Defparameter operation - common-lisp

Concerning the operation of defparameter--it looks like it only affects the symbol-value cell of the parameter:
> (defparameter *p* 16)
*P*
> (symbol-plist '*p*)
NIL
> (setf (symbol-plist '*p*) '(prop1 val1 prop2 val2))
(PROP1 VAL1 PROP2 VAL2)
> (defparameter *p* 16)
*P*
> (symbol-plist '*p*)
(PROP1 VAL1 PROP2 VAL2)
so the plist has not been reset to nil. Same kind of operation with the symbol-function cell, etc. This seems rather awkward, since reloading/compiling a file with defparameters leaves the prior settings hanging around. Is there any convenient way to reset everything at once?

Yes, you can reset everything completely by uninterning the symbol:
[1]> (defparameter foo 10)
FOO
[2]> (defun foo (x) (1+ x))
FOO
[3]> (setf (get 'foo 'x) 'y)
Y
[4]> (symbol-plist 'foo)
(X Y SYSTEM::DEFINITION
((DEFUN FOO (X) (1+ X)) . #(NIL NIL NIL NIL ((DECLARATION OPTIMIZE DECLARATION)))))
[5]> (setq old-foo 'foo)
FOO
[6]> (unintern 'foo)
T
now the symbol is uninterned and when we re-create it with read, it has a "clean slate":
[7]> (fboundp 'foo)
NIL
[8]> (boundp 'foo)
NIL
but we can still access its slots via old-foo:
[9]> old-foo
#:FOO
[10]> (fdefinition old-foo)
#<FUNCTION #:FOO (X) (DECLARE (SYSTEM::IN-DEFUN #:FOO)) (BLOCK #:FOO (1+ X))>
PS. As Barry mentions in a comment, this might have unintended consequences and should be used with care.
Specifically, this is a fine method while actively developing code.
However, there should be no need for this "clean-up" operation in production.

To add to what sds told you, if you unintern a symbol and introduce a fresh symbol with the same name, then you need to rebuild all the code that referenced the uninterned symbol so that it can refer to the new one. That's why it is impractical to use in production code, where recompiling things over and over would probably not be a good idea.
Is there any convenient way to reset everything at once?
Just define a function and reset what you need to when appropriate:
Or, if you want it to happen only once after you call defparameter, you can do it too.
Note that I would recommend clearing only the properties of the plist you control instead of getting rid of all its content. The environment might store some useful stuff there, like debugging information.
(defparameter *p* 16)
(eval-when (:compile-time :load-time :execute)
(dolist (property '(prop-a prop-b prop-c))
(setf (get '*p* property) nil)))
Edit:
I forgot about REMF:
(remf (symbol-plist '*p*) 'property)
The above removes the entry, whereas (setf (get ...) nil) binds the indicator to NIL.

Related

In Common Lisp, how to use lexical scope and funcall to make another function be passed as an argument?

