I'm writing a Lisp (code at GitHub) and I want to implement local bindings. Currently I have two syntaxes:
(let <var> <val> <expr>)
for binding a single variable or function, and
(with (<var1> <val1> ... <varN> <valN>) <expr>)
to bind multiple values at once.
At present, the bindings are evaluated sequentially, and each new function binding retains a copy of the environment it was defined in, so <var2> can refer to <var1> but not vice-versa.
I would like to modify the code so that when binding multiple values at once you effectively have simultaneous binding. For example, I would like to be able to write (this is a trivial example, but it should illustrate the idea):
(define (h y)
(with ((f x) (if (eq? x 0) #t (g (- x 1)))
(g x) (if (eq? x 0) #f (f (- x 1))))
(f y))
At the moment this code doesn't run - g closes over f, but not the other way around.
Is there a canonical way to implement simultaneous binding in Lisp?
In SICP there's a section on internal definitions which covers this subject. In particular, the exercises 4.16, 4.18, 4.19 tell you how to implement different strategies for achieving simultaneous definitions.
The syntax is a bit different, but the idea in the book boils down to transforming this code:
(lambda <vars>
(define u <e1>)
(define v <e2>)
<e3>)
Into this code:
(lambda <vars>
(let ((u '*unassigned*)
(v '*unassigned*))
(set! u <e1>)
(set! v <e2>)
<e3>))
The same idea applies to your with special form. Take a look at the linked book for more implementation details.
In (with (a (+ 2 2))) we are binding a to the value of the expression (+ 2 2), so a becomes 4. But in (with ((f x) (+ x x))) we are doing something else: we are binding f to a function. This is a syntactic sugar for (with (f (lambda (x) (+ x x)))).
To handle this situation, you have to process the bindings in two passes. First collect all the variables and create the environment which contains all of them. Then evaluate the initializing experssions and store their values in the corresponding variables. The evaluation of these expressions takes place in that environment, so every expression has visibility over all the variables. The initialization is done by assignment. The variables can be initially nil or have some trap value which blows up if they are accessed.
Related
I'm in a process of implementing Hygienic macros in my Scheme implementation, I've just implemented syntax-rules, but I have this code:
(define odd?
(syntax-rules ()
((_ x) (not (even? x)))))
what should be the difference between that and this:
(define-syntax odd?
(syntax-rules ()
((_ x) (not (even? x)))))
from what I understand syntax-rules just return syntax transformer, why you can't just use define to assign that to symbol? Why I need to use define-syntax? What extra stuff that expression do?
Should first also work in scheme? Or only the second one?
Also what is the difference between let vs let-syntax and letrec vs letrec-syntax. Should (define|let|letrec)-syntax just typecheck if the value is syntax transformer?
EDIT:
I have this implementation, still using lisp macros:
;; -----------------------------------------------------------------------------
(define-macro (let-syntax vars . body)
`(let ,vars
,#(map (lambda (rule)
`(typecheck "let-syntax" ,(car rule) "syntax"))
vars)
,#body))
;; -----------------------------------------------------------------------------
(define-macro (letrec-syntax vars . body)
`(letrec ,vars
,#(map (lambda (rule)
`(typecheck "letrec-syntax" ,(car rule) "syntax"))
vars)
,#body))
;; -----------------------------------------------------------------------------
(define-macro (define-syntax name expr)
(let ((expr-name (gensym)))
`(define ,name
(let ((,expr-name ,expr))
(typecheck "define-syntax" ,expr-name "syntax")
,expr-name))))
This this code correct?
Should this code works?
(let ((let (lambda (x) x)))
(let-syntax ((odd? (syntax-rules ()
((_ x) (not (even? x))))))
(odd? 11)))
This question seems to imply some deep confusion about macros.
Let's imagine a language where syntax-rules returns some syntax transformer function (I am not sure this has to be true in RnRS Scheme, it is true in Racket I think), and where let and let-syntax were the same.
