In the following code how can i have the x and y variables to reflect the expressions given at macro call time?
(defmacro defrule (init-form &rest replication-patterns)
(let (rule-table)
`(destructuring-bind (p x y) ',init-form
#'(lambda (w h) (list x y)))))
When expanding a call like:
(defrule (70 (* 1/2 w) (+ h 3)))
it returns:
(DESTRUCTURING-BIND (P X Y) '(70 (* 1/2 W) (+ H 3))
#'(LAMBDA (W H) (LIST X Y)))
where the original expressions with W and H references are lost. I tried back-quoting the lambda function creation:
(defmacro defrule (init-form &rest replication-patterns)
(let (rule-table)
`(destructuring-bind (p x y) ',init-form
`#'(lambda (w h) (list ,x ,y)))))
But a same call:
(defrule (70 (* 1/2 w) (+ h 3)))
expands to:
(DESTRUCTURING-BIND
(P X Y)
'(70 (* 1/2 W) (+ H 3))
`#'(LAMBDA (W H) (LIST ,X ,Y)))
which returns a CONS:
#'(LAMBDA (W H) (LIST (* 1/2 W) (+ H 3)))
which can not be used by funcall and passed around like a function object easily. How can i return a function object with expressions i pass in as arguments for the x y part of the init-form with possible W H references being visible by the closure function?
You're getting a cons because you have the backquotes nested.
You don't need backquote around destructuring-bind, because you're destructuring at macro expansion time, and you can do the destructuring directly in the macro lambda list.
(defmacro defrule ((p x y) &rest replication-patterns)
(let (rule-table)
`#'(lambda (w h) (list ,x ,y))))
Looking at your code:
(defmacro defrule (init-form &rest replication-patterns)
(let (rule-table)
`(destructuring-bind (p x y) ',init-form
#'(lambda (w h) (list x y)))))
You want a macro, which expands into code, which then at runtime takes code and returns a closure?
That's probably not a good idea.
Keep in mind: it's the macro, which should manipulate code at macro-expansion time. At runtime, the code should be fixed. See Barmar's explanation how to improve your code.
Related
The Common Lisp HyperSpec covers the with-slots macro. However, the example is hard to grasp.
Is there an easier and more didactic example about it?
To better understand with-slots, one should look into defclass first.
No reader or writer functions are defined by default; their generation must be explicitly requested. However, slots can always be accessed using slot-value.
It means that, unless specific request, defclass don't create any accessor for the slots:
> (defclass point ()
(x y))
(let ((new-point (make-instance 'point)))
(setf (point-x new-point) 1))
Error: The function (COMMON-LISP:SETF COMMON-LISP-USER::POINT-X) is undefined.
In this case, one must use the slot-value function to access or modify the slot value.
(defclass point ()
(x y))
(let ((new-point (make-instance 'point)))
(setf (slot-value new-point 'x) 1))
Obviously, when there are several slots to update, the code become a little cumbersome:
(defmethod translate ((point-instance point) delta-x delta-y)
(setf (slot-value point-instance 'x)
(+ (slot-value point-instance 'x) delta-x))
(setf (slot-value point-instance 'y)
(+ (slot-value point-instance 'y) delta-y)))
For that reason, the with-slots macro can make the code easier to read:
(defmethod translate ((point-instance point) delta-x delta-y)
(with-slots (x y) point-instance
(setf x (+ x delta-x))
(setf y (+ y delta-y))))
Yes. This (great) tutorial of 2003 has a good one from the geometry domain.