I am using SBCL, Emacs, and Slime. Hence, I can do:
CL-USER> (defvar example #'(lambda (x) (* x 20)))
EXAMPLE
CL-USER> (funcall example 10)
200
Ok. It works as expected. Using the library Dexador, I can also so:
CL-USER> (ql:quickload :dexador)
To load "dexador":
Load 1 ASDF system:
dexador
; Loading "dexador"
.......
(:DEXADOR)
CL-USER> (dex:get "http://www.paulgraham.com")
"big HTML ommited"
200
#<HASH-TABLE :TEST EQUAL :COUNT 11 {10029F1443}>
#<QURI.URI.HTTP:URI-HTTP http://www.paulgraham.com>
#<SB-SYS:FD-STREAM for "socket 10.0.0.193:44936, peer: 74.6.52.135:80" {1002681F73}>
Now, I am trying to make the argument to be passed be a function! More specifically, the dex:get function. I tried different approaches, but none of them worked out:
CL-USER> (defvar example-failing #'(lambda (x) (x "http://www.paulgraham.com")))
; in: DEFVAR EXAMPLE-FAILING
; (LAMBDA (X) (X "http://www.paulgraham.com"))
;
; caught STYLE-WARNING:
; The variable X is defined but never used.
; in: DEFVAR EXAMPLE-FAILING
; (X "http://www.paulgraham.com")
;
; caught STYLE-WARNING:
; undefined function: COMMON-LISP-USER::X
;
; compilation unit finished
; Undefined function:
; X
; caught 2 STYLE-WARNING conditions
EXAMPLE-FAILING
CL-USER> (funcall example-failing dex:get)
; Evaluation aborted on #<UNBOUND-VARIABLE GET {1002C57103}>.
CL-USER> (funcall example-failing 'dex:get)
; Evaluation aborted on #<UNDEFINED-FUNCTION X {1002DEA263}>.
CL-USER> (funcall example-failing #'dex:get)
; Evaluation aborted on #<UNDEFINED-FUNCTION X {1002F906C3}>.
CL-USER> (funcall example-failing (function dex:get))
; Evaluation aborted on #<UNDEFINED-FUNCTION X {1003147F83}>.
I managed to do it with:
CL-USER> (defvar hacky-eval #'(lambda (x) (eval x)))
HACKY-EVAL
CL-USER> (funcall hacky-eval (dex:get "http://www.paulgraham.com"))
"big html omitted"
But, this feels as bad practice. Is there another way to fix this?
Thanks
I am confused by your question although not as confused as you seem to be. You already seem to know that, to call a function which is the value of a variable, you need either
funcall if you have all the arguments as individual things;
apply if you have only a list of arguments;
and that to get the function value of something you need (function thing) or equivalently #'thing[1].
But then you forget that in your function, and pay no attention to the copious warnings from SBCL.
So
(defvar *example* (lambda (f) (funcall f "http://www.paulgraham.com")))
...
(funcall *example* #'dex:get)
Note that none of this (and nothing in your question) relies on lexical scope: this would all have worked in any historical Lisp.
[1]: you don't need #' for (lambda ...) only because lambda is a macro which expands into (function (lambda ...)). Very old code sometimes uses the explicit #'(lambda ...) form since this macro did not always exist in CL.
Your code:
(defvar example-failing
#'(lambda (x)
(x "http://www.paulgraham.com")))
This makes no sense in Common Lisp. x is a variable. You can't use a variable as a function as in (x arg). In Common Lisp there are different namespaces for functions and variables. For example LET introduces a local variable and FLET introduces a local function.
The ways to call a function bound to a variable are:
(funcall x arg)
(apply x (list arg))
Thus correct examples would be:
(defvar example-failing
#'(lambda (x)
(apply x (list "http://www.paulgraham.com"))))
or
(defvar example-failing
#'(lambda (x)
(funcall x "http://www.paulgraham.com")))
Your solution is no solution
This is your example:
CL-USER> (defvar hacky-eval #'(lambda (x) (eval x)))
HACKY-EVAL
CL-USER> (funcall hacky-eval (dex:get "http://www.paulgraham.com"))
"big html omitted"
This does not work as you think.
(funcall hacky-eval (dex:get "http://www.paulgraham.com"))
is just the same as
(funcall hacky-eval "big html omitted")
and then
(eval "big html omitted")
and then
"big html omitted"
All your call to eval does is to evaluate a string to itself.
You really need to understand basic evaluation rules in Lisp:
(defun foo (arg)
(eval arg))
(foo (+ 3 4))
is simply the same as:
(defun foo (arg)
arg)
(foo (+ 3 4))
which is the same as
(identity (+ 3 4))
Note: if you pass just self evaluating data to EVAL, then all it does is to return the data
A function call (foo (+ 1 2)) works like this:
Lisp sees that FOO is a function
Lisp evaluates the arguments. (+ 1 2) -> 3
Lisp calls the function FOO with the evaluated argument: (funcall #'foo 3)
Lisp computes the function FOO: (EVAL 3) -> 3
Lisp returns the value(s) from FOO -> 3
Start by using a proper defun:
(defun request (url)
(dex:get url))
CL-USER> (request "http://…")
Now you want to use something else than dex:get? Well… write another function, because their argument handling, headers, return values… might be different.
(defun request-drakma (url)
(drakma:… url))
Maybe in later code you want to reference to either function?
(defun do-request (url &optional (get-fn #'request))
(funcall get-fn url))
(defvar example #'(lambda (x) (* x 20)))
Here you are giving a name to an anonymous function… just use defun^^

