Here is the short lambda macro definition taken from subr.el.
(defmacro lambda (&rest cdr)
(list 'function (cons 'lambda cdr)))
The fact, this is a recursive macro puts me in a fix. can someone explain, how this works & what it returns:
(list 'function (cons 'lambda cdr))
The reason this is not a recursive macro is that function acts like quote in that it prevents evaluation of its argument. Therefore, a recursive call to the lambda macro will not be made.
The difference between function and quote is that function permits byte compilation of its argument, while quote always preserves it verbatim. Thus, if you write (lambda () 1), it will be expanded to (function (lambda () 1)), and then replaced with byte code by the compiler.
It is not a recursive macro. Emacs Lisp macros have the form (SYMBOL...). It is only such a form that gets defined as a macro, and which is expanded. The occurrence of the symbol lambda in the macro definition body is not expanded. It is a ( followed by the symbol lambda and a possibly empty list of sexps, and then a ) that is matched and expanded.
The macro expansion returns the list (function (lambda CDR)), where CDR is the (unevaluated) list of sexps passed as args to the macro. When that list is evaluated it returns the list (lambda CDR), with the same (unevaluated) CDR.
Related
I'm having a problem passing a list to macro, where the list will be used to generate function name. For example, the code below causes an error.
(defmacro gen (str-lst)
`(defun ,(intern (string-upcase (car str-lst))) () (print "foo")))
(gen '("foo" "bar"))
The resulting error was:
*** - DEFUN/DEFMACRO: QUOTE is a special operator and may not be redefined. The following restarts are available: ABORT :R1
Abort main loop
How should I modify my code, and what is wrong with my code?
The thing makes me even more confused is that the code below, about which answer exits here, works fine.
(defmacro easy-one (str-lst)
`(mapc #'(lambda (str) (print str)) ,str-lst))
(easy-one '("foo" "bar"))
Don't quote the list. Macros don't evaluate their arguments, so you don't need to quote them to prevent them from being evaluated, like you do for ordinary functions.
(gen ("foo" "bar"))
When you quote it, you're executing
(get (quote ("foo" "bar")))
The value of str-list is the list (quote ("foo" "bar")), so (car str-list) is the symbol QUOTE. As a result, the macro expands to
(defun quote () (print "foo"))
That's why you get an error complaining that you're trying to redefine the built-in QUOTE.
The difference in your second example is that you're just substituting the parameter into the expansion, you're not using its value in the expansion code. So it expands to
(mapc #'(lambda (str) (print str)) '("foo" "bar")))
Here the list will be used when the expansion runs, not while the macro is being expanded. It needs to be quoted there, to prevent it from being evaluated as a function call.
You should use macroexpand to see how your macros are being expanded when debugging.
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 was told that I shouldn't quote lambda in, say,
(global-set-key (quote [f3]) '(lambda () (interactive) (other-window -1) ))
I tried that indeed if I don't quote lambda, it works equally well
(global-set-key (quote [f3]) (lambda () (interactive) (other-window -1) ))
However, I don't understand why the latter works (and is also being preferred, and now that the latter works, why the former also works).
If the lambda expression is defined as an other function, we would have called
(global-set-key (quote [f3]) 'my-function)
to prevent my-function to be evaluated immediately. I understand the lambda expression as an anonymous version of my-function. So why shouldn't lambda be quoted?
Thanks!
Using C-h f lambda <RET>:
A call of the form (lambda ARGS DOCSTRING INTERACTIVE BODY)
is self-quoting; the result of evaluating the lambda expression
is the expression itself.
So, this answers the question, why you don't need to quote the lambda expression. As to why you shouldn't do it... I think, this has to do with byte compilation. A quoted lambda expression is simply plain data. The byte code compiler has no choice but to include the expression as a constant list literal into its output. An unquoted lambda expression, on the other hand, can be compiled to byte code, resulting in faster execution.
List literals of the form (lambda (...) ...) are special-cased in emacs lisp evaluator and can be used as functions. That's why it works, regardless of whether you quote the lambda expression or not.
