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)))))
Related
I am trying to write an fx in lisp to tell if an object ends in nil.
(setq isList (lambda (listOfValues)
(if (null listOfValues) t)
( funcall isList (cdr listOfValues) )
)
)
However, I am having trouble checking if its nil in all cases. In particular, cdr would fail at last elt if it is not a list. How can I resolve this?
Before we get closer to answer your actual question, a few things. First, use defun to define functions, not "set a variable to a lambda", it will make you happier down the line. Second, Common Lisp style would vale been one of values, list-of-values, or just list (that would indicate we knew it was a list, so I would probably just have gone with values here), not "listOfValues" (case is typically smashed, and neither "listofvalues" nor "LISTOFVALUES" are easy to read).
So, back to the code. A list is composed of cons cells, of either atoms or other cons cells. We have two test functions, either consp or atom that would be useful in this case. We know that if we're looking at a cons, we need to recurse on its cdr, otherwise we're at the last element and can just check if we're looking at nil.
(defun is-proper-list (values)
(if (consp values)
(is-proper-list (cdr values))
(null values))) ;; We could do this test as (eql nil) as well
It can be done faster with
(defun listp (l)
(tailp nil l))
(tailp nil ...) tests, whether nil is the end of a given object after cdr-ing to the end.
tailp is a very special function. So don't use it without understanding it.
(tailp '(b c) '(a b c)) is e.g. NOT T, because '(b c) is not the same object like the ( ... b c). But in this case, because NIL is '() and is unique in Lisp, any nil is object-identical. Therefore one can use tailp here for this specific test, whether a given list ends with NIL.
listp is a lisp-convention conform name for this.
(predicate functions returning booleans ending with p for predicate.
Since no - used in the name, attach p without - otherwise attach -p).
(tailp (cdr '(a b c)) '(a b c)) ;; NIL
;; because the two lists are not object-identical
(setq l '(a b c))
(tailp (cddr l) l) ;; T ;; object-identical
I have a code which takes a list and returns all possible permutations by the parameter result.
But when I compile I have an error which says *** - =: (1+ INDEX) is not a number.
Is this message true or I messed up the code generally?
I am new to lisp I can looking for a fix and also open to suggestions from fucntional programmers.
;; Creates permutatiions of a given list and returns it via parameter
(defun create-permuations (source)
(setf result (list))
(create-permuations-helper source 0 '() result)
result)
(defmacro create-permuations-helper (source index cur result)
(if (= (list-length cur) index)
(cons cur result)
(loop for i from 0 to (list-length cur) do
(create-permuations-helper source (1+ index)
(append cur (list (nth i source))) result))))
99% of times when a compiler reports an error you can trust it to be true. Here Index is the list (1+ index), literally the 1+ symbol followed by the index symbol. This is so because you are using a macro, and macros operate on code.
In your macro, you do not return a form to be evaluated, you execute code during macro-expansion that depends on itself. That alone is an undefined behaviour. For example:
(defmacro a (x)
(if (plusp x)
(a (- x 1))
nil))
In the body of a, you want to expand code using a recursive call to itself. But the macro is not yet fully known and cannot be until the whole macro is defined.
Maybe the particular lisp implementation binds a to the macro function in body of the macro, which is a strange thing to do, or you evaluated the definition twice. The first time the compiler assumes a is an unknown function, then binds a to a macro, and the second time it tries to expand the macro.
Anyway macro are not supposed to be recursive.
In the example, since the macro does not evaluate its argument, the nested call to the macro is given the literal expression (- x 1), and not its actual value, which cannot be known anyway since x is unknown. You are crossing a level of abstraction here by trying to evaluate things at macroexpansion time.
But, macros can expand into code that refers to themselves.
(defmacro a (x)
(if (plusp x)
`(b (a ,(- x 1)))
nil))
Now, (a 2) expands into (b (a 1)), which itself macroexpands into (b (b (a 0))), and finally reaches a fixpoint which is (b (b nil)).
The difference is that the macro produces a piece of code and returns, which the compiler macroexpands again, whereas in the first example, the macro must already be expanded in the body of its own definition.
Possible implementation
One way to solve your problem is to define a local function that has access to a variable defined in your main function. Then, the local function can set it, and you do not need to pass a variable by reference (which is not possible to do):
(defun permut (list)
(let (result)
(labels ((recurse (stack list)
(if list
(dolist (x list)
(recurse (cons x stack)
(remove x list :count 1)))
(push stack result))))
(recurse nil list))
result))
Alternatively, you can split the process in two; first, define permut-helper, which is a higher-order function that takes a callback function; it generates permutations and calls the callback for each one:
(defun permut-helper (stack list callback)
(if list
(dolist (x list)
(permut-helper (cons x stack)
(remove x list :count 1)
callback))
(funcall callback stack)))
You call it with a function that pushes results into a list of permutations:
(defun permut (list)
(let (result)
(flet ((add-result (permutation)
(push permutation result)))
(permut-helper nil list #'add-result))
result))
Let us suppose we have a function func1 :
(defun func1 (&rest values)
; (do something with values...)
(loop for i in values collect i))
Now, we have a function func2 which calls func1 :
(defun func2 (&rest values)
; (do something with values...)
