I would like to have a clear understanding, what is 'Atom' in LISP?
Due to lispworks, 'atom - any object that is not a cons.'.
But this definition is not enough clear for me.
For example, in the code below:
(cadr
(caddar (cddddr L)))
Is 'L' an atom? On the one hand, L is not an atom, because it is cons, because it is the list (if we are talking about object, which is associated with the symbol L).
On the other hand, if we are talking about 'L' itself (not about its content, but about the symbol 'L'), it is an atom, because it is not a cons.
I've tried to call function 'atom',
(atom L) => NIL
(atom `L) => T
but still I have no clue... Please, help!
So the final question: in the code above, 'L' is an atom, or not?
P.S. I'm asking this question due to LISP course at my university, where we have a definition of 'simple expression' - it is an expression, which is atom or function call of one or two atomic parameters. Therefore I wonder if expression (cddddr L) is simple, which depends on whether 'L' is atomic parameter or not.
Your Lisp course's private definition of "simple expression" is almost certainly rooted purely in syntax. The idea of "atomic parameter" means that it's not a compound expression. It probably has nothing to do with the run-time value!
Thus, I'm guessing, these are simple expressions:
(+ 1 2)
42
"abc"
whereas these are not:
(+ 1 (* 3 4)) ;; (* 3 4) is not an atomic parameter
(+ a b c) ;; parameters atomic, but more than two
(foo) ;; not simple: fewer than one parameter, not "one or two"
In light of the last counterexample, it would probably behoove them to revise their definition.
On the one hand, L is not an atom, because it is cons, because it is the list (if we are talking about object, which is associated with the symbol L).
You are talking here about the meaning of the code being executed, its semantics. L here stands for a value, which is a list in your tests. At runtime you can inspect values and ask about their types.
On the other hand, if we are talking about 'L' itself (not about its content, but about the symbol 'L'), it is an atom, because it is not a cons.
Here you are looking at the types of the values that make up the syntax of your code, how it is being represented before even being evaluated (or compiled). You are manipulating a tree of symbols, one of them being L. In the source code, this is a symbol. It has no meaning by itself other than being a name.
Code is data
Lisp makes it easy to represent source code using values in the language itself, and easy to manipulate fragments of code at one point to build code that is executed later. This is often called homoiconicity, thought it is somewhat a touchy word because people don't always think the definition is precise enough to be useful. Another saying is "code is data", something that most language designers and programmers will agree to be true.
Lisp code can be built at runtime as follows (> is the prompt of the REPL, what follows is the result of evaluation):
> (list 'defun 'foo (list 'l) (list 'car 'l))
(DEFUN FOO (L) (CAR L))
The resulting form happens to be valid Common Lisp code, not just a generic list of values. If you evaluate it with (eval *), you will define a function named FOO that takes the first element of some list L.
NB. In Common Lisp the asterisk * is bound in the REPL to the last value being successfully returned.
Usually you don't build code like that, the Lisp reader turns a stream of characters into such a tree. For example:
> (read-from-string "(defun foo (l) (car l))")
(DEFUN FOO (L) (CAR L))
But the reader is called also implicitly in the REPL (that's the R in the acronym).
In such a tree of symbols, L is a symbol.
Evaluation model
When you call function FOO after it has been defined, you are evaluating the body of FOO in a context where L is bound to some value. And the rule for evaluating a symbol is to lookup the value it is bound to, and return that. This is the semantics of the code, which the runtime implements for you.
If you are using a simple interpreter, maybe the symbol L is present somewhere at runtime and its binding is looked up. Usually the code is not interpreted like that, it is possible to analyze it and transform it in an efficient way, during compilation. The result of compilation probably does not manipulate symbols anymore here, it just manipulates CPU registers and memory.
In all cases, asking for the type of L at this point is by definition of the semantics just asking the type for whatever value is bound to L in the context it appears.
An atom is anything that is not a cons cell
Really, the definition of atom is no more complex than that. A value in the language that is not a cons-cell is called an atom. This encompasses numbers, strings, everything.
Sometimes you evaluate a tree of symbols that happens to be code, but then the same rule applies.
Simple expressions
P.S. I'm asking this question due to LISP course at my university, where we have a definition of 'simple expression' - it is an expression, which is atom or function call of one or two atomic parameters. Therefore I wonder if expression (cddddr L) is simple, which depends on whether 'L' is atomic parameter or not.
In that course you are writing functions that analyze code. You are given a Lisp value and must decide if it is a simple expression or not.
You are not interested in any particular interpretation of the value being given, at no point you are going to traverse the value, see a a symbol and try to resolve it to a value: you are checking if the syntax is a valid simple expression or not.