I have a program that takes as inputs a chunk of data and a list of rules, applying both a set of standard rules and the rules given as input to the chunk of data. The size of both inputs may vary.
I want to be able to write a list of rules like this:
(rule-generating-macro
(rule-1-name rule-1-target
(rule-action-macro (progn actions more-actions)))
(rule-2-name rule-2-target
(rule-action-macro (or (action-2) (default-action))))
;; more rules
)
Right now, rules are more verbose -- they look more like
(defvar rule-list
`((rule-1-name rule-1-target
,#(rule-action-macro (progn actions more-actions)))
(rule-2-name rule-2-target
,#(rule-action-macro (or (action-2) (default-action))))
;; more rules
)
The latter form looks uglier to me, but I can't figure out how to write a macro that can handle a variable-length &rest argument, iterate over it, and return the transformed structure. Using a defun instead of a defmacro isn't really on the table because (as hopefully the example shows) I'm trying to control evaluation of the list of rules instead of evaluating the list when my program first sees it, and once you need to control evaluation, you're in defmacro territory. In this case, the thorny point is the rule-action-macro part - getting the interpreter to read that and use its expanded value has been problematic.
How can I create a macro that handles a variable-length argument so that I can write rule lists in a concise way?
defmacro will happily accept a &rest argument
(see Defining Macros for Emacs Lisp and Macro Lambda Lists for Common Lisp).
Then you can do pretty much anything you want with it in the macro body - e.g., iterate over it. Remember, macro is much more than just backquote!
E.g.:
(defmacro multidefvar (&rest vars)
(let ((forms (mapcar (lambda (var) `(defvar ,var)) vars)))
`(progn ,#forms)))
(macroexpand '(multidefvar a b c d))
==> (PROGN (DEFVAR A) (DEFVAR B) (DEFVAR C) (DEFVAR D))
I have following code which confuse me now, I hope some can tell me the difference and how to fix this.
(defmacro tm(a)
`(concat ,(symbol-name a)))
(defun tf(a)
(list (quote concat) (symbol-name a)))
I just think they should be the same effect, but actually they seem not.
I try to following call:
CL-USER> (tf 'foo)
(CONCAT "FOO")
CL-USER> (tm 'foo)
value 'FOO is not of the expected type SYMBOL.
[Condition of type TYPE-ERROR]
So, what's the problem?
What i want is:
(tm 'foo) ==> (CONCAT "FOO")
The first problem is that 'foo is expanded by the reader to (quote foo), which is not a symbol, but a list. The macro tries to expand (tm (quote foo)). The list (quote foo) is passed as the parameter a to the macro expansion function, which tries to get its symbol-name. A list is not a valid argument for symbol-name. Therefore, your macro expansion fails.
The second problem is that while (tm foo) (note: no quote) does expand to (concat "FOO"), this form will then be executed by the REPL, so that this is also not the same as your tf function. This is not surprising, of course, because macros do different things than functions.
First, note that
`(concat ,(symbol-name a))
and
(list (quote concat) (symbol-name a))
do the exact same thing. They are equivalent pieces of code (backquote syntax isn't restricted to macro bodies!): Both construct a list whose first element is the symbol CONCAT and whose second element is the symbol name of whatever the variable A refers to.
Clearly, this only makes sense if A refers to a symbol, which, as Svante has pointed out, isn't the case in the macro call example.
You could, of course, extract the symbol from the list (QUOTE FOO), but that prevents you from calling the macro like this:
(let ((x 'foo))
(tm x))
which raises the question of why you would event want to force the user of the macro to explicitly quote the symbol where it needs to be a literal constant anyway.
Second, the way macros work is this: They take pieces of code (such as (QUOTE FOO)) as arguments and produce a new piece of code that, upon macroexpansion, (more or less) replaces the macro call in the source code. It is often useful to reuse macro arguments within the generated code by putting them where they are going to be evaluated later, such as in
(defmacro tm2 (a)
`(print (symbol-name ,a)))
Think about what this piece of code does and whether or not my let example above works now. That should get you on the right track.
Finally, a piece of advice: Avoid macros when a function will do. It will make life much easier for both the implementer and the user.