So let's write this function:
(define (f v)
(let ([g v])
(g e (i 10)
(if (= i 0)
i
(e (- i 1))))))
Which we can turn into this, of course:
(define (f v n)
(v e (i n)
(if (<= i 0)
i
(e (- i 1)))))
And I will tell you in addition that there is no binding for e or i in the environment.
What is the interpreter meant to do with this definition? Could it compile it? Could it safely infer that i can't possibly make any sense since it is used as a function and then as a number? Can it safely do anything at all?
The answer is that no, it can't. Until it knows what the argument to the function is it can't do anything. And this means that each time f is called it has to make that decision again. In particular, v might be:
(syntax-rules ()
[(_ name (var init) form ...)
(letrec ([name (λ (var)
form ...)])
(name init))]))
Under which the definition of f does make some kind of sense.
And things get worse: much worse. How about this?
(define (f v1 v2 n)
(let ([v v1])
(v e (i n)
...
(set! v (if (eq? v v1) v2 v1))
...)))
What this means is that a system like this wouldn't know what the code it was meant to interpret meant until, the moment it was interpreting it, or even after that point, as you can see from the second function above.
So instead of this horror, Lisps do something sane: they divide the process of evaluating bits of code into phases where each phase happens, conceptually, before the next one.
Here's a sequence for some imagined Lisp (this is kind of close to what CL does, since most of my knowledge is of that, but it is not intended to represent any particular system):
there's a phase where the code is turned from some sequence of characters to some object, possibly with the assistance of user-defined code;
there's a phase where that object is rewritten into some other object by user- and system-defined code (macros) – the result of this phase is something which is expressed in terms of functions and some small number of primitive special things, traditionally called 'special forms' which are known to the processes of stage 3 and 4;
there may be a phase where the object from phase 2 is compiled, and that phase may involve another set of user-defined macros (compiler macros);
there is a phase where the resulting code is evaluated.
And for each unit of code these phases happen in order, each phase completes before the next one begins.
This means that each phase in which the user can intervene needs its own set of defining and binding forms: it needs to be possible to say that 'this thing controls what happens at phase 2' for instance.
That's what define-syntax, let-syntax &c do: they say that 'these bindings and definitions control what happens at phase 2'. You can't, for instance, use define or let to do that, because at phase 2, these operations don't yet have meaning: they gain meaning (possibly by themselves being macros which expand to some primitive thing) only at phase 3. At phase 2 they are just bits of syntax which the macro is ingesting and spitting out.
I’m trying to get get my macro to do an extra evaluation of its result before returning it. Can this be done without eval?
I'm trying to solve the problem in exercise 4 below:
Define a macro nth-expr that takes an integer n and an arbitrary number of expressions, evaluates the nth expression and returns its value. This exercise is easy to solve, if you assume that the first argument is a literal integer.
4. As exercise 3, but assume that the first argument is an expression to be evaluated.
It's easy to get the macro to pick the right expression:
(defmacro nth-expr% (n &rest es)
`(nth ,n ',es))
CL-USER> (defvar i 1)
I
CL-USER> (nth-expr% (1+ i) (+ 2 3) (- 4 3) (+ 3 1))
(+ 3 1)
The expression (+ 3 1) is the one we want, but we want the macro to evaluate it to 4 before returning it.
It can of course be done with eval:
(defmacro nth-expr%% (n &rest es)
`(eval (nth ,n ',es)))
CL-USER> (nth-expr%% (1+ i) (+ 2 3) (- 4 3) (+ 3 1))
4
But is there another way?
It feels like the solution should be to put the body of nth-expr% in a helper macro and have the top level macro only contain an unquoted call to this helper:
(defmacro helper (n es)
`(nth ,n ',es))
(defmacro nth-expr (n &rest es) ; doesn't work!
(helper n es))
The idea is that the call to helper would return (+ 3 1), and this would then be the expansion of the call to nth-expr, which at run-time would evaluate to 4. It blows up, of course, because N and ES get treated like literals.
That's not that easy.
Using eval is not good, since eval does not evaluate the code in the local lexical environment.