Memoization with a closure example from Land of Lisp

On page 329 of Land of Lisp, Conrad Barski explains the technique of memoization with the following example code
(let ((old-neighbors (symbol-function 'neighbors))
(previous (make-hash-table)))
(defun neighbors (pos)
(or (gethash pos previous)
(setf (gethash pos previous) (funcall old-neighbors pos)))))
The idea is nice: when I call the neighbors function, I save the result into a hash table, so that the next time calling neighbors with the same value of pos, I can just look-up the result without having to compute it again.
So I was wondering, whether it would not be easier to rename the function neighbors to old-neighbors by editing and recompiling its source code (given on page 314 of the book). Then the memoization example could be simplified to
(let ((previous (make-hash-table)))
(defun neighbors (pos)
(or (gethash pos previous)
(setf (gethash pos previous) (funcall old-neighbors pos)))))
or, by turning previous into a global variable *previous-neighbors* beforehand, even into
(defun neighbors (pos)
(or (gethash pos *previous-neighbors*)
(setf (gethash pos *previous-neighbors*) (funcall old-neighbors pos))))
thus rendering the closure unnecessary.
So my question is: what is the reason for doing it this way?
Reasons I could imagine:
It's didactical, showing what could be done with a closure (which had been explained just before) and providing an example of symbol-function.
This technique is applicable even in situations, where you cannot or may not change the source code of neighbors.
I am missing something.
You have made some good observations.
Generally the style to use closures like that is more likely to be found in Scheme code - where Scheme developers often use functions to return functions.
Generally as you have detected it makes little sense to have a memoize function foo calling an function old-foo. Using global variables reduces encapsulation (-> information hiding), but increases the ability to debug a program, since one can more easily inspect the memoized values.
A similar, but potentially more useful, pattern would be this:
(defun foo (bar)
<does some expensive computation>)
(memoize 'foo)
Where ˋmemoizeˋ would be something like this
(defun memoize (symbol)
(let ((original-function (symbol-function symbol))
(values (make-hash-table)))
(setf (symbol-function symbol)
(lambda (arg &rest args)
(or (gethash arg values)
(setf (gethash arg values)
(apply original-function arg args)))))))
The advantage is that you don't need to write the memoizing code for each function. You only need one function memoize. In this case the closure also makes sense - though you could also have a global table storing memoize tables.
Note the limitations of above: the comparison uses EQL and only the first argument of the function to memoize.
There are also more extensive tools to provide similar functionality.
See for example:
https://github.com/fare/fare-memoization
https://github.com/sharplispers/cormanlisp/blob/master/examples/memoize.lisp
https://github.com/AccelerationNet/function-cache
Using my memoize from above:
CL-USER 22 > (defun foo (n)
(sleep 3)
(expt 2 n))
FOO
CL-USER 23 > (memoize 'foo)
#<Closure 1 subfunction of MEMOIZE 40600008EC>
The first call with arg 10 runs three seconds:
CL-USER 24 > (foo 10)
1024
The second call with arg 10 runs faster:
CL-USER 25 > (foo 10)
1024
The first call with arg 2 runs three seconds:
CL-USER 26 > (foo 2)
4
The second call with arg 2 runs faster:
CL-USER 27 > (foo 2)
4
The third call with arg 10 runs fast:
CL-USER 28 > (foo 10)
1024

