I've been trying to write a recursive loop in clojure that will print me out the very last number in the list. The point is not that I need to get the last number (for which I'm sure there's a built in function for that) but that I want to better understand recursion and macros in clojure. So I have this macro...
(defmacro loop-do [the-list]
`(if (= (count '~the-list) 1)
(println (first '~the-list))
(loop-do (rest '~the-list))))
But I get a stackoverflow error. What am I doing wrong?
How will people use your macro?
Somewhere, someone will call:
(loop-do list)
As a piece of code, those are only two symbols in a list. The first one is recognized as your macro, and the second one, list, is a symbol that represents a variable that will be bound at runtime. But your macro only knows that this is a symbol.
The same goes for:
(loop-do (compute-something))
The argument is a form, but you do not want to get the last element of that form, only the last element of the list obtained after evaluating the code.
So: you only know that in your macro, the-list will be bound to an expression that, at runtime, will have to be a list. You cannot use the-list as-if it was a list itself: neither (count 'list) nor (count '(compute-something)) does what you want.
You could expand into (count list) or (count (compute-something)), though, but the result would only be computed at runtime. The job of the macro is only to produce code.
Recursive macros
Macros are not recursive: they expand into recursive calls.
(and a b c)
might expand as:
(let [a0 a] (if a0 a0 (and b c)))
The macroexpansion process is a fixpoint that should terminate, but the macro does not call itself (what would that mean, would you expand the code while defining the macro?). A macro that is "recursive" as-in "expands into recursive invocations" should have a base case where it does not expand into itself (independently of what will, or will not, happen at runtime).
(loop-do x)
... will be replaced by:
(loop-do (rest 'x))
... and that will be expanded again.
That's why the comments say the size actually grows, and that's why you have a stackoverflow error: macroexpansion never finds a fixpoint.
Debugging macros
You have a stackoverflow error. How do you debug that?
Use macroexpand-1, which only performs one pass of macroexpansion:
(macroexpand-1 '(loop-do x))
=> (if (clojure.core/= (clojure.core/count (quote x)) 1)
(clojure.core/println (clojure.core/first (quote x)))
(user/loop-do (clojure.core/rest (quote x))))
You can see that the generated code still contains a call to usr/loop-do , but that the argument is (clojure.core/rest (quote x)). That's the symptom you should be looking for.
Related
SBCL generates spurious style warnings about undefined functions. (The functions are defined, just later in the file.) I want to solve this problem once and for all. Fortunately, there is a way to do this:
(declaim (sb-ext:muffle-conditions style-warning))
The downside is that CCL, for obvious reasons, barfs on a program containing the above. I try to solve this problem with a conditional:
(#+sbcl (declaim (sb-ext:muffle-conditions style-warning)))
but now SBCL is unhappy: "illegal function call".
How do you put such a declaim into a portable program?
Note that while the existing answer is right, disabling warnings is not a good practice. In your case, it is probably not necessary.
Common Lisp has a notion of compilation unit, where multiple definitions are grouped together. This gives a chance for the compiler/interpreter to take care of cross-references among functions (an interpreter could collect warnings and keep only those that are not found later, for example).
For example, in file #P"/tmp/foo.pl":
(defun mut-rec-foo (x)
(when (plusp x)
(mut-rec-bar (1- x))))
(defun mut-rec-bar (x)
(print x)
(mut-rec-foo (1- x)))
Do not evaluate anything in the file; instead do:
(compile-file #P"/tmp/foo.pl")
; compiling (DEFUN MUT-REC-FOO ...)
; compiling (DEFUN MUT-REC-BAR ...)
; /tmp/foo.fasl written
; compilation finished in 0:00:00.002
No warning. You can then call (load #P"/tmp/foo.fasl") to have the definitions in your current lisp environment, without warnings.
Typically, ASDF and by extension Quicklisp use COMPILE-FILE, so your problem should disappear as soon as you bundle your files into a system.
You can also do:
(with-compilation-unit ()
(defun mut-rec-foo/bis (x)
(when (plusp x)
(mut-rec-bar/bis (1- x))))
(defun mut-rec-bar/bis (x)
(print x)
(mut-rec-foo/bis (1- x))))
Evaluating the whole block shows no warning for *EVALUATOR-MODE* being both :COMPILE or :INTERPRET.
What you witnessed happens when you evaluate each expression one after the other (or maybe one region after another one). There, the compiler has no way to know that the function already exists. Silencing the warning is the worse option, because you might actually have made an error.
