calculate (* n 1) as the value for any argument n.
When no recursion is needed, cl-flet is fine to use:
(defun my-square (number)
"Show the square of the argument."
(interactive "nSquare of: ")
(cl-flet ((square (n) (* n n)))
(message "Square of %d is %d" number (square number))))
Both cl-labels and cl-flet provide lexically-scoped functions, visible only to the code written within the bodies of those macro calls; and in particular not to the code of any other functions which we may be calling.
If you are defining a helper function such as in the examples above, lexical scoping is probably what you want, as there's a reasonable chance you'll only be calling your helper within the macro body.
If you are trying to temporarily override an existing function, however, there's a pretty fair chance you'll need the functions you're calling to see the override. In such cases you need the override to have dynamic scope.
In the past flet was the way to provide dynamic scope for temporary functions, but flet is now deprecated in favour of using cl-letf with a 'place' of (symbol-function 'FUNC)
In the following examples, the multiplication function is overridden, and the dynamic scope means that my-square and my-factorial will see and use our temporary definition of multiplication.
(defun my-bad-square ()
"Maths gone wrong."
(interactive)
(cl-letf (((symbol-function '*) '+))
(call-interactively 'my-square)))
(defun my-bad-factorial ()
"More maths gone wrong."
(interactive)
(cl-letf (((symbol-function '*)
(lambda (x y) (- x y))))
(call-interactively 'my-factorial)))
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))
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.
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 want to make a macro for binding variables to values given a var-list and a val-list.
This is my code for it -
(defmacro let-bind (vars vals &body body)
`(let ,(loop for x in vars
for y in vals
collect `(,x ,y))
,#body))
While it works correct if called like (let-bind (a b) (1 2) ...), it doesn't seem to work when called like
(defvar vars '(a b))
(defvar vals '(1 2))
(let-bind vars vals ..)
Then I saw some effects for other of my macros too. I am a learner and cannot find what is wrong.
Basic problem: a macro sees code, not values. A function sees values, not code.
CL-USER 2 > (defvar *vars* '(a b))
*VARS*
CL-USER 3 > (defvar *vals* '(1 2))
*VALS*
CL-USER 4 > (defmacro let-bind (vars vals &body body)
(format t "~%the value of vars is: ~a~%" vars)
`(let ,(loop for x in vars
for y in vals
collect `(,x ,y))
,#body))
LET-BIND
CL-USER 5 > (let-bind *vars* *vals* t)
the value of vars is: *VARS*
Error: *VARS* (of type SYMBOL) is not of type LIST.
1 (abort) Return to top loop level 0.
You can see that the value of vars is *vars*. This is a symbol. Because the macro variables are bound to code fragments - not their values.
Thus in your macro you try to iterate over the symbol *vars*. But *vars* is a symbol and not a list.
You can now try to evaluate the symbol *vars* at macro expansion time. But that won't work also in general, since at macro expansion time *vars* may not have a value.
Your macro expands into a let form, but let expects at compile time real variables. You can't compute the variables for let at a later point in time. This would work only in some interpreted code where macros would be expanded at runtime - over and over.
If you’ve read the other answers then you know that you can’t read a runtime value from a compiletime macro (or rather, you can’t know the value it will have at runtime at compiletime as you can’t see the future). So let’s ask a different question: how can you bind the variables in your list known at runtime.
In the case where your list isn’t really variable and you just want to give it a single name you could use macroexpand:
(defun symbol-list-of (x env)
(etypecase x
(list x)
(symbol (macroexpand x env))))
(defmacro let-bind (vars vals &body body &environment env)
(let* ((vars (symbol-list-of vars env))
(syms (loop for () in vars collect gensym)))
`(destructuring-bind ,syms ,vals
(let ,(loop for sym in syms for bar in vars collect (list var sym)) ,#body))))
This would somewhat do what you want. It will symbol-macroexpand the first argument and evaluate the second.
What if you want to evaluate the first argument? Well we could try generating something that uses eval. As eval will evaluate in the null lexical environment (ie can’t refer to any external local variables), we would need to have eval generate a function to bind variables and then call another function. That is a function like (lambda (f) (let (...) (funcall f)). You would evaluate the expression to get that function and then call it with a function which does he body (but was not made by eval and so captures the enclosing scope). Note that this would mean that you could only bind dynamic variables.
What if you want to bind lexical variables? Well there is no way to go from symbol to the memory location of a variable at runtime in Common Lisp. A debugger might know how to do this. There is no way to get a list of variables in scope in a macro, although the compiler knows this. So you can’t generate a function to set a lexically bound symbol. And it would be even harder to do if you wanted to shadow the binding although you could maybe do it with some symbol-macrolet trickery if you knew every variable in scope.
But maybe there is a better way to do this for special variables and it turns out there is. It’s an obscure special form called progv. It has the same signature that you want let-bind to have except it works. link.
So I can do this:
(defparameter *some-function* ... ; returns lambda later
or this:
(defun some-function ...
With either, I can use funcall:
(funcall 'some-function ...
or
(funcall *some-function* ...
With the defun version I can also do this:
(some-function ...
I cannot do that with the defparameter function.
defparameter provides easier technique for re-assigning some-function to a different function (or anything else, including non-function data) later.
But other than these two points, what are other considerations of using one over another?
This is an odd one to answer as we are somewhat comparing apples with oranges.
For those new to lisp who are looking at this, defparameter is for defining a dynamic variable whereas defun is for defining a function.
If you are worried about being able to programmatically reassign a function without using defun check out the following:
CL-USER> (defun jam () (print 'some-jam))
JAM
CL-USER> (jam)
SOME-JAM
CL-USER> (setf (symbol-function 'jam) (lambda () (print 'some-ham)))
#<FUNCTION (LAMBDA ()) {1004C033DB}>
CL-USER> (jam)
SOME-HAM
So defparameter doesn’t have an advantage when it comes to reassigning a function. Also if you want to redefine the function you could look into the compile command.