avoiding side effects implementing OR macro function in lisp

I'm trying to implement the macro-function OR in Lisp
My attempt:
(defmacro or2 (test &rest args)
`(if ,test ,test (if (list ,#args) (or2 ,#args) nil)) )
However, if I test with something like this:
(or2 (print 1) 2 )
1
1
1
Whereas with the default OR:
(or (print 1) 2)
1
1
I understand that this is because of my two ,test at the beginning of my if clause, but I don't see how i could avoid it. How could I avoid applying twice the test effects ?
How would you solve the problem of side-effects if you had to code it by hand?
(or2 (print 1) 2)
Intermediate variable
Most probably, you would do this:
(let ((value (print 1)))
(if value value 2))
You need to define a local variable which holds the value of the first expression, so that later you can reference the variable instead of re-evaluating the same expression more than once.
But what if you already have a variable named value in the lexical context where you expand the code? What if, instead of 2, you were referencing that other value? This problem is named variable capture.
Gensym
In Common Lisp, you introduce a fresh symbol, that is guaranteed to not be already bound to anything, using GENSYM.
(let ((symbol (gensym)))
`(let ((,symbol ,test))
(if ,symbol ,symbol ...)))
Recursive expansion
(list ,#args)
The above is the same as writing directly ,args.
But you are confusing macroexpansion and execution times. If you inject args directly in the code, it will be evaluated (most likely, this is going to fail as a bad function call). What you want instead is to test if args is non-null during macroexpansion.
Besides, you should probably first test if your list of expression contains more than one element, in order to simplify the generated code.
Roughly speaking, you have to take into account the following cases:
(or2) is nil
(or2 exp) is the same as exp
(or2 exp &rest args) is the same as the following, where var is a fresh symbol:
`(let ((,var ,exp))
(if ,var ,var (or2 ,#args)))
Please make use of macroexpand-1:
(macroexpand-1 '(or2 (print 1) 2))
; ==> (if (print 1) (print 1) (if (list 2) (or2 2) nil)) ;
; ==> t
With macros you wish the order of evaluation to be expected and you wish expressions to only be evaluated once. Thus the expansion should have been something like this:
(let ((tmp (print 1)))
(if tmp
tmp
(or2 2)))
And tmp should be a symbol generated by gensym. Also when args is nil you should expand or2 to only test:
(defmacro or2 (test &rest args)
(if (endp args)
test
(let ((tmp (gensym "tmp")))
`(let ((,tmp ,test))
(if ,tmp
,tmp
(or2 ,#args))))))
you can make use of macros to simplify this:
(defmacro or2 (test &rest args)
(if (endp args)
test
(once-only (test)
`(if ,test
,test
(or2 ,#args)))))