If you know in advance that a function will exist, but not in your compilation unit (maybe it is only defined at runtime), the you can declaim that fact, as follows:
(declaim (ftype function my-function))
The above says that my-function must be assumed to be fbound to an object of type function. You could also give more information by refining what kind of function you claim it to be:
(declaim (ftype (function (number) (values string &optional)) num-to-string))
... for a function that accepts a number and returns exactly one value, a string.
(declaim (ftype (function () nil) forever-loop))
... for a function that accepts nothing and never return a value (loop or signals an error).
Omit the outer pair of parentheses:
#+sbcl (declaim (sb-ext:muffle-conditions style-warning))
As you are using declaim, I assume, that the declaration appears at the top-level of a compilation unit. If you need to group multiple top-level statements, you can wrap them all with a progn (which doesn't change the "top-level"-ness).
The reason SBCL did complain is, that its reader reads
((declaim (sb-ext:muffle-conditions style-warning)))
(as the :SBCL feature is present), which is simply a syntax error. CCL does not complain, because its reader reads
()
which is simply another way to spell nil.
Consider this question. Here the basic problem is the code:
(progv '(op arg) '(1+ 1)
(eval '(op arg)))
The problem here is that progv binds the value to the variable as variable's symbol-value not symbol-function. But, that's obvious because we didn't explicitly suggest which values are functions.
The Plan
So, in order to solve this problem, I thought of manually dynamically binding the variables, to their values based on the type of values. If the values are fboundp then they should be bound to the symbol-function of the variable. A restriction, is that match-if can't be a macro. It has to be a function, because it is called by a funcall.
Macro : functioner:
(defmacro functioner (var val)
`(if (and (symbolp ',val)
(fboundp ',val))
(setf (symbol-function ',var) #',val)
(setf ,var ,val)))
Function: match-if:
(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))))
Here, the let part declares all the variables lexically (supposedly). Then declare declares them special. Then functioner binds the variables and their values aptly. Then the code in the pattern is evaluated. If the code part is true, then only the pattern-matcher function pat-match is invoked.
The Problem
The problem is that in the function, all it's arguments are evaluated. Thus bindings in the let and declare parts will be replaced by something like :
((v1 . val1)(v2 . val2)(v3 . val3))
not
'((v1 . val1)(v2 . val2)(v3 . val3))
So, it's treated as code, not a list.
So, I'm stuck here. And macros won't help me on this one.
Any help appreciated.
Not the answer you are looking for, but PROGV is a special operator; it is granted the ability to modify the dynamic bindings of variables at runtime; AFAIK, you can't simply hack it to operate on "dynamic function bindings".
The point of progv is to use list of symbols and values that are evaluated, meaning that you can generate symbols at runtime and bind them dynamically to the corresponding values.
You might be able to find a solution with eval but note that if you macroexpand into (eval ...), then you loose the surrounding lexical context, which is generally not what you want ("eval" operates on the null lexical environment). I speculate that you could also have a custom code walker which works on top-level forms but reorganizes them, when it finds your special operator, to bring the context back in, producing something like (eval '(let (...) ...)).
Why doesn't this work?
( ((lambda () (lambda (x) (funcall #'1+ x)))) 2)
; yields Compile-time error: illegal function call
I ran into a situation like this and it later turned out that a funcall fixes it, i.e.
(funcall ((lambda () (lambda (x) (funcall #'1+ x)))) 2) ; => 3
I'm confused because it seems like the first one should work, because I actually have a function I'm calling, not just a symbol that may belong to either namespace (i.e. (type-of ((lambda () #'1+))) ; => FUNCTION). I thought it would be kind of like how you don't need to funcall a lambda for example, e.g.((lambda (x) x) :HI) ; => :HI. What am I missing?
Common Lisp uses the word form for everything which can be evaluated.
A form is either
a symbol like foo
a compound form, a list, see below
or a self-evaluating object (like numbers, characters, arrays, strings, ...).
A compound form is either
a special form (<special-operator> ...)
a lambda form like (lambda (...) ...)
a macro form (<macroname> ...)
or a function form (<functionname> ...).
Above is the set of compound forms. The ANSI Common Lisp specification provides no way to add a new type of forms or a different syntax. The interface of what forms the functions like EVAL or COMPILE accept is not extensible.
So something like
(((lambda (foo)
(lambda (bar)
(list foo bar)))
1)
2)
is not valid Common Lisp. This is not meaningful in Common Lisp:
( <not a lambda form,
not a special operator,
not a macro name
and not a function name>
2)
Note that Common Lisp allows lambda forms, special operators, macro names and function names as the first element in a compound form. But it does not allow variables and it does not allow other compound forms as the first element in a compound form.