Create a class to represent points in 3-dimensions:
(defclass point ()
(x y z))
Create a variable to instantiate the class and a function to set the values:
(defvar my-point
(make-instance 'point))
(defun set-point-values (point x y z)
(setf (slot-value point 'x) x
(slot-value point 'y) y
(slot-value point 'z) z))
In the REPL, do:
CL-USER 17 > (set-point-values my-point 3 4 12)
12
Now, think about a function to compute the distance between points. A brute force way would be:
(defun brute-force-distance-from-origin (point)
(let ((x (slot-value point 'x))
(y (slot-value point 'y))
(z (slot-value point 'z)))
(sqrt (+ (* x x)
(* y y)
(* z z)))))
Using the with-slots macro:
(defun distance-from-origin (point)
(with-slots (x y z) point (sqrt (+ (* x x)
(* y y)
(* z z)))))
Calling the function in the REPL works as expected:
CL-USER> (distance-from-origin my-point)
13.0
I'm trying to write a macro in Common Lisp that takes any number of expressions and builds a list containing each expression followed by its evaluation in a single line. For example, if I name my macro as
(defmacro list-builder (&rest exp)
...)
and I run
(let ((X 1) (Y 2)) (list-builder (+ X Y) (- Y X) X))
I want it to return:
'((+ X Y) 3 (- Y X) 1 X 1)
The best I've been able to do so far is get a list of the expressions using the code
(defmacro list-builder (&rest exp)
`',#`(',exp ,exp))
INPUT: (let ((X 1) (Y 2)) (list-builder (+ X Y) (+ Y X) X))
'((+ X Y) (+ Y X) X)
Strictly speaking, the macro itself cannot do that; what the macro must do is generate code in which the argument expressions are embedded in such a way that they are evaluated, and also in such a way that they are quoted.
Given (list-builder (+ x y) (+ y x) x) we would like to generate this code: (list '(+ x y) (+ x y) '(+ y x) (+ y x) 'x x).
We can split the macro into an top-level wrapper defined with defmacro and an expander function that does the bulk of the work of producing the list arguments; The macro's body just sticks the list symbol on it and returns it.
Macro helper functions have to be wrapped with a little eval-when dance in Common Lisp to make sure they are available in all conceivable situations that the macro might be processed:
(eval-when (:compile-toplevel :load-toplevel :execute)
(defun list-builder-expander (exprs)
(cond
((null exprs) nil)
((atom exprs) (error "list-builder: dotted syntax unsupported":))
(t (list* `',(car exprs) (car exprs)
(list-builder-expander (cdr exprs)))))))
(defmacro list-builder (&rest exprs)
(cons 'list (list-builder-expander exprs)))
A "slick" implementation, all in one defmacro, inside a single backquote expression, might go like this:
(defmacro list-builder (&rest exprs)
`(list ,#(mapcan (lambda (expr) (list `',expr expr)) exprs)))
The "dotted syntax unsupported" check we implemented before now becomes an error out of mapcan.
The lambda turns each expression E into the list ((quote E) E). mapcan catenates these lists together to form the arguments for list, which are then spliced into the (list ...) form with ,#.
The form `',expr follows from applying the quote shorthand to `(quote ,expr).
Of course, a lisp macro can do that. Since lisp macros provide full control over evaluation of their arguments.
You have to use macro helper functions only in cases in which you want to use recursion. Since macros have problems to call themselves recursively.
But by looping over the &rest rest argument, you can generate variadic macros (macros with arbitrary number of arguments) and still control the evaluation of each of its arguments.
After some trial and error cycles (macro construction is an incremental procedure, since macros are complex structures), I obtained the
"simpler" solution:
(defmacro list-builder (&rest rest)
`(list ,#(loop for x in `,rest
nconcing (list `',x x))))
Test by:
(let ((X 1)
(Y 2))
(list-builder (+ X Y) (- Y X) X))
;; ((+ X Y) 3 (- Y X) 1 X 1)
Sometimes, in loop constructs, instead of collect/collecting, use nconc/nconcing in combination with (list ...) to have more control over how the elements are consed together.
The
(list `',x x)
ensures, that the second x gets evaluated, while the first
`',x
places the content of x into the expression, while its quoting prevents the evluation of the expression placed for x.
The outer list in combination with the splicing of the loop construct into it,
finally captures (prevents) the intrinsic very final evaluation of the macro body.
