I have a struct spider:
(defstruct spider omegas values k)
And an instance spider:
(set '*spider* (make-spider
:omegas '()
:values (list *input*)
:k '(#'omegashift #'dec #'dupval '((0 . #'dec) (1 . #'inc) (2 . #'dec)))))
But when I run the expression: (listp (car (spider-k *spider*)) on Emacs and SBCL (and SLIME is involved but I'm not sure what it is.)
The REPL returns T. This is obviously confusing as (car (spider-k *spider*) correctly returns #'OMEGASHIFT and (listp (function OMEGASHIFT)) properly returns NIL.
Why is (listp (car (spider-k *spider*)) true? Shouldn't it be false?
#'omegashift is a reader-macro that expands to the list (function omegashift).
When you evaluate (function omegashift) you get a function, but you're not evaluating it because you quoted the list. So you're just getting the list that the reader-macro expands to.
You'd see the same thing if you did (listp (car '('foo))). 'foo expands to the list (quote foo). This evaluates to the symbol foo, but the quote before the list prevents evaluation.
To get functions instead of lists, you need to evaluate the function expressions. You can do this by calling the function list rather than quoting a list.
(setq *spider* (make-spider
:omegas '()
:values (list *input*)
:k (list #'omegashift #'dec #'dupval (list (cons 0 #'dec) (cons 1 #'inc) (cons 2 #'dec)))))
You can also use backquote to simplify this:
(setq *spider* (make-spider
:omegas '()
:values (list *input*)
:k `(,#'omegashift ,#'dec ,#'dupval '((0 . ,#'dec) (1 . ,#'inc) (2 . ,#'dec)))))
Inside a backquoted expression, you use comma to mark the sub-expressions that you want to evaluate.
BTW, you should use setq to assign variables, not set with a quoted symbol. They're equivalent for global variables, but you can't use set with a local variable.
Related
I am trying to learn Common Lisp with the book Common Lisp: A gentle introduction to Symbolic Computation. In addition, I am using SBCL, Emacs, and Slime.
In the last chapter, on Macros, the author presents examples to re-write the built-in incf macro. He teaches the concept with two different approaches: using back-quote notation and without it. Such as:
(defmacro list-notation-my-incf (x)
(list 'setq x (list '+ x 1)))
(defmacro backquote-notation-my-incf (x)
`(setq ,x (+ ,x 1)))
Later, the author introduces another example:
In the example below, TWO-FROM-ONE is a macro that takes a function
name and another object as arguments; it expands into a call to the
function with two arguments, both of which are the quoted object.
He only uses back-quote character to do it:
(defmacro two-from-one (func object)
`(,func ',object ',object))
And it works as expected:
CL-USER> (two-from-one cons stack-overflow)
(STACK-OVERFLOW . STACK-OVERFLOW)
Using slime-macroexpad-1, I have:
(CONS 'STACK-OVERFLOW 'STACK-OVERFLOW)
As an exercise that I created for myself, I tried doing the same, but avoiding the back-quote notation. Unfortunately, I could not make it work:
(defmacro list-two-from-one (func object)
(list func (quote object) (quote object)))
Slime throws the error:
The variable OBJECT is unbound.
[Condition of type UNBOUND-VARIABLE]
Doing a macro expansion, I see:
(CONS OBJECT OBJECT)
If I try a different approach, it seems to be closer, but it does not work either:
(defmacro list-two-from-one (func object)
(list func object object))
Throws the error:
The variable STACK-OVERFLOW is unbound.
[Condition of type UNBOUND-VARIABLE]
And, finally, the macro expansion indicates:
(CONS STACK-OVERFLOW STACK-OVERFLOW)
I feel stuck. How do I successfully re-write the macro without using back-quote notation?
Thanks.
What you are looking for is something like
(defmacro list-two-from-one (func object)
(list func (list 'quote object) (list 'quote object)))
basically, the body of a macro should return the code, that, when evaluated, produces the desired result.
I.e., the macro body should produce (CONS 'STACK-OVERFLOW 'STACK-OVERFLOW).
Since 'a is the same as (quote a), you want your macro to produce
(CONS (QUOTE STACK-OVERFLOW) (QUOTE STACK-OVERFLOW))
which is what my defmacro above returns.
