What is the difference between
(function (lambda ...))
and
(lambda ...)
and
'(lambda ...)
?
It seems three are interchangeable in a lot of cases.
They are pretty interchangeable. The answer is that function enables the lambda to be byte compiled, whereas the other two do not (and are equivalent). Note: this does not mean that function actually byte compile the lambda.
How might one figure that out? A little Emacs lisp introspection provides some clues. To start: C-h f function RET:
function is a special form in 'C
source code'.
(function arg)
Like 'quote', but preferred for
objects which are functions. In byte
compilation, 'function' causes its
argument to be compiled. 'quote'
cannot do that.
Ok, so that's the difference between (function (lambda ...)) and '(lambda ...), the first tells the byte compiler that it may safely compile the expression. Whereas the 'ed expressions may not necessarily be compiled (for they might just be a list of numbers.
What about just the bare (lambda ...)? C-h f lambda RET shows:
lambda is a Lisp macro in `subr.el'.
(lambda args [docstring] [interactive]
body)
Return a lambda expression. A call of
the form (lambda args docstring
interactive body) is self-quoting; the
result of evaluating the lambda
expression is the expression itself.
The lambda expression may then be
treated as a function, i.e., stored as
the function value of a symbol, passed
to 'funcall' or 'mapcar', etc.
Therefore, (lambda ...) and '(lambda ...) are equivalent.
Also, there is the notation #'(lambda ...), which is syntactic sugar for (function (lambda ...)).
For more information on functions in Emacs lisp, read the Functions info pages.
Just to check all this, you can type the following into the *scratch* buffer and evaluate the expressions:
(caddr '(lambda (x) (+ x x)))
(+ x x)
(caddr (lambda (x) (+ x x)))
(+ x x)
(caddr (function (lambda (x) (+ x x))))
(+ x x)
(equal '(lambda (x) (+ x x))
(function (lambda (x) (+ x x))))
t
(equal '(lambda (x) (+ x x))
(lambda (x) (+ x x)))
t
So, all three variants of using lambda just build up lists that may be used as functions (one of which may be byte compiled).
Well (quote (lambda...)) and (lambda...) are not equivalent (when byte compiling). Quoted lambdas are not byte compiled whereas everything else is.
For example:
(defun foo (a)
(byte-code-function-p a))
(defun bar ()
(foo '(lambda () (ignore 'me))))
(defun bar2 ()
(foo (lambda () (ignore 'me))))
(defun bar3 ()
(foo (function (lambda () (ignore 'me)))))
(defun bar4 ()
(foo #'(lambda () (ignore 'me))))
(byte-compile 'bar)
(byte-compile 'bar2)
(byte-compile 'bar3)
(byte-compile 'bar4)
(bar) ; -> nil
(bar2) ; -> t
(bar3) ; -> t
(bar4) ; -> t
You usually don't want to quote a lambda unless the function you are going to pass the lambda to is doing something else with it than just funcall it.
Related
What is the fundamental difference in the functions defined using defun and setf as below and is one method preferred over another outside of style considerations?
Using defun:
* (defun myfirst (l)
(car l) )
MYFIRST
* (myfirst '(A B C))
A
Using setf:
* (setf (fdefinition 'myfirst) #'(lambda (l) (car l)))
#<FUNCTION (LAMBDA (L)) {10021B477B}>
* (myfirst '(A B C))
A
If, as according to Wikipedia:
named functions are created by storing a lambda expression in a symbol using the defun macro
Using setf to create a variable in a different way requires the use of funcall:
* (defvar myfirst)
MYFIRST
* (setf myfirst (lambda (l) (car l)))
#<Interpreted Function (LAMBDA (X) (+ X X)) {48035001}>
* (funcall myfirst '(A B C))
A
My understanding is that this type of variable is different than the previous in that this variable is not found in the same namespace as the defun bound symbol as described in Why multiple namespaces?.
First of all, one should never underestimate the importance of style.
We write code not just for computers to run, but, much more importantly, for people to read.
Making code readable and understandable for people is a very important aspect of software development.
Second, yes, there is a big difference between (setf fdefinition) and defun.
The "small" differences are that defun can also set the doc string of the function name (actually, depending on how your imeplementation works, it might do that with lambda also), and creates a named block (seen in the macroexpansions below) which you would otherwise have to create yourself if you want to.
The big difference is that the compiler "knows" about defun and will process it appropriately.
