I am trying to create a macro for while loop in DrRacket. Here is what I wrote:
(require mzlib/defmacro)
(define-macro my-while
(lambda (condition body)
(list 'local (list (list 'define (list 'while-loop)
(list 'if condition
(list body (list 'while-loop))
'(void))))
'(while-loop))))
(define x 0)
(my-while (< x 10)
(begin
(display x)
(newline)
(set! x (+ x 1))))
The output of this program is:
0
1
2
3
4
5
6
7
8
9
error: procedure application: expected procedure, given: #<void>; arguments were: #<void>
Can someone help me with this? Why wouldn't this macro just terminate and return void. It seems that when the condition is not true, the system tries to apply the void as an argument to some procedure.
Ouch:
Using this style of while loop encourages excessive use of imperative programming.
Using define-macro creates unhygienic macros, which is a nightmare in Scheme.
While I don't encourage writing an imperative-style loop macro, for your reference, here's a non-define-macro version of the same macro:
(define-syntax-rule (my-while condition body ...)
(let loop ()
(when condition
body ...
(loop))))
It uses syntax-rules, which creates hygienic macros, and is much, much easier to read than what you have.
Now, for the actual answer for your question, first, let's write your original macro out in a more readable way:
(define-macro my-while
(lambda (condition body)
`(local ((define (while-loop)
(if ,condition
(,body (while-loop))
(void))))
(while-loop))))
Once you write it out this way, you can see where the real problem is: in the (,body (while-loop)) line, which should instead have been (begin ,body (while-loop)).
Why use a macro when a plain old function will do?
;; fun-while : (-> Boolean) (-> Any) -> Void
(define (fun-while condition body)
(when (condition)
(body)
(fun-while condition body))
Of course, this requires you to pass in repeatable actions that can be called (this is why condition and body are surrounded with parens in the body of fun-while), so you do need a macro if you want prettier syntax. But once you have a function that has the desired behavior, putting some sugar on top is trivial for this case:
(define-syntax-rule (my-while condition body ...)
(fun-while (lambda () condition)
(lambda () body ...)))
Now, as has been said, this encourages imperative style, which is frowned upon. Instead of mutation, try making the state explicit instead:
;; pure-while : forall State.
;; (State -> Boolean) ; the "condition" that inspects the state
;; (State -> State) ; the "body" that moves from one state to the next
;; -> ; curried
;; State ; the current state
;; -> State ; produces the ending state
(define ((pure-while condition make-next) current-state)
(if (condition current-state)
(pure-while condition make-next (make-next current-state))
current-state))
You'll notice that the first two arguments are now functions from State to something, and the result of applying to 2 arguments is also a function from State -> State. This is a recurring pattern that, as a Haskeller, I'd call the "State Monad". Discussion of putting sugar on top of this concept is a little beyond the scope of this conversation, though, so I'll just stop there.
Another version of while uses a do loop:
(define-syntax while
(syntax-rules ()
((while pred? stmt ...)
(do () ((not pred?))
stmt ...))))
Because it's been a while:
a while macro for Racket 6.0
#lang racket
(define-syntax while
(syntax-rules ()
((_ pred? stmt ...)
(do () ((not pred?))
stmt ...))))
Related
Currently, i have been reading the book the scheme programming language written by Kent Dybvig.
In section 5.7, it implements memorized lazy eval in scheme using the scheme macro system.
The source code as
(define-syntax delay
(syntax-rules ()
[(_ expr) (make-promise (lambda () expr))]))
(define make-promise
(lambda (p)
(let ([val #f] [set? #f])
(lambda ()
(unless set?
(let ([x (p)])
(unless set?
(set! val x)
(set! set? #t))))
val))))
(define force
(lambda (promise)
(promise)))
But i can not understand why the variable set? need to be tested twice in the procedure make-promise.