Your macro should expand to:
CL-USER 10 > (macroexpand '(two-from-one2 cons stack-overflow))
(CONS (QUOTE STACK-OVERFLOW) (QUOTE STACK-OVERFLOW))
So create lists with quote like this:
(defmacro two-from-one2 (func object)
(list func (list 'quote object) (list 'quote object)))
Test:
CL-USER 9 > (two-from-one2 cons stack-overflow)
(STACK-OVERFLOW . STACK-OVERFLOW)
How does lisp quote work internally?
For example:
(quote (+ 1 (* 1 2)) )
seems to be equivalent to
(list '+ 1 (list '* 1 2))
which means it is some how symbolizing the Head values recursively. Is this function a built in?
Run (equal (quote (+ 1 (* 1 2))) (list '+ 1 (list '* 1 2))) if you don't believe me.
How does it work?
quote is really really simple to implement. It does mostly nothing. The quote special operator just returns the enclosed object like it is. Nothing more. No evaluation. The object is not changed in any way.
Evaluation of quoted forms
Probably a good time to read McCarthy, from 1960:
Recursive Functions of Symbolic Expressions and Their Computation by Machine, Part I
Pages 16/17 explain evaluation with eval. Here:
eq [car [e]; QUOTE] → cadr [e];
or in s-expression notation:
(cond
...
((eq (car e) 'quote)
(cadr e))
...)
Above code implements the evaluation rule for QUOTE: If the expression is a list and the first element of the list is the symbol QUOTE, then return the second element of the list.
Equivalence of a quoted list with a list created by LIST
(equal (quote (+ 1 (* 1 2)))
(list '+ 1 (list '* 1 2)))
The result is T. This means that both result lists are structurally equivalent.
(eq (quote (+ 1 (* 1 2)))
(list '+ 1 (list '* 1 2)))
The result is NIL. This means that the first cons cell of the linked lists are not the same objects. EQ tests whether we really have the same cons cell object.
QUOTE returns a literal data object. The consequences of modifying this object is undefined. So, don't do it.
LIST returns a new freshly consed list each time it is called. The fresh new list will not share any cons cells with any earlier allocated list.
So the main difference is that QUOTE is a built-in operator, which returns literal and unevaluated data. Whereas LIST is a function which creates a new,fresh list with its arguments as contents.
See the effects with respect to EQ and EQUAL:
CL-USER 6 >
(flet ((foo () (quote (+ 1 (* 1 2))))
(bar () (list '+ 1 (list '* 1 2))))
(list (list :eq-foo-foo (eq (foo) (foo)))
(list :eq-foo-bar (eq (foo) (bar)))
(list :eq-bar-bar (eq (foo) (bar)))
(list :equal-foo-foo (equal (foo) (foo)))
(list :equal-foo-bar (equal (foo) (bar)))
(list :equal-bar-bar (equal (foo) (bar)))))
((:EQ-FOO-FOO T)
(:EQ-FOO-BAR NIL)
(:EQ-BAR-BAR NIL)
(:EQUAL-FOO-FOO T)
(:EQUAL-FOO-BAR T)
(:EQUAL-BAR-BAR T))
is quote a function?
quote can't be a function, since it returns its enclosed data unevaluated. Thus it is a special evaluation rule.
If quote were a function, it's arguments were evaluated. But that's exactly what is NOT what quote is supposed to do.
why does Lisp need QUOTE?
Lisp usually uses s-expressions to write Lisp code. So s-expressions have a both purpose to denote data and we use it to write programs. In a Lisp program lists are used for function calls, macro forms and special forms. symbols are used as variables:
(+ n 42)
Here (+ n 42) is a list and n is a symbol. But we also want to use lists as data in our programs and we want to use symbols as data. Thus we have to quote them, so that Lisp will not see them as programs, but as data:
(append '(+ n) '(42)) evaluates to (+ n 42)
Thus in a Lisp program, lists and variables are by default part of the language elements, for example as function calls and variables. If we want to use lists and symbols as literal data, we have to quote them, to prevent the evaluator treating them as Lisp code to evaluate.
quote does nothing more than return its argument unevaluated. But what is an unevaluated argument?