Remember, if we allow an expression to be evaluated to determine the number of another expression to execute, then we don't know this number at macro expansion time - since the expression could be based on a value that needs to be computed - for example based on some variable:
(nth-expression
foo
(bar)
(baz))
So we might want to think about code which does that:
(case foo
(0 (bar))
(1 (baz)))
CASE is evaluating foo and then uses the result to find a clause which has the same value in its head. The consequent forms of that clause then will be evaluated.
Now we need to write code which expands the former into the latter.
This would be a very simple version:
(defmacro nth-expression (n-form &body expressions)
`(case ,n-form
,#(loop for e in expressions
and i from 0
collect `(,i ,e))))
Question: what might be drawbacks of using CASE like that?
Knuto: Rainer Joswig may be asking you to think about how the case statement works. Namely, that after evaluating the keyform (ie, the first argument), it will be compared sequentially to the key in each clause until a match is found. The comparisons could be time consuming if there are many clauses. You can discover this by carefully reading the entry for case in the Hyperspec (as he more than once has insisted I do):
The keyform or keyplace is evaluated to produce the test-key. Each of
the normal-clauses is then considered in turn.
Also note that constructing many case clauses will add to the time to expand and compile the macro at compile time.
Regarding your use of eval in nth-expr%%, you can still achieve the effect of an eval by switching to apply:
(defmacro nth-expr%% (n &rest es)
`(let ((ne (nth ,n ',es)))
(apply (car ne) (cdr ne))))
But see Plugging the Leaks at http://www.gigamonkeys.com/book/macros-defining-your-own.html about a more robust treatment.
In general, a more efficient way to process the expressions is as a simple vector, rather than a list. (The problem statement does not rule out a vector representation.) While nth and case involve searching through the expressions one-by-one, a function like aref or svref can directly index into it. Assuming a vector of expressions is passed to the macro along with an index, perhaps first requiring (coerce expressions 'simple-vector) if a list, then the result can be computed in constant time no matter how many expressions there are:
(defmacro nth-expr%%% (n es)
`(let ((ne (svref ',es ,n)))
(apply (car ne) (cdr ne))))
so that now
(defvar i 1)
(nth-expr%%% (1+ i) #((+ 2 3) (- 4 3) (+ 3 1))) -> 4
This question already has answers here:
What is the difference between Lisp-1 and Lisp-2?
(2 answers)
Closed 6 years ago.
I am attempting to understand the lambda notion found within Emacs Lisp.
In ielm, executing:
((lambda (x) (* x x)) 5)
gives us 25, and
(let ((x 4)) (* x x))
gives us 16. However when I do:
(let ((f (lambda (x) (* x x)))) (f 7))
it does not give me 49, but instead informs me:
*** Eval error *** Symbol's function definition is void: f
Don't know why, I am sure the syntax is right and f is defined in the let?
Using cl-flet to define let-ed function
We can actually do this without using funcall. The cl module includes standard functions from Common Lisp. We first import it:
(require 'cl)
Thereafter we can use cl-flet to define our function:
(cl-flet ((f (x) (* x x)))
(f 7))
I'd be surprised if this isn't a duplicate, but I can't find it readily here on Stack Overflow. In "Lisp-2" languages (e.g., Emacs Lisp and Common Lisp), there are separate namespaces for functions and variables. A function call looks like:
((lambda ...) ...) ; call the lambda function
or
(f ...) ; call the function binding of f
If you want to call the function that is the value of a variable, then you need to use funcall or apply:
(apply f ...)
(funcall f ...)
