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Macros enable you to define new control constructs and other language features. A macro is defined much like a function, but instead of telling how to compute a value, it tells how to compute another Lisp expression which will in turn compute the value. We call this expression the expansion of the macro.
Macros can do this because they operate on the unevaluated expressions for the arguments, not on the argument values as functions do. They can therefore construct an expansion containing these argument expressions or parts of them.
If you are using a macro to do something an ordinary function could do, just for the sake of speed, consider using an inline function instead. See section Inline Functions.
Suppose we would like to define a Lisp construct to increment a
variable value, much like the
++ operator in C. We would like to
(inc x) and have the effect of
(setq x (1+ x)).
Here's a macro definition that does the job:
(defmacro inc (var) (list 'setq var (list '1+ var)))
When this is called with
(inc x), the argument
x---not the value of
x. The body
of the macro uses this to construct the expansion, which is
x (1+ x)). Once the macro definition returns this expansion, Lisp
proceeds to evaluate it, thus incrementing
A macro call looks just like a function call in that it is a list which starts with the name of the macro. The rest of the elements of the list are the arguments of the macro.
Evaluation of the macro call begins like evaluation of a function call except for one crucial difference: the macro arguments are the actual expressions appearing in the macro call. They are not evaluated before they are given to the macro definition. By contrast, the arguments of a function are results of evaluating the elements of the function call list.
Having obtained the arguments, Lisp invokes the macro definition just
as a function is invoked. The argument variables of the macro are bound
to the argument values from the macro call, or to a list of them in the
case of a
&rest argument. And the macro body executes and
returns its value just as a function body does.
The second crucial difference between macros and functions is that the value returned by the macro body is not the value of the macro call. Instead, it is an alternate expression for computing that value, also known as the expansion of the macro. The Lisp interpreter proceeds to evaluate the expansion as soon as it comes back from the macro.
Since the expansion is evaluated in the normal manner, it may contain calls to other macros. It may even be a call to the same macro, though this is unusual.
You can see the expansion of a given macro call by calling
Function: macroexpand form &optional environment
This function expands form, if it is a macro call. If the result
is another macro call, it is expanded in turn, until something which is
not a macro call results. That is the value returned by
macroexpand. If form is not a macro call to begin with, it
is returned as given.
macroexpand does not look at the subexpressions of
form (although some macro definitions may do so). Even if they
are macro calls themselves,
macroexpand does not expand them.
macroexpand does not expand calls to inline functions.
Normally there is no need for that, since a call to an inline function is
no harder to understand than a call to an ordinary function.
If environment is provided, it specifies an alist of macro definitions that shadow the currently defined macros. This is used by byte compilation.
(defmacro inc (var) (list 'setq var (list '1+ var))) => inc (macroexpand '(inc r)) => (setq r (1+ r)) (defmacro inc2 (var1 var2) (list 'progn (list 'inc var1) (list 'inc var2))) => inc2 (macroexpand '(inc2 r s)) => (progn (inc r) (inc s)) ;
incnot expanded here.
You might ask why we take the trouble to compute an expansion for a macro and then evaluate the expansion. Why not have the macro body produce the desired results directly? The reason has to do with compilation.
When a macro call appears in a Lisp program being compiled, the Lisp compiler calls the macro definition just as the interpreter would, and receives an expansion. But instead of evaluating this expansion, it compiles the expansion as if it had appeared directly in the program. As a result, the compiled code produces the value and side effects intended for the macro, but executes at full compiled speed. This would not work if the macro body computed the value and side effects itself--they would be computed at compile time, which is not useful.
In order for compilation of macro calls to work, the macros must be
defined in Lisp when the calls to them are compiled. The compiler has a
special feature to help you do this: if a file being compiled contains a
defmacro form, the macro is defined temporarily for the rest of
the compilation of that file. To use this feature, you must define the
macro in the same file where it is used and before its first use.
While byte-compiling a file, any
require calls at top-level are
executed. One way to ensure that necessary macro definitions are
available during compilation is to require the file that defines them.
See section Features.
A Lisp macro is a list whose CAR is
macro. Its CDR should
be a function; expansion of the macro works by applying the function
apply) to the list of unevaluated argument-expressions
from the macro call.