(defmacro list-builder (&rest args)
`(let ((lst ',args)
(acc nil))
(dolist (v lst)
(push v acc)
(push (eval v) acc))
(nreverse acc)))
We could create the list builder macro to take rest parameters as you did (I simply renamed them as args for pseudo code). I'd create a quoted list (lst) of the expressions within the list, and an empty list (acc) to store the expressions and whatever they evaluate to later. Then we can use dolist to iterate through our list and push each expression to the list, followed by whatever it evaluates to by running eval on the expression. Then we can finally use nreverse to get the correct order for the list.
We can then call it:
(let ((x 1)
(y 2))
(declare (special x))
(declare (special y))
(list-builder (+ x y) (- y x) x))
The result will be:
((+ X Y) 3 (- Y X) 1 X 1)
CL-USER>
I want to execute a function with 2 local variables, but the values of these of these variables should depend on some condition. For example, let's say I have 2 variables x and y, and I want to swap them inside let if y > x. The swap should be temporary, I don't want to mutate state with rotatef. My code would look something like:
(setq x 2)
(setq y 1)
(let (if (> x y) ((x y) (y x)) ((x x) (y y)))
(cons x y)) ; should return (1 . 2)
But the expression inside let is not valid Lisp. How do I conditionally assign values to local variables? The work around is to put the body in flet and call it with different arguments, but it look clumsy:
(flet ((body (x y) (cons x y)))
(if (< x y)
(body x y)
(body y x)))
Multiple-value-bind and values
There are lots of alternatives, some of which have already been pointed out in other answers. I think that the question in the title ("Conditional variable binding in Common Lisp") is a nice case for multiple-value-bind and values. I've used different variable names in the following just to make it clear where x and y are, and where the original values are coming from. The names can be the same, though; this just shadows them inside.
(let ((a 3)
(b 2))
(multiple-value-bind (x y)
(if (< a b)
(values a b)
(values b a))
(cons x y)))
;=> (2 . 3)
Then, using a bit of macrology, we can make this a bit cleaner, much like coredump did:
(defmacro if-let (test bindings &body body)
"* Syntax:
let ({var | (var [then-form [else-form]])}*) declaration* form* => result*
* Description:
Similar to LET, but each binding instead of an init-form can have a
then-form and and else-form. Both are optional, and default to NIL.
The test is evaluated, then variables are bound to the results of the
then-forms or the else-forms, as by LET."
(let ((bindings (mapcar #'(lambda (binding)
(destructuring-bind (variable &optional then else)
(if (listp binding) binding (list binding))
(list variable then else)))
bindings)))
`(multiple-value-bind ,(mapcar 'first bindings)
(if ,test
(values ,#(mapcar 'second bindings))
(values ,#(mapcar 'third bindings)))
,#body)))
(pprint (macroexpand-1 '(if-let (< x y) ((x x y)
(y y x))
(cons x y))))
; (MULTIPLE-VALUE-BIND (X Y)
; (IF (< X Y)
; (VALUES X Y)
; (VALUES Y X))
; (CONS X Y))
(let ((a 3) (b 2))
(if-let (< a b)
((x a b)
(y b a))
(cons x y)))
;=> (2 . 3)
Comparison with progv
In terms of use, this has some similarities with sindikat's answer, but multiple-value-bind establishes bindings just like let does: lexical by default, but a global or local special declaration will make the bindings dynamic. On the other hand, progv establishes dynamic bindings. This means that if the bindings are entirely introduced by progv, you won't see much difference (except in trying to return closures), but that you can't shadow bindings. We can see this without having to do any conditional work at all. Here are two sample snippets. In the first, we see that the inner reference to x actually refers to the lexical binding, not the dynamic one established by progv. To refer to the one established by progv, you actually need to declare the inner reference to be special. progv doesn't accept declarations, but we can use locally.
(let ((x 1))
(progv '(x) '(2)
x))
;=> 1
(let ((x 1))
(progv '(x) '(2)
(locally (declare (special x))
x)))
;=> 2
multiple-value-bind actually does the binding the way we'd expect:
(let ((x 1))
(multiple-value-bind (x) (values 2)
x))
;=> 2
It's probably better to use a binding construct like multiple-value-bind that establishes lexical bindings by default, just like let does.