E.g., if your file is
(defun foo (x)
(+ (* x x) x 1))
(defun bar (x)
(+ (foo 1 2 x) x))
then the compiler will probably warn you that you call foo in bar with the wrong number of arguments:
WARNING: in BAR in lines 3..4 : FOO was called with 3 arguments, but it requires 1
argument.
[FOO was defined in lines 1..2 ]
If you replace the defun foo with (setf (fdefinition 'foo) (lambda ...)), the compiler is unlikely to handle it as carefully. Moreover, you will probably get a warning along the lines of
The following functions were used but not defined:
FOO
You might want to examine what defun does in your implementation by macroexpanding it:
(macroexpand-1 '(defun foo (x) "doc" (print x)))
CLISP expands it to
(LET NIL (SYSTEM::REMOVE-OLD-DEFINITIONS 'FOO)
(SYSTEM::EVAL-WHEN-COMPILE
(SYSTEM::C-DEFUN 'FOO (SYSTEM::LAMBDA-LIST-TO-SIGNATURE '(X))))
(SYSTEM::%PUTD 'FOO
(FUNCTION FOO
(LAMBDA (X) "doc" (DECLARE (SYSTEM::IN-DEFUN FOO)) (BLOCK FOO (PRINT X)))))
(EVAL-WHEN (EVAL)
(SYSTEM::%PUT 'FOO 'SYSTEM::DEFINITION
(CONS '(DEFUN FOO (X) "doc" (PRINT X)) (THE-ENVIRONMENT))))
'FOO)
SBCL does:
(PROGN
(EVAL-WHEN (:COMPILE-TOPLEVEL) (SB-C:%COMPILER-DEFUN 'FOO NIL T))
(SB-IMPL::%DEFUN 'FOO
(SB-INT:NAMED-LAMBDA FOO
(X)
"doc"
(BLOCK FOO (PRINT X)))
(SB-C:SOURCE-LOCATION)))
The point here is that defun has a lot "under the hood", and for a reason. setf fdefinition is, on the other hand, more of "what you see is what you get", i.e., no magic involved.
This does not mean that setf fdefinition has no place in a modern lisp codebase. You can use it, e.g., to implement a "poor man's trace" (UNTESTED):
(defun trace (symbol)
(setf (get symbol 'old-def) (fdefinition symbol)
(fdefinition symbol)
(lambda (&rest args)
(print (cons symbol args))
(apply (get symbol 'old-def) args))))
(defun untrace (symbol)
(setf (fdefinition symbol) (get symbol 'old-def))
(remprop symbol 'odd-def))
I'm trying to understand the following two snippets of code:
(defun make-adder1 (n) `(lambda (x) (+ ,n x)))
(defun make-adder2 (n) (lexical-let ((n n)) (lambda (x) (+ n x))))
These both seem to produce callables:
(funcall (make-adder1 3) 5) ;; returns 8
(funcall (make-adder2 3) 5) ;; returns 8
These both work. I have two main questions:
1) I don't understand the disparity in "quoting level" between the two approaches. In the first case, the lambda expression is quoted, which means the "symbol itself" is returned instead of the value. In the second case, it seems like the statement with the lambda will get evaluated, so the value of the lambda will be returned. Yet, these both work with funcall. When using funcall on a defun'ed function, it has to be quoted. Is lexical-let doing some kind of quoting automatically? Isn't this, kind of surprising?
2) Reading other posts on this topic, I'm given to understand that the first approach will break down under certain circumstances and deviate from what one would expect from working with lambdas and higher order functions in other languages, because elisp has dynamic scoping by default. Can someone give a concrete example of code that makes this difference apparent and explain it?
In the first example there is no variable n in the resulting function, which is just (lambda (x) (+ 3 x)). It does not need lexical binding because there is no free variable in the lambda, i.e., no variable that needs to be kept in a binding of a closure. If you don't need the variable n to be available, as a variable in uses of the function, i.e., if its value at function definition time (=3) is all you need, then the first example is all you need.
(fset 'ad1 (make-adder1 3))
(symbol-function 'ad1)
returns:
(lambda (x) (+ 3 x))
The second example creates what is, in effect, a function that creates and applies a complicated closure.
(fset 'ad2 (make-adder2 3))
(symbol-function 'ad2)
returns
(lambda (&rest --cl-rest--)
(apply (quote (closure ((--cl-n-- . --n--) (n . 3) t)
(G69710 x)
(+ (symbol-value G69710) x)))
(quote --n--)
--cl-rest--))
A third option is to use a lexical-binding file-local variable and use the most straightforward definition. This creates a simple closure.