It is possible to use an anonymous Lisp macro just like an anonymous
function, but this is never done, because it does not make sense to pass
an anonymous macro to mapping functions such as
practice, all Lisp macros have names, and they are usually defined with
the special form
Special Form: defmacro name argument-list body-forms...
defmacro defines the symbol name as a macro that looks
(macro lambda argument-list . body-forms)
This macro object is stored in the function cell of name. The
value returned by evaluating the
defmacro form is name, but
usually we ignore this value.
The shape and meaning of argument-list is the same as in a
function, and the keywords
&optional may be used
(see section Advanced Features of Argument Lists). Macros may have a documentation string, but
interactive declaration is ignored since macros cannot be
It could prove rather awkward to write macros of significant size,
simply due to the number of times the function
list needs to be
called. To make writing these forms easier, a macro ``'
(often called backquote) exists.
Backquote allows you to quote a list, but selectively evaluate
elements of that list. In the simplest case, it is identical to the
quote (see section Quoting). For example, these
two forms yield identical results:
(` (a list of (+ 2 3) elements)) => (a list of (+ 2 3) elements) (quote (a list of (+ 2 3) elements)) => (a list of (+ 2 3) elements)
By inserting a special marker, `,', inside of the argument to backquote, it is possible to evaluate desired portions of the argument:
(list 'a 'list 'of (+ 2 3) 'elements) => (a list of 5 elements) (` (a list of (, (+ 2 3)) elements)) => (a list of 5 elements)
It is also possible to have an evaluated list spliced into the
resulting list by using the special marker `,@'. The elements of
the spliced list become elements at the same level as the other elements
of the resulting list. The equivalent code without using
often unreadable. Here are some examples:
(setq some-list '(2 3)) => (2 3) (cons 1 (append some-list '(4) some-list)) => (1 2 3 4 2 3) (` (1 (,@ some-list) 4 (,@ some-list))) => (1 2 3 4 2 3) (setq list '(hack foo bar)) => (hack foo bar) (cons 'use (cons 'the (cons 'words (append (cdr list) '(as elements))))) => (use the words foo bar as elements) (` (use the words (,@ (cdr list)) as elements (,@ nil))) => (use the words foo bar as elements)
The reason for
(,@ nil) is to avoid a bug in Emacs version 18.
The bug occurs when a call to
,@ is followed only by constant
(` (use the words (,@ (cdr list)) as elements))
would not work, though it really ought to.
(,@ nil) avoids the
problem by being a nonconstant element that does not affect the result.
Macro: ` list
This macro returns list as
quote would, except that the
list is copied each time this expression is evaluated, and any sublist
of the form
(, subexp) is replaced by the value of
subexp. Any sublist of the form
is replaced by evaluating listexp and splicing its elements
into the containing list in place of this sublist. (A single sublist
can in this way be replaced by any number of new elements in the
There are certain contexts in which `,' would not be recognized and should not be used:
;; Use of a `,' expression as the CDR of a list. (` (a . (, 1))) ; Not
(a . 1)=> (a \, 1) ;; Use of `,' in a vector. (` [a (, 1) c]) ; Not
[a 1 c]error--> Wrong type argument ;; Use of a `,' as the entire argument of ``'. (` (, 2)) ; Not 2 => (\, 2)
Common Lisp note: in Common Lisp, `,' and `,@' are implemented as reader macros, so they do not require parentheses. Emacs Lisp implements them as functions because reader macros are not supported (to save space).
The basic facts of macro expansion have all been described above, but there consequences are often counterintuitive. This section describes some important consequences that can lead to trouble, and rules to follow to avoid trouble.
When defining a macro you must pay attention to the number of times the arguments will be evaluated when the expansion is executed. The following macro (used to facilitate iteration) illustrates the problem. This macro allows us to write a simple "for" loop such as one might find in Pascal.