This is the reason from the book
The second test of the variable set? in make-promise is necessary in the event that, as a result of
applying p, the promise is recursively forced. Since a promise must always return the same value, the result of
the first application of p to complete is returned.
which i can not understand
Could anyone explain it to me? Thanks!
The key is that force may reenter itself. Maybe an example can help you understand this:
(define x 5)
(letrec ([f (delay
(if (zero? x)
0
(begin
(set! x (- x 1))
(+ (force f) 1))))])
(force f))
The result will be 0, because the inner force call returns 0.
If without the second test, the result will be 5. In this situation, every (force f) returns different values.
What exactly is different between these implementations of 'when'?
(define-syntax when
(syntax-rules ()
((_ pred b1 ...)
(if pred (begin b1 ...)))))
vs.
(define (my-when pred b1 ...)
(if pred (begin b1 ...)))
For example, when 'my-when' is used in this for loop macro:
(define-syntax for
(syntax-rules ()
((_ (i from to) b1 ...)
(let loop((i from))
(my-when (< i to)
b1 ...
(loop (+ i 1)))))))
an error occurs:
(for (i 0 10) (display i))
; Aborting!: maximum recursion depth exceeded
I do not think 'when' can be implemented as a function, but I do not know why...
Scheme has strict semantics.
This means that all of a function's parameters are evaluated before the function is applied to them.
Macros take source code and produce source code - they don't evaluate any of their parameters.
(Or, well, I suppose they do, but their parameters are syntax - language elements - rather than what you normally think of as values, such as numbers or strings. Macro programming is meta-programming. It's important to be aware of which level you're programming at.)
In your example this means that when my-when is a function, (loop (+ i 1)) must be evaluated before my-when can be applied to it.
This leads to an infinite recursion.
When it's a macro, the my-when form is first replaced with the equivalent if-form
(if (< i to)
(begin
b1 ...
(loop (+ i 1))))
and then the whole thing is evaluated, which means that (loop (+ i 1)) only gets evaluated when the condition is true.
If you implement when as a procedure like you did, then all arguments are evaluated. In your for implementation, the evaluation would be processed like this:
evaluate (< i to)
evaluate expansion result of b1 ...
evaluate (loop (+ i 1)) <- here goes into infinite loop!
evaluate my-when
Item 1-3 can be reverse or undefined order depending on your implementation but the point is nr. 4. If my-when is implemented as a macro, then the macro is the first one to be evaluated.
If you really need to implement with a procedure, then you need to use sort of delaying trick such as thunk. For example:
(define (my-when pred body) (if (pred) (body)))
(my-when (lambda () (< i 10)) (lambda () (display i) (loop (+ i 1))))
I have various functions and I want to call each function with the same value. For instance,
I have these functions:
(defun OP1 (arg) ( + 1 arg) )
(defun OP2 (arg) ( + 2 arg) )
(defun OP3 (arg) ( + 3 arg) )
And a list containing the name of each function:
(defconstant *OPERATORS* '(OP1 OP2 OP3))
So far, I'm trying:
(defun TEST (argument) (dolist (n *OPERATORS*) (n argument) ) )
I've tried using eval, mapcar, and apply, but these haven't worked.
This is just a simplified example; the program that I'm writing has eight functions that are needed to expand nodes in a search tree, but for the moment, this example should suffice.
Other answers have provided some idiomatic solutions with mapcar. One pointed out that you might want a list of functions (which *operators* isn't) instead of a list of symbols (which *operators* is), but it's OK in Common Lisp to funcall a symbol. It's probably more common to use some kind of mapping construction (e.g., mapcar) for this, but since you've provided code using dolist, I think it's worth looking at how you can do this iteratively, too. Let's cover the (probably more idiomatic) solution with mapping first, though.