When a Lisp program is defined, it is either read from textual source into s-expression form or constructed directly in terms of s-expressions. A macro would be an example of generating s-expressions. Either way there is a data structure comprising (mostly) symbols and conses that represents the program.
Most Lisp expressions will call upon evaluation and compilation machinery to interpret this data structure as terms in a program. quote is treated specially and passed these uninterpreted symbols and conses as its argument. In short, quote does almost nothing - the value it returns already exists and is simply passed through.
You can observe the difference between passing through and fresh construction by using eq to test the identity of the return value of quote:
(defun f () '(1 2))
(defun g () (list 1 2))
(eq (f) (f)) => T
(eq (g) (g)) => NIL
As you can see, quote returns the same conses each time through.
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'm reading Peter Norvig's Paradigms of AI. In chapter 6.2, the author uses code like below (not the original code, I picked out the troubling part):
Code Snippet:
(progv '(op arg) '(1+ 1)
(eval '(op arg)))
As the author's original intent, this code should return 2, but in sbcl 1.1.1, the interpreter is apparently not looking up op in the environment, throwing out op: undefined function.
Is this implementation specific? Since the code must have been tested on some other lisp.
p.s Original code
You probably mean
(progv '(op arg) '(1+ 1)
(eval '(funcall op arg)))
Edit(2013-08-21):
PAIP was written in pre-ANSI-Common-Lisp era, so it's possible the code
there contains a few noncompliances wrt the standard. We can make
the examples work with the following revision:
(defun match-if (pattern input bindings)
"Test an arbitrary expression involving variables.
The pattern looks like ((?if code) . rest)."
(and (eval (reduce (lambda (code binding)
(destructuring-bind (var . val) binding
(subst val var code)))
bindings :initial-value (second (first pattern))))
(pat-match (rest pattern) input bindings)))
;; CL-USER> (pat-match '(?x ?op ?y is ?z (?if (eql (?op ?x ?y) ?z))) '(3 + 4 is 7))
;; ((?Z . 7) (?Y . 4) (?OP . +) (?X . 3) (T . T))
;; CL-USER> (pat-match '(?x ?op ?y (?if (?op ?x ?y))) '(3 > 4))
;; NIL
Elements in first positions are not looked up as values, but as functions and there is no concept of dynamic binding in the function namespace.
I'd say after a quick look that the original code was designed to evaluate in a context like
(progv '(x y) '(12 34)
(eval '(> (+ x y) 99)))
i.e. evaluating a formula providing substitution for variables, not for function names.
The other answers so far are right, in that the actual form being evaluated is not the variables being bound by progv (simply (op arg)), but none have mentioned what is being evaluated. In fact, the comments in the code you linked to provide a (very) short explanation (this is the only code in that file that uses progv):
(defun match-if (pattern input bindings)
"Test an arbitrary expression involving variables.
The pattern looks like ((?if code) . rest)."
;; *** fix, rjf 10/1/92 (used to eval binding values)
(and (progv (mapcar #'car bindings)
(mapcar #'cdr bindings)
(eval (second (first pattern))))
(pat-match (rest pattern) input bindings)))
The idea is that a call to match-if gets called like
(match-if '((?if code) . rest) input ((v1 val1) (v2 val2) ...))
and eval is called with (second (first pattern)), which the value of code. However, eval is called within the progv that binds v1, v2, &c., to the corresponding val1, val2, &c., so that if any of those variables appear free in code, then they are bound when code is evaluated.
Problem
The problem that I see here is that, by the code we can't tell if the value is to be saved as the variable's symbol-value or symbol-function. Thus when you put a + as a value to some corresponding variable, say v, then it'll always be saved as the symbol-value of var, not it's symbol-function.
Therefore when you'll try to use it as, say (v 1 2) , it won't work. Because there is no function named v in the functions' namespace(see this).
So, what to do?
A probable solution can be explicit checking for the value that is to be bound to a variable. If the value is a function, then it should be bound to the variable's function value. This checking can be done via fboundp.
So, we can make a macro functioner and a modified version of match-if. functioner checks if the value is a function, and sets it aptly. match-if does the dynamic local bindings, and allows other code in the scope of the bound variables.