The difference between apply and funcall is well documented in other places, but the quick difference is that apply expects an argument list (in some Lisps, a "spreadable argument list"), whereas funcall takes the arguments directly. E.g.,
(let ((f (lambda (a b) (+ a b))))
(funcall f 1 2) ; arguments directly
(apply f '(1 2))) ; arguments in a list
In a "Lisp-1", (like Scheme), you don't need funcall, since there's only one space for bindings. In Scheme you can do:
(let ((f (lambda (a b) (+ a b))))
(f 1 2))
The difference between Lisp-1 and Lisp-2 languages is described more in What is the difference between Lisp-1 and Lisp-2?. The big difference is when the system sees a function call, how it figures out what function to call. In a Lisp-1, variables have values, and that's all. So when the system sees something like:
(f ...)
f is a variable, and the only possible thing to call is the value of f. In a Lisp-2, the systems tries to call the function value of f, which doesn't have to have anything do with the value of the variable f. That can be a bit confusing at first, but it's actually pretty handy in many cases, because it makes it harder to accidentally obscure functions with common names. E.g., in a Lisp-1, you'll see lots of people use lst as an argument name instead of list, because if they named the argument list, then they couldn't use the (standard) function list within the function.
I am trying to solve the last part of question 4.4 of the Structure and Interpretation of computer programming; the task is to implement or as a syntactic transformation. Only elementary syntactic forms are defined; quote, if, begin, cond, define, apply and lambda.
(or a b ... c) is equal to the first true value or false if no value is true.
The way I want to approach it is to transform for example (or a b c) into
(if a a (if b b (if c c false)))
the problem with this is that a, b, and c would be evaluated twice, which could give incorrect results if any of them had side-effects. So I want something like a let
(let ((syma a))
(if syma syma (let ((symb b))
(if symb symb (let ((symc c))
(if (symc symc false)) )) )) )
and this in turn could be implemented via lambda as in Exercise 4.6. The problem now is determining symbols syma, symb and symc; if for example the expression b contains a reference to the variable syma, then the let will destroy the binding. Thus we must have that syma is a symbol not in b or c.
Now we hit a snag; the only way I can see out of this hole is to have symbols that cannot have been in any expression passed to eval. (This includes symbols that might have been passed in by other syntactic transformations).
However because I don't have direct access to the environment at the expression I'm not sure if there is any reasonable way of producing such symbols; I think Common Lisp has the function gensym for this purpose (which would mean sticking state in the metacircular interpreter, endangering any concurrent use).
Am I missing something? Is there a way to implement or without using gensym? I know that Scheme has it's own hygenic macro system, but I haven't grokked how it works and I'm not sure whether it's got a gensym underneath.
I think what you might want to do here is to transform to a syntactic expansion where the evaluation of the various forms aren't nested. You could do this, e.g., by wrapping each form as a lambda function and then the approach that you're using is fine. E.g., you can do turn something like
(or a b c)
into
(let ((l1 (lambda () a))
(l2 (lambda () b))
(l3 (lambda () c)))
(let ((v1 (l1)))
(if v1 v1
(let ((v2 (l2)))
(if v2 v2
(let ((v3 (l3)))
(if v3 v3
false)))))))
(Actually, the evaluation of the lambda function calls are still nested in the ifs and lets, but the definition of the lambda functions are in a location such that calling them in the nested ifs and lets doesn't cause any difficulty with captured bindings.) This doesn't address the issue of how you get the variables l1–l3 and v1–v3, but that doesn't matter so much, none of them are in scope for the bodies of the lambda functions, so you don't need to worry about whether they appear in the body or not. In fact, you can use the same variable for all the results:
(let ((l1 (lambda () a))
(l2 (lambda () b))
(l3 (lambda () c)))
(let ((v (l1)))
(if v v
(let ((v (l2)))
(if v v
(let ((v (l3)))
(if v v
false)))))))
At this point, you're really just doing loop unrolling of a more general form like:
(define (functional-or . functions)
(if (null? functions)
false
(let ((v ((first functions))))
(if v v
(functional-or (rest functions))))))
and the expansion of (or a b c) is simply
(functional-or (lambda () a) (lambda () b) (lambda () c))
This approach is also used in an answer to Why (apply and '(1 2 3)) doesn't work while (and 1 2 3) works in R5RS?. And none of this required any GENSYMing!