Pass a "variable name" to defvar

I've been struggling with this for two days now, and I can't find the answer.
What I want is to define three variables, a, b, and c each with a value of 0.
Naive:
(dolist (lbl '(a b c)) (defvar lbl 0))
Doesn't do what I want. a, b, and c remain undefined, and lbl now has a value of 0.
I think I may understand why this can't work: defvar is a macro, not a function, and as such I am passing it the form lbl, not the current value of label (which is a, b, c in turn). I think.
But in the resulting macroexpansion, shouldn't lbl eventually be linked-up(?) or evaluated(?) to the value I'm intending? Obviously not, either because it can't be done or I'm doing it wrong.
I want to understand:
How to make this work: (dolist (lbl '(a b c)) (defvar lbl 0))
What's going wrong under the hood. I have a feeling it has something to do with symbols or the mechanics of the quote operator.
Here are a few options:
With eval, by building a defvar expression:
(dolist (lbl '(a b c))
(eval `(defvar ,lbl 0))
With proclaim and setf of symbol-value (note: set is deprecated, since 1994 for what it's worth):
(dolist (lbl '(a b c))
(proclaim `(special ,lbl))
(setf (symbol-value lbl) 0))
This is actually mostly what defvar does (see notes in the linked page), but each Lisp implementation usually also records source file location, as they do for other defining macros.
Under the hood, defvar is a macro that makes the variable special (i.e. with dynamic extent bindings in the current dynamic environment; note: there's no portable undoing for this!), and optionally initializes it if it's not yet bound.
The fact that it's a macro means it doesn't evaluate its arguments, so it can take the variable name literally, and it does so. As such, (defvar lbl 0) will define the variable lbl, not the symbol stored in a lbl variable.
The fact that it optionally initializes the variable means that the initializing expression will not even be evaluated if the variable is boundp. So, its secondary effects won't happen if the variable is already initialized. This might or might not be expected.
Note that this expression isn't actually evaluated at macro-expansion time, it's left for evaluation when the expansion is evaluated, which in a REPL means right after macro expansion (and possibly after compilation, depending on the Lisp implementation; read more about evaluation and compilation, it's quite interesting).
Similar:
(dolist (lbl '(a b c))
(let ((lbl 0))
(print lbl)))
Why is lbl 0 and not some of a, b, c?
Because LET binds the symbol lbl and not its value.
Similar with (DEFVAR FOO 3).
Imagine following code:
(DEFVAR FOO 3)
(LET ((FOO 3)) ...)
Now, if we compile this code, the Lisp compiler recognizes the DEFVAR declaration and now knows that FOO is a special global variable. Thus in the let form FOO will be dynamically bound.
Compare this code:
(dolist (v '(FOO)) (eval `(DEFVAR ,v 3)))
(LET ((FOO 3)) ...)
The compiler won't see the DEFVAR and does not know that it should be a global special variable. In the LET form, FOO will have a lexical binding.
Thus DEFVAR needs to be a macro which knows the symbol at compile time (!) and which expands into a form that informs the compiler that the symbol is a special global variable. The form also sets the value when executed.
Thus the best way to create multiple DEFVAR declarations from a list of variables is to write a macro, which expands into a PROGN form with multiple DEFVARs. Inside the PROGN, the compiler will still recognize them.
CL-USER 21 > (pprint (macroexpand '(defvar* (a b c) 0)))
(PROGN (DEFVAR A 0) (DEFVAR B 0) (DEFVAR C 0))
Implemented as:
(defmacro defvar* (vars initial-value)
`(progn
,#(loop for var in vars
do (check-type var symbol)
collect `(defvar ,var ,initial-value))))
Note that it makes sense to check that the variables are really provided as symbols.
defvar is a special form which makes sure the symbol of it's first argument is a bound variable. If the variable is not bound the evaluated expression of the second argument becomes the bound variables value. Thus:
(defvar *x* 10) ; if *x* was not bound it's now 10
(defvar *x* 20) ; since *x* is defined nothing happens
Notice that *x* is not evaluated but is used unevaluated. In order to get the same functionality by using a variable that evaluates to a symbol which you want to exist as a variable in global scope you need to do something like this:
(defvar b 10)
(dolist (lbl '(a b c))
(when (not (boundp lbl))
(setf (symbol-value lbl) 0)))
Still, neither of the ones not already bound becomes special like with defvar, but at least you get the same behaviour:
(list a b c) ; => (0 10 0)
Perhaps you should just do:
(defvar *a* 0)
(defvar *b* 0)
(defvar *c* 0)
If you have a lot of variables you need to do this with you can do:
(defmacro defvars (lst value)
(loop :for e :in lst
:collect `(defvar ,e ,value) :into result
:finally (return (cons 'progn result))))
(defparameter *w* 10)
(defvars (*q* *w* *e*) 1)
(list *q* *w* *e* ; ==> (1 10 1)
Also, it's really important to earmuff your global variables. Once special it will follow dynamic binding. eg.
(defun test ()
(let ((*b* 15))
(test2)))
(defun test2 ()
*b*)
(test) ; ==> 15
Reimplementing DEFVAR
You can approximate the behavior of defvar with a function like this:
(defun %defvar (symbol value documentation)
"Define a global special variable.
symbol---a symbol
value---nil or a function of zero arguments
documentation---nil or a documentation string
returns symbol
Proclaim SYMBOL globally as a special variable. If VALUE is non-nil,
then if SYMBOL is not already bound, SYMBOL is assigned the value
returned by calling VALUE. DOCUMENATION is assigned as the
documentation of type variable to for SYMBOL."
(prog1 symbol
;; make it globally special
(proclaim (list 'special symbol))
;; if a value is provided, and symbol isn't
;; already bound, set its value to the result
;; of calling the value-function
(when (not (null value))
(unless (boundp symbol)
(setf (symbol-value symbol)
(funcall value))))
;; set the documentation
(setf (documentation symbol 'variable) documentation)))
Then you can do, e.g.,
CL-USER> (%defvar '*the-answer* (lambda () 42) "the answer")
*THE-ANSWER*
CL-USER> *the-answer*
42
CL-USER> (documentation '*the-answer* 'variable)
"the answer"
And with your original code, you could do something like:
(dolist (lbl '(a b c)) (%defvar lbl (lambda () 0)))
Now, how does this relate to what defvar actually does? Well, you could now implement a defvar like macro by doing:
(defmacro define-var (symbol &optional (value nil value-p) documentation)
`(%defvar
',symbol
,(if value-p `(lambda () ,value) 'nil)
,documentation))
This expands as we'd expect:
CL-USER> (macroexpand-1 '(define-var *the-answer* 42 "the answer"))
(%DEFVAR '*THE-ANSWER* (LAMBDA () 42) "the answer")
You can actually use macroexpand to look at what an implementation does, too. E.g., in SBCL:
CL-USER> (macroexpand-1 '(defvar *the-answer* 42 "the answer"))
(PROGN
(EVAL-WHEN (:COMPILE-TOPLEVEL) (SB-IMPL::%COMPILER-DEFVAR '*THE-ANSWER*))
(SB-IMPL::%DEFVAR '*THE-ANSWER* (UNLESS (BOUNDP '*THE-ANSWER*) 42) 'T
"the answer" 'T (SB-C:SOURCE-LOCATION)))
This isn't too much different from what we wrote above, though it's handling the non-evaluation of the form when the variable is already bound in a slightly different way, and it's also got some handling for recording a source location. The general idea is the same, though.
Why things don't get "linked up"
But in the resulting macroexpansion, shouldn't lbl eventually be
linked-up(?) or evaluated(?) to the value I'm intending?
The original code is:
(dolist (lbl '(a b c)) (defvar lbl 0))
We can macroexpand this to see what it becomes (in SBCL):
CL-USER> (macroexpand '(dolist (lbl '(a b c)) (defvar lbl 0)))
(BLOCK NIL
(LET ((#:N-LIST1022 '(A B C)))
(TAGBODY
#:START1023
(UNLESS (ENDP #:N-LIST1022)
(LET ((LBL (TRULY-THE (MEMBER C B A) (CAR #:N-LIST1022))))
(SETQ #:N-LIST1022 (CDR #:N-LIST1022))
(TAGBODY (DEFVAR LBL 0)))
(GO #:START1023))))
NIL)
T
Now, we can still see LBL in two places, including in (defvar LBL 0). So why don't things get "matched up"? To see that, we need to remember that the defvar inside the let will also be macroexpanded. To what? This:
CL-USER> (macroexpand '(DEFVAR LBL 0))
(PROGN
(EVAL-WHEN (:COMPILE-TOPLEVEL) (SB-IMPL::%COMPILER-DEFVAR 'LBL))
(SB-IMPL::%DEFVAR 'LBL (UNLESS (BOUNDP 'LBL) 0) 'T NIL 'NIL
(SB-C:SOURCE-LOCATION)))
But now we see that SBCL's internals are getting the symbol named "LBL"; the call (sb-impl::%defvar 'lbl …) is calling the function sb-impl::%defvar with the symbol lbl, and there's no connection between that symbol and the lexical variable that happens to be represented in the source by the same symbol. After all, if you write:
CL-USER> (let ((a 89))
(list 'a a))
(A 89)
You want to be able to get the symbol a and the number 89, right? The macroexpansion of defvar includes a call to a function with the quotation of one of the arguments to macro.