(defmacro functioner (var val)
`(if (and (symbolp ',val)
(fboundp ',val))
(setf (symbol-function ',var) #',val)
(setf ,var ,val)))
(defun match-if (pattern input bindings)
(eval `(and (let ,(mapcar #'(lambda (x) (list (car x))) bindings)
(declare (special ,# (mapcar #'car bindings)))
(loop for i in ',bindings
do (eval `(functioner ,(first i) ,(rest i))))
(eval (second (first ',pattern))))
(pat-match (rest ',pattern) ',input ',bindings))))
Could someone explain to me what's going on in this very simple code snippet?
(defun test-a ()
(let ((x '(nil)))
(setcar x (cons 1 (car x)))
x))
Upon a calling (test-a) for the first time, I get the expected result: ((1)).
But to my surprise, calling it once more, I get ((1 1)), ((1 1 1)) and so on.
Why is this happening? Am I wrong to expect (test-a) to always return ((1))?
Also note that after re-evaluating the definition of test-a, the return result resets.
Also consider that this function works as I expect:
(defun test-b ()
(let ((x '(nil)))
(setq x (cons (cons 1 (car x))
(cdr x)))))
(test-b) always returns ((1)).
Why aren't test-a and test-b equivalent?
The Bad
test-a is self-modifying code. This is extremely dangerous. While the variable x disappears at the end of the let form, its initial value persists in the function object, and that is the value you are modifying. Remember that in Lisp a function is a first class object, which can be passed around (just like a number or a list), and, sometimes, modified. This is exactly what you are doing here: the initial value for x is a part of the function object and you are modifying it.
Let us actually see what is happening:
(symbol-function 'test-a)
=> (lambda nil (let ((x (quote (nil)))) (setcar x (cons 1 (car x))) x))
(test-a)
=> ((1))
(symbol-function 'test-a)
=> (lambda nil (let ((x (quote ((1))))) (setcar x (cons 1 (car x))) x))
(test-a)
=> ((1 1))
(symbol-function 'test-a)
=> (lambda nil (let ((x (quote ((1 1))))) (setcar x (cons 1 (car x))) x))
(test-a)
=> ((1 1 1))
(symbol-function 'test-a)
=> (lambda nil (let ((x (quote ((1 1 1))))) (setcar x (cons 1 (car x))) x))
The Good
test-b returns a fresh cons cell and thus is safe. The initial value of x is never modified. The difference between (setcar x ...) and (setq x ...) is that the former modifies the object already stored in the variable x while the latter stores a new object in x. The difference is similar to x.setField(42) vs. x = new MyObject(42) in C++.
The Bottom Line
In general, it is best to treat quoted data like '(1) as constants - do not modify them:
quote returns the argument, without evaluating it. (quote x) yields x.
Warning: quote does not construct its return value, but just returns
the value that was pre-constructed by the Lisp reader (see info node
Printed Representation). This means that (a . b) is not
identical to (cons 'a 'b): the former does not cons. Quoting should
be reserved for constants that will never be modified by side-effects,
unless you like self-modifying code. See the common pitfall in info
node Rearrangement for an example of unexpected results when
a quoted object is modified.
If you need to modify a list, create it with list or cons or copy-list instead of quote.
See more examples.
PS1. This has been duplicated on Emacs.
PS2. See also Why does this function return a different value every time? for an identical Common Lisp issue.
PS3. See also Issue CONSTANT-MODIFICATION.
I found the culprit is indeed 'quote. Here's its doc-string:
Return the argument, without evaluating it.
...
Warning: `quote' does not construct its return value, but just returns
the value that was pre-constructed by the Lisp reader
...
Quoting should be reserved for constants that will
never be modified by side-effects, unless you like self-modifying code.
I also rewrote for convenience
(setq test-a
(lambda () ((lambda (x) (setcar x (cons 1 (car x))) x) (quote (nil)))))
and then used
(funcall test-a)
to see how 'test-a was changing.
It looks like the '(nil) in your (let) is only evaluated once. When you (setcar), each call is modifying the same list in-place. You can make (test-a) work if you replace the '(nil) with (list (list)), although I presume there's a more elegant way to do it.
(test-b) constructs a totally new list from cons cells each time, which is why it works differently.