In SICP you are given two ways of implementing or. One that handles them as special forms which is trivial and one as derived expressions. I'm unsure if they actually thought you would see this as a problem, but you can do it by implementing gensym or altering variable? and how you make derived variables like this:
;; a unique tag to identify special variables
(define id (vector 'id))
;; a way to make such variable
(define (make-var x)
(list id x))
;; redefine variable? to handle macro-variables
(define (variable? exp)
(or (symbol? exp)
(tagged-list? exp id)))
;; makes combinations so that you don't evaluate
;; every part twice in case of side effects (set!)
(define (or->combination terms)
(if (null? terms)
'false
(let ((tmp (make-var 'tmp)))
(list (make-lambda (list tmp)
(list (make-if tmp
tmp
(or->combination (cdr terms)))))
(car terms)))))
;; My original version
;; This might not be good since it uses backquotes not introduced
;; until chapter 5 and uses features from exercise 4.6
;; Though, might be easier to read for some so I'll leave it.
(define (or->combination terms)
(if (null? terms)
'false
(let ((tmp (make-var 'tmp)))
`(let ((,tmp ,(car terms)))
(if ,tmp
,tmp
,(or->combination (cdr terms)))))))
How it works is that make-var creates a new list every time it is called, even with the same argument. Since it has id as it's first element variable? will identify it as a variable. Since it's a list it will only match in variable lookup with eq? if it is the same list, so several nested or->combination tmp-vars will all be seen as different by lookup-variable-value since (eq? (list) (list)) => #f and special variables being lists they will never shadow any symbol in code.
This is influenced by eiod, by Al Petrofsky, which implements syntax-rules in a similar manner. Unless you look at others implementations as spoilers you should give it a read.
For the life of me, I can't understand continuations. I think the problem stems from the fact that I don't understand is what they are for. All the examples that I've found in books or online are very trivial. They make me wonder, why anyone would even want continuations?
Here's a typical impractical example, from TSPL, which I believe is quite recognized book on the subject. In english, they describe the continuation as "what to do" with the result of a computation. OK, that's sort of understandable.
Then, the second example given:
(call/cc
(lambda (k)
(* 5 (k 4)))) => 4
How does this make any sense?? k isn't even defined! How can this code be evaluated, when (k 4) can't even be computed? Not to mention, how does call/cc know to rip out the argument 4 to the inner most expression and return it? What happens to (* 5 .. ?? If this outermost expression is discarded, why even write it?
Then, a "less" trivial example stated is how to use call/cc to provide a nonlocal exit from a recursion. That sounds like flow control directive, ie like break/return in an imperative language, and not a computation.
And what is the purpose of going through these motions? If somebody needs the result of computation, why not just store it and recall later, as needed.
Forget about call/cc for a moment. Every expression/statement, in any programming language, has a continuation - which is, what you do with the result. In C, for example,
x = (1 + (2 * 3));
printf ("Done");
has the continuation of the math assignment being printf(...); the continuation of (2 * 3) is 'add 1; assign to x; printf(...)'. Conceptually the continuation is there whether or not you have access to it. Think for a moment what information you need for the continuation - the information is 1) the heap memory state (in general), 2) the stack, 3) any registers and 4) the program counter.
So continuations exist but usually they are only implicit and can't be accessed.
In Scheme, and a few other languages, you have access to the continuation. Essentially, behind your back, the compiler+runtime bundles up all the information needed for a continuation, stores it (generally in the heap) and gives you a handle to it. The handle you get is the function 'k' - if you call that function you will continue exactly after the call/cc point. Importantly, you can call that function multiple times and you will always continue after the call/cc point.
Let's look at some examples:
> (+ 2 (call/cc (lambda (cont) 3)))
5
In the above, the result of call/cc is the result of the lambda which is 3. The continuation wasn't invoked.
Now let's invoke the continuation:
> (+ 2 (call/cc (lambda (cont) (cont 10) 3)))
12
By invoking the continuation we skip anything after the invocation and continue right at the call/cc point. With (cont 10) the continuation returns 10 which is added to 2 for 12.
Now let's save the continuation.