Tracing a closure

Is it possible to trace a closure in CL? For ex., can I trace foo-3 below?
(defun foo (n)
(lambda (i) (incf n i)))
FOO
(setf foo-3 (foo 3))
#<CLOSURE :LAMBDA (I) (INCF N I)>
(funcall foo-3 2)
5
(funcall foo-3 2)
7
(trace ???)
I don't think this is possible: as far as I know, the trace macro generally works by replacing the function at a given symbol by a wrapper that calls the original and also prints out the tracing bit.
If you're interested in the (complicated) implementation details, the SBCL code is in src/code/ntrace.lisp (you probably want to look at the trace-1 function).
Of course, if all you want to do is print something out when foo-3 is called, you could always put a print statement inside the lambda form in foo...
It is indeed possible to do so. Trace looks for functions in the function-namespace, so make sure to not mix values and functions.
(setf (symbol-function 'test)
(let ((n 0))
(lambda (x)
(incf n x))))
=>
#<Interpreted Closure TEST>
(trace test)
...
(test 4)
=>
0[2]: (TEST 4)
0[2]: returned 4
4
(test 3)
=>
0[2]: (TEST 3)
0[2]: returned 7
7
I think the problem here is that trace requires a function name, rather than there being a problem with tracing closures. Continuing from your example above, you can call foo-3 from a named function, and trace that:
(defun call-foo-3 (i)
(funcall foo-3 i))
(trace call-foo-3)
(call-foo-3 2)
0: (CALL-FOO-3 2)
0: CALL-FOO-3 returned 15

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