;;; foo.el --- toto -*- lexical-binding: t -*-
(defun make-adder3 (n) (lambda (x) (+ n x)))
(fset 'ad3 (make-adder3 3))
(symbol-function 'ad3)
returns:
(closure ((n . 3) t) (x) (+ n x))
(symbol-function 'make-adder1)
returns:
(lambda (n)
(list (quote lambda)
(quote (x))
(cons (quote +) (cons n (quote (x))))))
(symbol-function 'make-adder2)
returns:
(closure (t)
(n)
(let ((--cl-n-- (make-symbol "--n--")))
(let* ((v --cl-n--)) (set v n))
(list (quote lambda)
(quote (&rest --cl-rest--))
(list (quote apply)
(list (quote quote)
(function
(lambda (G69709 x)
(+ (symbol-value G69709) x))))
(list (quote quote) --cl-n--)
(quote --cl-rest--)))))
(symbol-function 'make-adder3)
returns
(closure (t) (n) (function (lambda (x) (+ n x))))
Assume that the macro would take the boolean types a and b . If a is nil, then the macro should return nil (without ever evaluating b), otherwise it returns b. How do you do this?
This really depends on what you can use.
E.g., is or available? if? cond?
Here is one example:
(defmacro and (a b)
`(if ,a ,b nil)
EDIT. In response to a comment, or is more complicated because we have to avoid double evaluation:
(defmacro or (a b)
(let ((v (gensym "OR")))
`(let ((,v ,a))
(if ,v ,v ,b))))
sds's answer is nice and concise, but it has two limitations:
It only works with two arguments, whereas the built in and and or take any number of arguments. It's not too hard to update the solution to take any number of arguments, but it would be a bit more complicated.
More importantly, it's based very directly in terms of delayed operations that are already present in the language. I.e., it takes advantage of the fact that if doesn't evaluate the then or else parts until it has first evaluated the condition.
It might be a good exercise, then, to note that when a macro needs to delay evaluation of some forms, it's often the simplest strategy (in terms of implementation, but not necessarily the most efficient) to use a macro that expands to a function call that takes a function. For instance, a naive implementation of with-open-file might be:
(defun %call-with-open-file (pathname function)
(funcall function (open pathname)))
(defmacro my-with-open-file ((var pathname) &body body)
`(%call-with-open-file
,pathname
(lambda (,var)
,#body)))
Using a technique like this, you can easily get a binary and (and or):
(defun %and (a b)
(if (funcall a)
(funcall b)
nil))
(defmacro my-and (a b)
`(%and (lambda () ,a)
(lambda () ,b)))
CL-USER> (my-and t (print "hello"))
"hello" ; printed output
"hello" ; return value
CL-USER> (my-and nil (print "hello"))
NIL
or is similar:
(defun %or (a b)
(let ((aa (funcall a)))
(if aa
aa
(funcall b))))
(defmacro my-or (a b)
`(%or (lambda () ,a)
(lambda () ,b)))
To handle the n-ary case (since and and or actually take any number of arguments), you could write a function that takes a list of lambda functions and calls each of them until you get to one that would short circuit (or else reaches the end). Common Lisp actually already has functions like that: every and some. With this approach, you could implement and in terms of every by wrapping all the arguments in lambda functions:
(defmacro my-and (&rest args)
`(every #'funcall
(list ,#(mapcar #'(lambda (form)
`(lambda () ,form))
args))))
For instance, with this implementation,
(my-and (listp '()) (evenp 3) (null 'x))
expands to:
(EVERY #'FUNCALL
(LIST (LAMBDA () (LISTP 'NIL))
(LAMBDA () (EVENP 3))
(LAMBDA () (NULL 'X))))
Since all the forms are now wrapped in lambda functions, they won't get called until every gets that far.
The only difference is that and is specially defined to return the value of the last argument if all the preceding ones are true (e.g., (and t t 3) returns 3, not t, whereas the specific return value of every is not specified (except that it would be a true value).