Note also that the definitions in your course might be a bit different than the one from any particular exising flavor (this is not necessarily Common Lisp, or Scheme, but a toy LISP dialect). Follow in priority the definitions from your course.
Imagine we have a predicate which tells us if an object is not a number:
(not-number-p 3) -> NIL
(not-number-p "string") -> T
(let ((foo "another string))
(not-number-p foo)) -> T
(not-number '(1 2 3)) -> T
(not-number (first '(1 2 3)) -> NIL
We can define that as:
(defun not-number-p (object)
(not (numberp object))
Above is just the opposite of NUMBERP.
NUMBERP -> T if object is a number
NOT-NUMBER-P -> NIL if object is a number
Now imagine we have a predicate NOT-CONS-P, which tells us if an object is not a CONS cell.
(not-cons-p '(1 . 2)) -> NIL
(let ((c '(1 . 2)))
(not-cons-p c)) -> NIL
(not-cons-p 3) -> T
(let ((n 4))
(not-cons-p n)) -> T
(not-cons-p NIL) -> T
(not-cons-p 'NIL) -> T
(not-cons-p 'a-symbol) -> T
(not-cons-p #\space) -> T
The function NOT-CONS-P can be defined as:
(defun not-cons-p (object)
(if (consp object)
NIL
T))
Or shorter:
(defun not-cons-p (object)
(not (consp object))
The function NOT-CONS-P is traditionally called ATOM in Lisp.
In Common Lisp every object which is not a cons cell is called an atom. The function ATOM is a predicate.
See the Common Lisp HyperSpec: Function Atom
Your question:
(cadr (caddar (cddddr L)))
Is 'L' an atom?
How would we know that? L is a variable. What is the value of L?
(let ((L 10))
(atom L)) -> T
(let ((L (cons 1 2)))
(atom L) -> NIL
(atom l) answers this question:
-> is the value of L an atom
(atom l) does not answer this question:
-> is L an atom? L is a variable and in a function call the value of L is passed to the function ATOM.
If you want to ask if the symbol L is an atom, then you need to quote the symbol:
(atom 'L) -> T
(atom (quote L)) -> T
symbols are atoms. Actually everything is an atom, with the exception of cons cells.
Related
I am taking a programming language principle class where the professor talks about macro using Lisp (Precisly, Elisp). However, he didn't teach us how to write this language. As a result, I am trying to learn myself.
However, there are something that I just can't understand regarding the use of "tick" (') in Lisp. I do, however, understand the basic use. For example, if we have (cdr '(a b c)) it will give us a list (b c). Without the tick symbol, (a b c) would be evaluated as a function.
In the code written by my professor, I have noticed a strange use of "tick" symbol that I can't really understand the usage:
; not sure why there is a "'" in front of if and it is being added to a list ??
(defmacro myand(x y) (list 'if x y nil))
; Another example. Tick symbol before "greater than" symbol ? and made into a list?
(list 'if (list '> x y))
The uses of list and tick symbol just doesn't really make sense to me. Can anyone explain what is going on here? I suppose this is something special to Lisp
The 'tick symbol' is ': '<x> is syntactic sugar for (quote <x>) for any <x>. And (quote <x>) is simply <x> for any object <x>.
quote is needed in Lisp because programs in Lisp often want to reason about things which would have meaning as parts of Lisp programs, and need to say 'this isn't part of this program: it's data'. So, for instance, if I have a list of the form
((a . 1) (b . 2) ... (x . 26)) and I want to look up a symbol in it, I might write a function like this:
(defun ass (key alist)
(if (null? alist)
'()
(if (eql (car (car alist)) key)
(car alist)
(ass key (cdr alist)))))
Then when I wanted to look up, say x, I would have to say (ass (quote x) ...) because if I said (ass x ...) then x would be treated as a variable, and I don't want that: I want the symbol x.
In your example you are looking at programs which write Lisp programs – macros in other words – and these programs need to spend a lot of their time manipulating what will be Lisp source code as data. So in your myand macro, what you want is that
(myand x y)
should be rewritten as
(if x y nil)
Which is the same thing. So to construct this you need a list of four elements: the symbol if, the two variables and the symbol nil. Well you can make a list with list:
(list 'if x y 'nil)
Except that it turns out that the value of the symbol nil is just itself (nil is rather special in many Lisps in fact), so you can not bother quoting that: (list 'if x y nil) (I personally would have quoted it in this case, to make it clear what was literal and what was being evaluated, but that's probably not common).
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.
How would you implement your own set! function in Scheme? A set! function is a destructive procedure that changes a value that is defined taking into account the previous value.
As noted in the comments, set! is a primitive in Scheme that must be provided by the implementation. Similarly, you can't implement the assignment operator, =, in most programming languages. In Common Lisp, setf can be extended (using setf-expanders) to allow (setf form value) to work on new kinds of forms.