Mapping
You have a fixed argument, argument, and you want to be able to take a function function and call it with that `argument. We can abstract this as a function:
(lambda (function)
(funcall function argument))
Now, we want to call this function with each of the operations that you've defined. This is simple to do with mapcar:
(defun test (argument)
(mapcar (lambda (function)
(funcall function argument))
*operators*))
Instead of operators, you could also write '(op1 op2 op3) or (list 'op1 'op2 'op3), which are lists of symbols, or (list #'op1 #'op2 #'op3) which is a list of functions. All of these work because funcall takes a function designator as its first argument, and a function designator is
an object that denotes a function and that is one of: a symbol (denoting the function named by that symbol in the global environment), or a function (denoting itself).
Iteratively
You can do this using dolist. The [documentation for actually shows that dolist has a few more tricks up its sleeve. The full syntax is from the documentation
dolist (var list-form [result-form]) declaration* {tag | statement}*
We don't need to worry about declarations here, and we won't be using any tags, but notice that optional result-form. You can specify a form to produce the value that dolist returns; you don't have to accept its default nil. The common idiom for collecting values into a list in an iterative loop is to push each value into a new list, and then return the reverse of that list. Since the new list doesn't share structure with anything else, we usually reverse it destructively using nreverse. Your loop would become
(defun test (argument)
(let ((results '()))
(dolist (op *operators* (nreverse results))
(push (funcall op argument) results))))
Stylistically, I don't like that let that just introduces a single value, and would probably use an &aux variable in the function (but this is a matter of taste, not correctness):
(defun test (argument &aux (results '()))
(dolist (op *operators* (nreverse results))
(push (funcall op argument) results)))
You could also conveniently use loop for this:
(defun test2 (argument)
(loop for op in *operators*
collect (funcall op argument)))
You can also do somewhat succinctly, but perhaps less readably, using do:
(defun test3a (argument)
(do ((results '() (list* (funcall (first operators) argument) results))
(operators *operators* (rest operators)))
((endp operators) (nreverse results))))
This says that on the first iteration, results and operators are initialized with '() and *operators*, respectively. The loop terminates when operators is the empty list, and whenever it terminates, the return value is (nreverse results). On successive iterations, results is a assigned new value, (list* (funcall (first operators) argument) results), which is just like pushing the next value onto results, and operators is updated to (rest operators).
FUNCALL works with symbols.
From the department of silly tricks.
(defconstant *operators* '(op1 op2 o3))
(defun test (&rest arg)
(setf (cdr arg) arg)
(mapcar #'funcall *operators* arg))
There's a library, which is almost mandatory in any anywhat complex project: Alexandria. It has many useful functions, and there's also something that would make your code prettier / less verbose and more conscious.
Say, you wanted to call a number of functions with the same value. Here's how you'd do it:
(ql:quickload "alexandria")
(use-package :alexandria)
(defun example-rcurry (value)
"Calls `listp', `string' and `numberp' with VALUE and returns
a list of results"
(let ((predicates '(listp stringp numberp)))
(mapcar (rcurry #'funcall value) predicates)))
(example-rcurry 42) ;; (NIL NIL T)
(example-rcurry "42") ;; (NIL T NIL)
(defun example-compose (value)
"Calls `complexp' with the result of calling `sqrt'
with the result of calling `parse-integer' on VALUE"
(let ((predicates '(complexp sqrt parse-integer)))
(funcall (apply #'compose predicates) value)))
(example-compose "0") ;; NIL
(example-compose "-1") ;; T
Functions rcurry and compose are from Alexandria package.
I am having trouble with Lisp's backquote read macro. Whenever I try to write a macro that seems to require the use of embedded backquotes (e.g., ``(w ,x ,,y) from Paul Graham's ANSI Common Lisp, page 399), I cannot figure out how to write my code in a way that compiles. Typically, my code receives a whole chain of errors preceded with "Comma not inside a backquote." Can someone provide some guidelines for how I can write code that will evaluate properly?