If you don't want to use progv, as mentioned by sindikat, you always can wtite something like that:
(defmacro let-if (if-condition then-bindings else-bindings &body body)
`(if ,if-condition
(let ,then-bindings
,#body)
(let ,else-bindings
,#body)))
So expression like
(let-if (> x y) ((x y) (y x)) ((x x) (y y))
(cons x y))
Will expand into:
(IF (> X Y)
(LET ((X Y) (Y X))
(CONS X Y))
(LET ((X X) (Y Y))
(CONS X Y)))
rotatef
How about:
CL-USER> (defvar x 2)
X
CL-USER> (defvar y 1)
Y
CL-USER> (let ((x x) ; these variables shadow previously defined
(y y)) ; X and Y in body of LET
(when (> x y)
(rotatef x y))
(cons x y))
(1 . 2)
CL-USER> x ; here the original variables are intact
2 ; ^
CL-USER> y ; ^
1 ; ^
However, I think that in every such practical case there are lispier ways to solve problem without macros. Answer by msandiford is probably the best from functional point of view.
psetf
Although rotatef is really efficient method (it probably would be compiled to about three machine instructions swapping pointers in memory), it is not general.
Rainer Joswing posted just a great solution as a comment shortly after posting of the question. To my shame, I checked macro psetf only few minutes ago, and this should be very efficient and general solution.
Macro psetf first evaluates its even arguments, then assigns evaluated values to variables at odd positions just like setf does.
So we can write:
(let ((x x)
(y y))
(when (> x y)
(psetf x y y x))
...)
And that's it, one can conditionally rebind anything to anything. I think it's way better than using macros. Because:
I don't think it's such a common situation;
Some macros in the posted answers repeat their body code, which may be really big: thus you get bigger compiled file (it's fair price for using macro, but not in this case);
Every custom macro does make code harder to understand for other people.
One solution is to use progv instead of let, its first argument is a list of symbols to bind values to, second argument is a list of values, rest is body.
(progv '(x y) (if (< x y) (list x y) (list y x))
(cons x y)) ; outputs (1 . 2)
Another alternative might be:
(let ((x (min x y))
(y (max x y)))
(cons x y))
My suggestion would be one of destructuring-bind or multiple-value-bind.
If you anticipate needing to do this a lot, I would suggest using a macro to generate the bindings. I've provided a possible macro (untested).
(defmacro cond-let (test-expr var-bindings &body body)
"Execute BODY with the VAR-BINDINGS in place, with the bound values depending on
the trueness of TEST-EXPR.
VAR-BINDINGS is a list of (<var> <true-value> <false-value>) with missing values
being replaced by NIL."
(let ((var-list (mapcar #'car var-bindings))
(then-values (mapcar #'(lambda (l)
(when (cdr l)
(nth 1 l)))
var-bindings))
(else-values (mapcar #'(lambda (l)
(when (cddr l))
(nth 2 l)))
var-bindings))
`(destructuring-bind ,var-list
(if ,test-expr
(list ,#then-values)
(list ,#else-values)))))
Question:
((lambda (x y) (x y)) (lambda (x) (* x x)) (* 3 3))
This was #1 on the midterm, I put "81 9" he thought I forgot to cross one out lawl, so I cross out 81, and he goes aww. Anyways, I dont understand why it's 81.
I understand why (lambda (x) (* x x)) (* 3 3) = 81, but the first lambda I dont understand what the x and y values are there, and what the [body] (x y) does.
So I was hoping someone could explain to me why the first part doesn't seem like it does anything.
This needs some indentation to clarify
((lambda (x y) (x y))
(lambda (x) (* x x))
(* 3 3))
(lambda (x y) (x y)); call x with y as only parameter.
(lambda (x) (* x x)); evaluate to the square of its parameter.
(* 3 3); evaluate to 9
So the whole thing means: "call the square function with the 9 as parameter".