Because Scheme's set! only modifies variable bindings (like Common Lisp's setq), it is still worth asking how can we implement functions like set-car! and other structural modifiers. This could be seen, in one sense, as a generalization of assignment to variables, but because lexical variables (along with closures) are sufficient to represent arbitrarily complex structures, it can also be seen as a more specialized case. In Scheme (aside from built in primitives like arrays), mutation of object fields is a specialization, because objects can be implemented by lexical closures, and implemented in terms of set!. This is a typical exercise given when showing how structures, e.g., cons cells, can be implemented using lexical closures alone. Here's an example that shows the implementation of single-value mutable cells::
(define (make-cell value)
(lambda (op)
(case op
((update)
(lambda (new-value)
(set! value new-value)))
((retrieve)
(lambda ()
value)))))
(define (set-value! cell new-value)
((cell 'update) new-value))
(define (get-value cell)
((cell 'retrieve)))
Given these definitions, we can create a cell, that starts with the value 4, update the value to 8 using our set-value!, and retrieve the new value:
(let ((c (make-cell 4)))
(set-value! c 8)
(get-value c))
=> 8
As has been mentioned, set! is a primitive and can't be implemented as a procedure. To really understand how it works under the hood, I suggest you take a look at the inner workings of a Lisp interpreter. Here's a great one to start: the metacircular evaluator in SICP, in particular the section titled "Assignments and definitions". Here's an excerpt of the parts relevant for the question:
(define (eval exp env)
(cond ...
((assignment? exp) (eval-assignment exp env))
...
(else (error "Unknown expression type -- EVAL" exp))))
(define (assignment? exp)
(tagged-list? exp 'set!))
(define (eval-assignment exp env)
(set-variable-value! (assignment-variable exp)
(eval (assignment-value exp) env)
env)
'ok)
(define (set-variable-value! var val env)
(define (env-loop env)
(define (scan vars vals)
(cond ((null? vars)
(env-loop (enclosing-environment env)))
((eq? var (car vars))
(set-car! vals val))
(else (scan (cdr vars) (cdr vals)))))
(if (eq? env the-empty-environment)
(error "Unbound variable -- SET!" var)
(let ((frame (first-frame env)))
(scan (frame-variables frame)
(frame-values frame)))))
(env-loop env))
In the end, a set! operation is just a mutation of a binding's value in the environment. Because a modification at this level is off-limits for a "normal" procedure, it has to be implemented as a special form.
Can't, can't, can't. Everyone is so negative! You can definitely do this in Racket. All you need to do is to define your own "lambda" macro, that introduces mutable cells in place of all arguments, and introduces identifier macros for all of those arguments, so that when they're used as regular varrefs they work correctly. And a set! macro that prevents the expansion of those identifier macros, so that they can be mutated.
Piece of cake!
set! modifies a binding between a symbol and a location (anyting really). Compilers and Interpreters would treat set! differently.
An interpreter would have an environment which is a mapping between symbols and values. Strictly set! changes the value of the first occurrence of the symbol to point to the result of the evaluation of the second operand. In many implementations you cannot set! something that is not already bound. In Scheme it is expected that the variable is already bound. Reading SICP or Lisp in small pieces and playing with examples would make you a master implementer of interpreters.
In a compilers situation you don't really need a symbol table. You can keep evaluated operands on a stack and set! need either change what the stack location pointed to or if it's a free variable in a closure you can use assignment conversion. E.g. it could be boxed to a box or a cons.
(define (gen-counter start delta)
(lambda ()
(let ((cur start))
(set! start (+ cur delta))
cur)))
Might be translated to:
(define (gen-counter start delta)
(let ((start (cons start '()))
(lambda ()
(let ((cur (car start)))
(set-car! start (+ cur delta))
cur)))))
You might want to read Control-Flow Analysis of Higher-Order Languages where this method is used together with lots of information on compiler technique.
If you are implementing a simple interpreter, then it is not all that hard. Your environment will map from identifiers to their values (or syntactic keywords to their transformers). The identifier->value mapping will need to account for possible value changes. As such:
(define (make-cell value)
`(CELL ,value))
(define cell-value cadr)
(define (cell-value-set! cell value)
(set-car! (cdr cell) value))
...
((set-stmt? e)
(let ((name (cadr e))
(value (interpret (caddr e) env)))
(let ((cell (env-lookup name)))
(assert (cell? cell))
(cell-value-set! cell value)
'return-value-for-set!)))
...
With two other changes being that when you bind an identifier to a value (like in a let or lambda application) you need to extend the environment with something like:
(env-extend name (cell value) env)
and also when retrieving a value you'll need to use cell-value.