As an example, I currently need a macro which takes a form that describes a rule in the form of '(function-name column-index value) and generates a predicate lambda body to determine whether the element indexed by column-index for a particular row satisfies the rule. If I called this macro with the rule '(< 1 2), I would want a lambda body that looks like the following to be generated:
(lambda (row)
(< (svref row 1) 2))
The best stab I can make at this is as follows:
(defmacro row-satisfies-rule (rule)
(let ((x (gensym)))
`(let ((,x ,rule))
(lambda (row)
(`,(car ,x) (svref row `,(cadr ,x)) `,(caddr ,x))))))
Upon evaluation, SBCL spews the following error report:
; in: ROW-SATISFIES-RULE '(< 1 2)
; ((CAR #:G1121) (SVREF ROW (CADR #:G1121)) (CADDR #:G1121))
;
; caught ERROR:
; illegal function call
; (LAMBDA (ROW) ((CAR #:G1121) (SVREF ROW (CADR #:G1121)) (CADDR #:G1121)))
; ==>
; #'(LAMBDA (ROW) ((CAR #:G1121) (SVREF ROW (CADR #:G1121)) (CADDR #:G1121)))
;
; caught STYLE-WARNING:
; The variable ROW is defined but never used.
; (LET ((#:G1121 '(< 1 2)))
; (LAMBDA (ROW) ((CAR #:G1121) (SVREF ROW (CADR #:G1121)) (CADDR #:G1121))))
;
; caught STYLE-WARNING:
; The variable #:G1121 is defined but never used.
;
; compilation unit finished
; caught 1 ERROR condition
; caught 2 STYLE-WARNING conditions
#<FUNCTION (LAMBDA (ROW)) {2497F245}>
How can I write macros to generate the code I need, and in particular, how do I implement row-satisfies-rule?
Using the ideas from Ivijay and discipulus, I have modified the macro so that it compiles and works, even allowing forms to be passed as the arguments. It runs a bit differently from my originally planned macro since I determined that including row as an argument made for smoother code. However, it is ugly as sin. Does anyone know how to clean it up so it performs the same without the call to eval?
(defmacro row-satisfies-rule-p (row rule)
(let ((x (gensym))
(y (gensym)))
`(let ((,x ,row)
(,y ,rule))
(destructuring-bind (a b c) ,y
(eval `(,a (svref ,,x ,b) ,c))))))
Also, an explanation of clean, Lispy ways to get macros to generate code to properly evaluate the arguments at runtime would be greatly appreciated.
First of all, Lisp macros have "destructuring" argument lists. This is a nice feature that means instead of having an argument list (rule) and then taking it apart with (car rule) (cadr rule) (caddr rule), you can simply make the argument list ((function-name column-index value)). That way the macro expects a list of three elements as an argument, and each element of the list is then bound to the corresponding symbol in the arguemnt list. You can use this or not, but it's usually more convenient.
Next, `, doesn't actually do anything, because the backquote tells Lisp not to evaluate the following expression and the comma tells it to evaluate it after all. I think you meant just ,(car x), which evaluates (car x). This isn't a problem anyway if you use destructuring arguments.
And since you're not introducing any new variables in the macro expansion, I don't think (gensym) is necessary in this case.
So we can rewrite the macro like this:
(defmacro row-satisfies-rule ((function-name column-index value))
`(lambda (row)
(,function-name (svref row ,column-index) ,value)))
Which expands just how you wanted:
(macroexpand-1 '(row-satisfies-rule (< 1 2)))
=> (LAMBDA (ROW) (< (SVREF ROW 1) 2))
Hope this helps!
If you need the argument to be evaluated to get the rule set, then here's a nice way to do it:
(defmacro row-satisfies-rule (rule)
(destructuring-bind (function-name column-index value) (eval rule)
`(lambda (row)
(,function-name (svref row ,column-index) ,value))))
Here's an example:
(let ((rules '((< 1 2) (> 3 4))))
(macroexpand-1 '(row-satisfies-rule (car rules))))
=> (LAMBDA (ROW) (< (SVREF ROW 1) 2))
just like before.