EDIT: The same thing could be written as
((lambda (x) (* x x))
(* 3 3))
I guess the intent of the exercise is to highlight how evaluating a scheme form involves an implicit function application.
Let's look at this again...
((lambda (x y) (x y)) (lambda (x) (* x x)) (* 3 3))
To evaluate a form we evaluate each part of it in turn. We have three elements in our form. This one is on the first (function) position:
(lambda (x y) (x y))
This is a second element of a form and a first argument to the function:
(lambda (x) (* x x))
Last element of the form, so a second argument to the function.
(* 3 3)
Order of evaluation doesn't matter in this case, so let's just start from the left.
(lambda (x y) (x y))
Lambda creates a function, so this evaluates to a function that takes two arguments, x and y, and then applies x to y (in other words, calls x with a single argument y). Let's call this call-1.
(lambda (x) (* x x))
This evaluates to a function that takes a single argument and returns a square of this argument. So we can just call this square.
(* 3 3)
This obviously evaluates to 9.
OK, so after this first run of evaluation we have:
(call-1 square 9)
To evaluate this, we call call-1 with two arguments, square and 9. Applying call-1 gives us:
(square 9)
Since that's what call-1 does - it calls its first argument with its second argument. Now, square of 9 is 81, which is the value of the whole expression.
Perhaps translating that code to Common Lisp helps clarify its behaviour:
((lambda (x y) (funcall x y)) (lambda (x) (* x x)) (* 3 3))
Or even more explicitly:
(funcall (lambda (x y) (funcall x y))
(lambda (x) (* x x))
(* 3 3))
Indeed, that first lambda doesn't do anything useful, since it boils down to:
(funcall (lambda (x) (* x x)) (* 3 3))
which equals
(let ((x (* 3 3)))
(* x x))
equals
(let ((x 9))
(* x x))
equals
(* 9 9)
equals 81.
The answers posted so far are good, so rather than duplicating what they already said, perhaps here is another way you could look at the program:
(define (square x) (* x x))
(define (call-with arg fun) (fun arg))
(call-with (* 3 3) square)
Does it still look strange?
(define (repeated f n)
if (= n 0)
f
((compose repeated f) (lambda (x) (- n 1))))
I wrote this function, but how would I express this more clearly, using simple recursion with repeated?
I'm sorry, I forgot to define my compose function.
(define (compose f g) (lambda (x) (f (g x))))
And the function takes as inputs a procedure that computes f and a positive integer n and returns the procedure that computes the nth repeated application of f.
I'm assuming that (repeated f 3) should return a function g(x)=f(f(f(x))). If that's not what you want, please clarify. Anyways, that definition of repeated can be written as follows:
(define (repeated f n)
(lambda (x)
(if (= n 0)
x
((repeated f (- n 1)) (f x)))))
(define (square x)
(* x x))
(define y (repeated square 3))
(y 2) ; returns 256, which is (square (square (square 2)))
(define (repeated f n)
(lambda (x)
(let recur ((x x) (n n))
(if (= n 0)
args
(recur (f x) (sub1 n))))))
Write the function the way you normally would, except that the arguments are passed in two stages. It might be even clearer to define repeated this way:
(define repeated (lambda (f n) (lambda (x)
(define (recur x n)
(if (= n 0)
x
(recur (f x) (sub1 n))))
(recur x n))))
You don't have to use a 'let-loop' this way, and the lambdas make it obvious that you expect your arguments in two stages.
(Note:recur is not built in to Scheme as it is in Clojure, I just like the name)
> (define foonly (repeat sub1 10))
> (foonly 11)
1
> (foonly 9)
-1
The cool functional feature you want here is currying, not composition. Here's the Haskell with implicit currying:
repeated _ 0 x = x
repeated f n x = repeated f (pred n) (f x)
I hope this isn't a homework problem.
What is your function trying to do, just out of curiosity? Is it to run f, n times? If so, you can do this.
(define (repeated f n)
(for-each (lambda (i) (f)) (iota n)))