With this approach, implementing or (using some) is no more complicated than implementing and:
(defmacro my-or (&rest args)
`(some #'funcall ,#(mapcar #'(lambda (form)
`(lambda () ,form))
args)))
I'm reading Paul Graham's ANSI Common Lisp. In the chapter about macros he shows the following example:
(defmacro in (obj &rest choices)
(let ((insym (gensym)))
`(let ((,insym ,obj))
(or ,#(mapcar #'(lambda (c) `(eql ,insym ,c))
choices)))))
(Returns true if the first argument is equal to any of the other arguments)
He holds that it can't be written as a function. Wouldn't this function have the same functionality?
(defun in (obj &rest choices)
(reduce (lambda (x y)
(or x (eql y obj)))
choices
:initial-value nil))
The difference I see is that the macro will only evaluate arguments till it finds an eql argument. Is that it?
The point is, that the macro version evaluates the arguments lazily (it expands into an OR) stopping if a match is found. This cannot be achieved with a function, since a funcall will always evaluate all arguments first.
> (macroexpand '(in 42
(long-computation-1)
(long-computation-2)
(long-computation-3)))
(LET ((#:G799 42))
(OR (EQL #:G799 (LONG-COMPUTATION-1))
(EQL #:G799 (LONG-COMPUTATION-2))
(EQL #:G799 (LONG-COMPUTATION-3))))
To get the same effect you would need to write:
(defun in (obj &rest choices)
(reduce (lambda (x y)
(or x (eql (funcall y) obj)))
choices
:initial-value nil))
and use it this way:
(in 42
(function long-computation-1)
(function long-computation-2)
(function long-computation-3))
I started learning Common Lisp recently, and (just for fun) decided to rename the lambda macro.
My attempt was this:
> (defmacro λ (args &body body) `(lambda ,args ,#body))
It seems to expand correctly when by itself:
> (macroexpand-1 '(λ (x) (* x x)))
(LAMBDA (X) (* X X))
But when it's nested inside an expression, execution fails:
> ((λ (x) (* x x)) 2)
(Λ (X) (* X X)) is not a function name; try using a symbol instead
I am probably missing something obvious about macro expansion, but couldn't find out what it is.
Maybe you can help me out?
edit:
It does work with lambda:
> ((lambda (x) (* x x)) 2)
4
edit 2:
One way to make it work (as suggested by Rainer):
> (set-macro-character #\λ (lambda (stream char) (quote lambda)))
(tested in Clozure CL)
In Common Lisp LAMBDA is two different things: a macro and a symbol which can be used in a LAMBDA expression.
The LAMBDA expression:
(function (lambda (x) (foo x)))
shorter written as
#'(lambda (x) (foo x))
An applied lambda expression is also valid:
((lambda (x) (+ x x)) 4)
Above both forms are part of the core syntax of Common Lisp.
Late in the definition of Common Lisp a macro called LAMBDA has been added. Confusingly enough, but with good intentions. ;-) It is documented as Macro LAMBDA.
(lambda (x) (+ x x))
expands into
(function (lambda (x) (+ x x))
It makes Common Lisp code look slightly more like Scheme code and then it is not necessary to write
(mapcar #'(lambda (x) (+ x x)) some-list)
With the LAMBDA macro we can write
(mapcar (lambda (x) (+ x x)) some-list)
Your example fails because
((my-lambda (x) (* x x)) 2)
is not valid Common Lisp syntax.
Common Lisp expects either
a data object
a variable
a function call in the form (function args...)
a function call in the form ((lambda (arglist ...) body) args...)
a macro form like (macro-name forms...)
a special form using one of the built-in special operators like FUNCTION, LET, ...
defined in the list of special operators in Common Lisp
As you can see a syntax of
((macro-name forms...) forms...)
is not a part of Common Lisp.
It is possible to read the character λ as LAMBDA:
(defun λ-reader (stream char)
(declare (ignore char stream))
'LAMBDA)
(set-macro-character #\λ #'λ-reader)
Example:
CL-USER 1 > ((λ (x) (* x x)) 3)
9
CL-USER 2 > '(λ (x) (* x x))
(LAMBDA (X) (* X X))
You might also think of LAMBDA as an operator which, given a term and a list of free variables, returns a function. This p.o.v. takes LAMBDA out of the family of basic functions and elementary macros -- at least as far as the interpreter is concerned.
(defun lambda-char (stream char)
"A lambda with only ONE arg _"
(declare (ignore char))
(let ((codes (read stream nil)))
`(lambda (_) ,codes)))
(set-macro-character #\λ #'lambda-char t)
λ(+ 1 2 _) ; => (lambda (_) (+ 1 2 _))
Maybe this is more concise, with ONLY ONE arg of _