If you want to include row in the macro and have it give you your answer straightaway instead of making a function to do that, try this:
(defmacro row-satisfies-rule-p (row rule)
(destructuring-bind (function-name column-index value) rule
`(,function-name (svref ,row ,column-index) ,value)))
Or if you need to evaluate the rule argument (e.g. passing '(< 1 2) or (car rules) instead of (< 1 2)) then just use (destructuring-bind (function-name column-index value) (eval rule)
Actually, a function seems more appropriate than a macro for what you're trying to do. Simply
(defun row-satisfies-rule-p (row rule)
(destructuring-bind (function-name column-index value) rule
(funcall function-name (svref row column-index) value)))
works the same way as the macro and is much neater, without all the backquoting mess to worry about.
In general, it's bad Lisp style to use macros for things that can be accomplished by functions.
One thing to understand is that the backquote feature is completely unrelated to macros. It can be used for list creation. Since source code usually consists of lists, it may be handy in macros.
CL-USER 4 > `((+ 1 2) ,(+ 2 3))
((+ 1 2) 5)
The backquote introduces a quoted list. The comma does the unquote: the expression after the comma is evaluated and the result inserted. The comma belongs to the backquote: the comma is only valid inside a backquote expression.
Note also that this is strictly a feature of the Lisp reader.
Above is basically similar to:
CL-USER 5 > (list '(+ 1 2) (+ 2 3))
((+ 1 2) 5)
This creates a new list with the first expression (not evaluated, because quoted) and the result of the second expression.
Why does Lisp provide backquote notation?
Because it provides a simple template mechanism when one wants to create lists where most of the elements are not evaluated, but a few are. Additionally the backquoted list looks similar to the result list.
you don't need nested backquotes to solve this problem. Also, when it's a macro, you don't have to quote your arguments. So (row-satisfies-rule (< 1 2)) is lispier than (row-satisfies-rule '(< 1 2)).
(defmacro row-satisfies-rule (rule)
(destructuring-bind (function-name column-index value) rule
`(lambda (row)
(,function-name (svref row ,column-index) ,value))))
will solve the problem for all calls in the first form. Solving the problem when in the second form is left as an exercise.
In this page there is a comment after the post that gives a very short implementation of amb as a procedure:
(define (amb-backtrack)
(error "no solution found"))
(define (amb . args)
(call/cc (lambda (return)
(let ((backtrack amb-backtrack))
(map (lambda (x)
(call/cc (lambda (k)
(set! amb-backtrack k)
(return x))))
args)
(backtrack 'fail)))))
But I usually see amb implemented as a macro -- in the schemers.org FAQ, and also in Dorai Sitaram's book:
(define amb-fail '*)
(define initialize-amb-fail
(lambda ()
(set! amb-fail
(lambda ()
(error "amb tree exhausted")))))
(initialize-amb-fail)
(define-macro amb
(lambda alts...
`(let ((+prev-amb-fail amb-fail))
(call/cc
(lambda (+sk)
,#(map (lambda (alt)
`(call/cc
(lambda (+fk)
(set! amb-fail
(lambda ()
(set! amb-fail +prev-amb-fail)
(+fk 'fail)))
(+sk ,alt))))
alts...)
(+prev-amb-fail))))))
So -- the macro version is longer, and a little harder to understand. I could not see any advantages of it over the procedure version, and of course I would rather use a procedure than a macro. Is there anything I missed?
The difference is that a procedure call always evaluates all the arguments.
(amb 1 (very-expensive-computation))
will, with the procedure version of amb, perform the very-expensive-computation, then yield 1. If yielding 1 suffices for all further computation, then you've wasted a lot of time on a computation whose result value is never used. Even worse things happen, as #Eli Barzilay mentions in a comment, when amb is used to model an infinite non-determinism such as generating all natural numbers.
The macro version avoids this and its behavior is therefore closer to that of non-deterministic programming languages such as Prolog.