Of course, only mutable identifiers need a cell, but for a simple interpreter allocating a cell for all identifier values is fine.
In scheme there are 2 "global" environments.
There is an environment where you store your highest defined symbols and there is another environment where are stored the primitive functions and the global variables that represent parameters of the system.
When you write something like (set! VAR VAL), the interpreter will look for the binding of VAR.
You cannot use set! on a variable that was not bound. The binding is done either by define or by lambda or by the system for the primitive operators.
Binding means allocating a location in some environment. So the binding function has the signature symbol -> address.
Coming back to set!, the interpreter will look in the environment where VAR is bound. First it looks in local environment (created by lambda), then it its local parent environment, and so on, then in global environment, then in system environment (in that order) until it finds an environment frame that contains a binding for VAR.
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.
In elisp I can evaluate or as a function just like +.
(or nil 0 nil) ==> 0
(+ 1 0 1) ==> 2
I can use apply to apply + to a list
(apply '+ '(1 0 1)) ==> 2
So, I would think or would work the same way, but it doesn't.
(apply 'or '(nil 0 nil)) ==> error: (invalid-function or)
I imagine this comes from some internal magic used to implement the short-circuit evaluation. How can I use apply to execute the or operation over a list?
P.S. my desired application is to find out whether any elements on the command line match a particular pattern, so the important part of what I am writing is:
(apply 'or (mapcar (lambda (x) (string-match-p "pattern" x)) command-line-args))
But it doesn't work
The problem is that or is a macro (which is the "internal magic" in question), and you're right that that's done so it can do short-circuiting. If or was a function, then calling it would need to follow the usual rules for evaluating a function call: all the arguments would need to get evaluated before the call is made.
See also this question -- it's about Scheme but it's the exact same issue.
As for a solution, you should probably use some, as in:
(some (lambda (x) (string-match-p "pattern" x)) command-line-args)
Note: this uses common lisp that is not included in emacs by default. Just use (require 'cl)
If it makes you feel any better, you're in good company! This is the third question in the "Common Pitfalls" section of the Lisp FAQ:
Here's the simple, but not necessarily satisfying, answer: AND and OR are
macros, not functions; APPLY and FUNCALL can only be used to invoke
functions, not macros and special operators.
...and Eli is of course right on the money with his suggestion to use SOME:
The Common Lisp functions EVERY and SOME can be used to get the
functionality you intend when trying to apply #'AND and #'OR.
(The FAQ and this answer are mostly about Common Lisp but in this case if you omit the # character the answer is the same.)
If you don't care performance, use (eval (cons 'or '(nil 0 nil)))
When I was trying to 'apply' a macro to an argument list, I got an error that the function is unbound, which means that, 'apply' only receives a function, instead of a macro, as its first argument.
In order to fix this, I wrote a new function 'apply-macro' as follows:
(defun apply-macro (macro arg-list)
(eval
`(,macro ,#(loop for arg in arg-list
collect `(quote ,arg)))))
For instance, I wrote a macro to concatenate multiple lists together:
(defmacro conc-lists (&rest lists)
`(funcall #'concatenate 'list ,#lists))
e.g.
(conc-lists '(a b) '(c d) '(e f)) ;;=> (A B C D E F)
Now try 'apply-macro':
(apply-macro 'conc-lists '((a b) (c d) (e f)))
It works and returns the same output.
In fact, it will be expanded into:
(eval
(conc-lists (quote (a b)) (quote (c d)) (quote (e f))))
You can also pass a form to a macro:
(apply-macro 'conc-lists (maplist #'list '(a b c)))
;;=> ((A B C) (B C) (C))
Get back to your question, it's solved:
(apply-macro 'or '(nil 0 nil)) ;;=> 0
I'm only guessing here, but I think or might be one of these 20-odd 'functions' in lisp that are not really functions, since they don't always evaluate all parameters.
This makes sense to make or one of these, since if you have found one TRUE, you can stop searching. In other words, or is not a function as a form of optimization. Still only guessing though.
Eli Barzilay's answer is correct and idiomatic. I want to provide an alternative answer based on dash.el, the library I use to write terse functional-style code, when I have to work with lists. or returns the first non-nil element, nil otherwise, due to short-circuiting. Therefore simply use -first:
(-first 'identity '(nil 0 1 nil)) ; 0
(-first 'identity '(nil nil)) ; nil
identity function simply returns its argument. Which is clever, because -first applies a predicate until it returns non-nil. identity returns non-nil if the argument is itself non-nil. If you simply want to test whether there is non-nil elements in a list, use -any? instead:
(-any? 'identity '(nil 0 1 nil)) ; t
(-any? 'identity '(nil nil)) ; nil