(defmacro for (var from init to final do &rest body) "Execute a simple \"for\" loop, e.g., (for i from 1 to 10 do (print i))." (list 'let (list (list var init)) (cons 'while (cons (list '<= var final) (append body (list (list 'inc var))))))) => for (for i from 1 to 3 do (setq square (* i i)) (princ (format "\n%d %d" i square))) ==> (let ((i 1)) (while (<= i 3) (setq square (* i i)) (princ (format "%d %d" i square)) (inc i))) -|1 1 -|2 4 -|3 9 => nil
do in this macro are
"syntactic sugar"; they are entirely ignored. The idea is that you
will write noise words (such as
in those positions in the macro call.)
This macro suffers from the defect that final is evaluated on
every iteration. If final is a constant, this is not a problem.
If it is a more complex form, say
this can slow down the execution significantly. If final has side
effects, executing it more than once is probably incorrect.
A well-designed macro definition takes steps to avoid this problem by
producing an expansion that evaluates the argument expressions exactly
once unless repeated evaluation is part of the intended purpose of the
macro. Here is a correct expansion for the
(let ((i 1) (max 3)) (while (<= i max) (setq square (* i i)) (princ (format "%d %d" i square)) (inc i)))
Here is a macro definition that creates this expansion:
(defmacro for (var from init to final do &rest body) "Execute a simple for loop: (for i from 1 to 10 do (print i))." (` (let (((, var) (, init)) (max (, final))) (while (<= (, var) max) (,@ body) (inc (, var))))))
Unfortunately, this introduces another problem.
The new definition of
for has a new problem: it introduces a
local variable named
max which the user does not expect. This
causes trouble in examples such as the following:
(let ((max 0)) (for x from 0 to 10 do (let ((this (frob x))) (if (< max this) (setq max this)))))
The references to
max inside the body of the
are supposed to refer to the user's binding of
max, really access
the binding made by
The way to correct this is to use an uninterned symbol instead of
max (see section Creating and Interning Symbols). The uninterned symbol can be
bound and referred to just like any other symbol, but since it is created
for, we know that it cannot appear in the user's program.
Since it is not interned, there is no way the user can put it into the
program later. It will never appear anywhere except where put by
for. Here is a definition of
for which works this way:
(defmacro for (var from init to final do &rest body) "Execute a simple for loop: (for i from 1 to 10 do (print i))." (let ((tempvar (make-symbol "max"))) (` (let (((, var) (, init)) ((, tempvar) (, final))) (while (<= (, var) (, tempvar)) (,@ body) (inc (, var)))))))
This creates an uninterned symbol named
max and puts it in the
expansion instead of the usual interned symbol
max that appears
in expressions ordinarily.
Another problem can happen if you evaluate any of the macro argument
expressions during the computation of the expansion, such as by calling
eval (see section Eval). If the argument is supposed to refer to the
user's variables, you may have trouble if the user happens to use a
variable with the same name as one of the macro arguments. Inside the
macro body, the macro argument binding is the most local binding of this
variable, so any references inside the form being evaluated do refer
to it. Here is an example:
(defmacro foo (a) (list 'setq (eval a) t)) => foo (setq x 'b) (foo x) ==> (setq b t) => t ; and
bhas been set. ;; but (setq a 'b) (foo a) ==> (setq 'b t) ; invalid! error--> Symbol's value is void: b
It makes a difference whether the user types
a conflicts with the macro argument variable
In general it is best to avoid calling
eval in a macro
definition at all.
Occasionally problems result from the fact that a macro call is expanded each time it is evaluated in an interpreted function, but is expanded only once (during compilation) for a compiled function. If the macro definition has side effects, they will work differently depending on how many times the macro is expanded.
In particular, constructing objects is a kind of side effect. If the macro is called once, then the objects are constructed only once. In other words, the same structure of objects is used each time the macro call is executed. In interpreted operation, the macro is reexpanded each time, producing a fresh collection of objects each time. Usually this does not matter--the objects have the same contents whether they are shared or not. But if the surrounding program does side effects on the objects, it makes a difference whether they are shared. Here is an example:
(defmacro new-object () (list 'quote (cons nil nil))) (defun initialize (condition) (let ((object (new-object))) (if condition (setcar object condition)) object))
initialize is interpreted, a new list
constructed each time
initialize is called. Thus, no side effect
survives between calls. If
initialize is compiled, then the
new-object is expanded during compilation, producing a
(nil) that is reused and altered each time
initialize is called.
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