> (define add-2 #f)
> (+ 2 (call/cc (lambda (cont) (set! add-2 cont) 3)))
5
> (add-2 10)
12
> (add-2 100)
102
By saving the continuation we can use it as we please to 'jump back to' whatever computation followed the call/cc point.
Often continuations are used for a non-local exit. Think of a function that is going to return a list unless there is some problem at which point '() will be returned.
(define (hairy-list-function list)
(call/cc
(lambda (cont)
;; process the list ...
(when (a-problem-arises? ...)
(cont '()))
;; continue processing the list ...
value-to-return)))
Here is text from my class notes: http://tmp.barzilay.org/cont.txt. It is based on a number of sources, and is much extended. It has motivations, basic explanations, more advanced explanations for how it's done, and a good number of examples that go from simple to advanced, and even some quick discussion of delimited continuations.
(I tried to play with putting the whole text here, but as I expected, 120k of text is not something that makes SO happy.
TL;DR: continuations are just captured GOTOs, with values, more or less.
The exampe you ask about,
(call/cc
(lambda (k)
;;;;;;;;;;;;;;;;
(* 5 (k 4)) ;; body of code
;;;;;;;;;;;;;;;;
)) => 4
can be approximately translated into e.g. Common Lisp, as
(prog (k retval)
(setq k (lambda (x) ;; capture the current continuation:
(setq retval x) ;; set! the return value
(go EXIT))) ;; and jump to exit point
(setq retval ;; get the value of the last expression,
(progn ;; as usual, in the
;;;;;;;;;;;;;;;;
(* 5 (funcall k 4)) ;; body of code
;;;;;;;;;;;;;;;;
))
EXIT ;; the goto label
(return retval))
This is just an illustration; in Common Lisp we can't jump back into the PROG tagbody after we've exited it the first time. But in Scheme, with real continuations, we can. If we set some global variable inside the body of function called by call/cc, say (setq qq k), in Scheme we can call it at any later time, from anywhere, re-entering into the same context (e.g. (qq 42)).
The point is, the body of call/cc form may contain an if or a condexpression. It can call the continuation only in some cases, and in others return normally, evaluating all expressions in the body of code and returning the last one's value, as usual. There can be deep recursion going on there. By calling the captured continuation an immediate exit is achieved.
So we see here that k is defined. It is defined by the call/cc call. When (call/cc g) is called, it calls its argument with the current continuation: (g the-current-continuation). the current-continuation is an "escape procedure" pointing at the return point of the call/cc form. To call it means to supply a value as if it were returned by the call/cc form itself.
So the above results in
((lambda(k) (* 5 (k 4))) the-current-continuation) ==>
(* 5 (the-current-continuation 4)) ==>
; to call the-current-continuation means to return the value from
; the call/cc form, so, jump to the return point, and return the value:
4
I won't try to explain all the places where continuations can be useful, but I hope that I can give brief examples of main place where I have found continuations useful in my own experience. Rather than speaking about Scheme's call/cc, I'd focus attention on continuation passing style. In some programming languages, variables can be dynamically scoped, and in languages without dynamically scoped, boilerplate with global variables (assuming that there are no issues of multi-threaded code, etc.) can be used. For instance, suppose there is a list of currently active logging streams, *logging-streams*, and that we want to call function in a dynamic environment where *logging-streams* is augmented with logging-stream-x. In Common Lisp we can do
(let ((*logging-streams* (cons logging-stream-x *logging-streams*)))
(function))
If we don't have dynamically scoped variables, as in Scheme, we can still do
(let ((old-streams *logging-streams*))
(set! *logging-streams* (cons logging-stream-x *logging-streams*)
(let ((result (function)))
(set! *logging-streams* old-streams)
result))
Now lets assume that we're actually given a cons-tree whose non-nil leaves are logging-streams, all of which should be in *logging-streams* when function is called. We've got two options:
We can flatten the tree, collect all the logging streams, extend *logging-streams*, and then call function.
We can, using continuation passing style, traverse the tree, gradually extending *logging-streams*, finally calling function when there is no more tree to traverse.