(func1 ???))
What should I put instead of ??? to "copy" all the parameters of func2's values to func1's values ?
For instance, I would have the following behavior :
(func2 1 2 3 4) ; result is (1 2 3 4) and not ((1 2 3 4)).
In an earlier question I tried to do something like this :
(defun func2 (&rest values)
(macrolet ((my-macro (v)
`(list ,#v)))
(func1 (my-macro values))))
But the defun cannot get the value because it is not runtime. In this answer, he suggested that I use apply, but this function takes a &rest parameter too, so it doesn't solve my problem...
If possible, I would rather avoid to change the prototype of both functions, and the behavior of func1.
In common lisp, it has to be
(apply #'func1 values) ;; since `func1` has to be looked up in function namespace
remember, Clojure and Racket/Scheme are Lisp1, and common lisp is Lisp2.
Alternative solution (just for the sake)
I was asking myself, how to get it done without apply - just for the sake.
The problem with
`(func2 ,#values)
is, that if e.g.
(func2 (list 1 2 3) (list 4) 5)
is called, the values variable is ((1 2 3) (4) 5)
But when it is spliced into (func1 ,#values), what is created is
(func1 (1 2 3) (4) 5). But if we compare this with the func2 call,
it should be rather (func1 (list 1 2 3) (list 4) 5) which is perhaps not possible, because when (func2 (list 1 2 3) (list 4) 5) is called -
in the lisp manner - the arguments of func2 are each evaluated, before they enter the function body of func2, so we end up with values as a list of already evaluated arguments, namely ((1 2 3) (4) 5).
So somehow, concerning the arguments for func1 in the last expression, we are one evaluation-step offbeat.
But there is a solution with quote, that we manage to quote each of the arguments before giving it to func1 in the last expression, to "synchronize" the func1 function call - to let the arguments' evaluation pause for one round.
So my first aim was to generate a new values list inside the func2 body where each of the values list's argument is quoted (this is done in the let-binding).
And then at the end to splice this quoted-values list into the last expression: (func1 '(1 2 3) '(4) '5) which can be regarded as equivalent to (func1 (list 1 2 3) (list 4) 5) for this kind of problems / for this kind of calls.
This was achieved by this code:
(defun func2 (&rest vals)
(let ((quoted-values (loop for x in vals
collect `',x)))
; do sth with vals here - the func2 function -
(eval `(func1 ,#quoted-values))))
This is kind of a macro (it creates code btw. it organizes new code) but executed and created in run-time - not in pre-compile time. Using an eval we execute that generated code on the fly.
And like macroexpand-1, we can look at the result - the code - to which the func1 expression "expands", by removing eval around it - I call it func2-1:
(defun func2-1 (&rest vals)
(let ((quoted-values (loop for x in vals
collect `',x)))
; do sth with vals here - the func2 function -
`(func1 ,#quoted-values)))
And if we run it, it returns the last expression as code immediately before it is evluated in the func2 version:
(func2-1 (list 1 2 3) (list 4) 5)
;; (FUNC1 '(1 2 3) '(4) '5) ;; the returned code
;; the quoted arguments - like desired!
And this happens if we call it using func2 (so with evaluation of the func1 all:
(func2 (list 1 2 3) (list 4) 5)
;; ((1 2 3) (4) 5) ;; the result of (FUNC1 '(1 2 3) '(4) '5)
So I would say this is exactly what you desired!
lists vs. spread arguments
In Common Lisp it is good style to pass lists as lists and not as spread arguments:
(foo (list 1 2 3)) ; better interface
(foo 1 2 3) ; interface is not so good
The language has been defined in a way that efficient function calling can be used by a compiler and this means that the number of arguments which can be passed to a function is limited. There is a standard variable which will tell us how many arguments a particular implementation supports:
This is LispWorks on my Mac:
CL-USER 13 > call-arguments-limit
2047
Some implementations allow much larger number of arguments. But this number can be as low as 50 - for example ABCL, Common Lisp on the JVM, allows only 50 arguments.
Computing with argument lists
But sometimes we want the arguments as a list and then we can use the &rest parameter:
(lambda (&rest args)
(print args))
This is slightly in-efficient, since a list will be consed for the arguments. Usually Lisp tries to avoid to cons lists for arguments - they will be passed in registers or on the stack - if possible.
If we know that the argument list will not be used, then we can give the compiler a hint to use stack allocation - if possible:
(lambda (&rest args)
(declare (dynamic-extent args))
(reduce #'+ args))
In above function, the list of arguments can be deallocated when leaving the function - because the argument list is no longer used then.
If you want to pass these arguments to another function you can use FUNCALL and usually more useful APPLY:
(lambda (&rest args)
(funcall #'write (first args) (second args) (third args)))
or more useful:
(lambda (&rest args)
(apply #'write args))
One can also add additional arguments to APPLY before the list to apply:
CL-USER 19 > ((lambda (&rest args)
(apply #'write
(first args) ; the object
:case :downcase ; additional args
(rest args))
(values))
'(defun foo () 'bar)
:pretty t
:right-margin 15)
(defun foo ()
'bar)
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.
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