Option 2 looks something like
(defparameter *logging-streams* '())
(defun extend-streams (stream-tree continuation)
(cond
;; a null leaf
((null stream-tree)
(funcall continuation))
;; a non-null leaf
((atom stream-tree)
(let ((*logging-streams* (cons stream-tree *logging-streams*)))
(funcall continuation)))
;; a cons cell
(t
(extend-streams (car stream-tree)
#'(lambda ()
(extend-streams (cdr stream-tree)
continuation))))))
With this definition, we have
CL-USER> (extend-streams
'((a b) (c (d e)))
#'(lambda ()
(print *logging-streams*)))
=> (E D C B A)
Now, was there anything useful about this? In this case, probably not. Some minor benefits might be that extend-streams is tail-recursive, so we don't have a lot of stack usage, though the intermediate closures make up for it in heap space. We do have the fact that the eventual continuation is executed in the dynamic scope of any intermediate stuff that extend-streams set up. In this case, that's not all that important, but in other cases it can be.
Being able to abstract away some of the control flow, and to have non-local exits, or to be able to pick up a computation somewhere from a while back, can be very handy. This can be useful in backtracking search, for instance. Here's a continuation passing style propositional calculus solver for formulas where a formula is a symbol (a propositional literal), or a list of the form (not formula), (and left right), or (or left right).
(defun fail ()
'(() () fail))
(defun satisfy (formula
&optional
(positives '())
(negatives '())
(succeed #'(lambda (ps ns retry) `(,ps ,ns ,retry)))
(retry 'fail))
;; succeed is a function of three arguments: a list of positive literals,
;; a list of negative literals. retry is a function of zero
;; arguments, and is used to `try again` from the last place that a
;; choice was made.
(if (symbolp formula)
(if (member formula negatives)
(funcall retry)
(funcall succeed (adjoin formula positives) negatives retry))
(destructuring-bind (op left &optional right) formula
(case op
((not)
(satisfy left negatives positives
#'(lambda (negatives positives retry)
(funcall succeed positives negatives retry))
retry))
((and)
(satisfy left positives negatives
#'(lambda (positives negatives retry)
(satisfy right positives negatives succeed retry))
retry))
((or)
(satisfy left positives negatives
succeed
#'(lambda ()
(satisfy right positives negatives
succeed retry))))))))
If a satisfying assignment is found, then succeed is called with three arguments: the list of positive literals, the list of negative literals, and function that can retry the search (i.e., attempt to find another solution). For instance:
CL-USER> (satisfy '(and p (not p)))
(NIL NIL FAIL)
CL-USER> (satisfy '(or p q))
((P) NIL #<CLOSURE (LAMBDA #) {1002B99469}>)
CL-USER> (satisfy '(and (or p q) (and (not p) r)))
((R Q) (P) FAIL)
The second case is interesting, in that the third result is not FAIL, but some callable function that will try to find another solution. In this case, we can see that (or p q) is satisfiable by making either p or q true:
CL-USER> (destructuring-bind (ps ns retry) (satisfy '(or p q))
(declare (ignore ps ns))
(funcall retry))
((Q) NIL FAIL)
That would have been very difficult to do if we weren't using a continuation passing style where we can save the alternative flow and come back to it later. Using this, we can do some clever things, like collect all the satisfying assignments:
(defun satisfy-all (formula &aux (assignments '()) retry)
(setf retry #'(lambda ()
(satisfy formula '() '()
#'(lambda (ps ns new-retry)
(push (list ps ns) assignments)
(setf retry new-retry))
'fail)))
(loop while (not (eq retry 'fail))
do (funcall retry)
finally (return assignments)))
CL-USER> (satisfy-all '(or p (or (and q (not r)) (or r s))))
(((S) NIL) ; make S true
((R) NIL) ; make R true
((Q) (R)) ; make Q true and R false
((P) NIL)) ; make P true
We could change the loop a bit and get just n assignments, up to some n, or variations on that theme. Often times continuation passing style is not needed, or can make code hard to maintain and understand, but in the cases where it is useful, it can make some otherwise very difficult things fairly easy.