Means this is not meaningful in Common Lisp:
( <a function form> 2)
Thus ((foo 1) 2) or (((foo 1) 2) 3) or ((((foo 1) 2) 3) 4) or (((((foo 1) 2) 3) 4) 5) is not legal in Common Lisp. You get the idea. To call function objects returned from function calls, we have to use (funcall (foo ...) ...). This makes calling returned function objects more obvious than just ((foo ...) ...).
Let's praise the designers of Common Lisp for this feature. Otherwise I might have to look at possibly meaningful code beginning with
(((((((((((( .....
and it would be very hard to figure out what it does. Basically that would be write-only code.
Your question:
Why must I funcall a function returned from another?
The short answer: because the syntax does not allow other ways, in Common Lisp.
The syntax of Common Lisp requires that, everytime you want to call a function through a compund form of the type:
(f a1 a2 ... an)
the first element of the list, f, must be a symbol denoting a function name, or a list denoting a lambda expression, i.e. (see the manual):
lambda expression n. a list which can be used in place of a function name in certain contexts to denote a function by directly describing its behavior rather than indirectly by referring to the name of an established function; its name derives from the fact that its first element is the symbol lambda.
So, this basically means that you cannot have as first element any expression that returns a function as value. In those cases, you must use funcall.
So, in your second example, the first argument of the funcall is ((lambda () (lambda (x) (funcall #'1+ x)))), which is a correct coumpound form, in which the first element of the list is the lambda expression (lambda () (lambda (x) (funcall #'1+ x))) (applied to an empty list of arguments).
In the first example, instead, you have as first element of the list an expression returning a function, so that you must use funcall.
I have read this article and then modified my code below：
(defparameter *my-fun* 1)
(defun my-func (v0)
(setf (symbol-function '*my-fun*)
(lambda (v1)
(+ v0 v1)))
'*my-fun*)
And call it in this way ((my-func 2) 3),but it also report "illegal function call". I think my code is keeping with lambda calculus,but where is wrong.
In my opinion, (my-func 2) returns symbol *my_fun*, and the function cell of *my-fun* points to a function object, so ((my-func 2) 3) => (*my-fun* 3) => ((lambda (v1) (+ 2 v1)) 3) => (+ 2 3) => 5
This works: First your identical definitions:
(defparameter *my-fun* 1)
(defun my-func (v0)
(setf (symbol-function '*my-fun*)
(lambda (v1)
(+ v0 v1)))
'*my-fun*)
However, invoke using funcall:
(funcall (my-func 2) 3)
I've just been reading up on the sharpsign colon reader macro and it sounded like it had a very similar effect to gensym
Sharpsign Colon: "introduces an uninterned symbol"
Gensym: "Creates and returns a fresh, uninterned symbol"
So a simple test
CL-USER> #:dave
; Evaluation aborted on #<UNBOUND-VARIABLE DAVE {1002FF77D3}>.
CL-USER> (defparameter #:dave 1)
#:DAVE
CL-USER> #:dave
; Evaluation aborted on #<UNBOUND-VARIABLE DAVE {100324B493}>.
Cool so that fails as it should.
Now for the macro test
(defmacro test (x)
(let ((blah '#:jim))
`(let ((,blah ,x))
(print ,blah))))
CL-USER> (test 10)
10
10
CL-USER>
Sweet so it can be used like in a gensym kind of way.
To me this looks cleaner than gensym with an apparently identical result. I'm sure I'm missing a vital detail so my question is, What it it?
Every time the macro is expanded, it will use the same symbol.
(defmacro foo () `(quote #:x))
(defmacro bar () `(quote ,(gensym)))
(eq (foo) (foo)) => t
(eq (bar) (bar)) => nil
Gensym will create a new symbol every time it is evaluated, but sharp colon will only create a new symbol when it is read.
While using sharp colon is unlikely to cause problems, there are a couple rare cases where using it would lead to nearly impossible to find bugs. It is better to be safe to begin with by always using gensym.
If you want to use something like sharp colon, you should look at the defmacro! macro from Let Over Lambda.
GENSYM is like MAKE-SYMBOL. The difference is that GENSYM supports fancy naming by counting up -> thus symbols kind of have unique names, which makes debugging a bit easier when having gensyms for example in macro expansions.
#:foo is a notation for the reader.
So you have a function which creates these and a literal notation. Note that, when *print-circle* is true, some kind of identity maybe preserved in s-expressions: #(#1=#:FOO #1#).
Generally this is similar to (a . b) and (cons 'a 'b), #(a b) and (vector 'a 'b)... One is literal data and the other one is a form which will create ('cons') fresh objects.
If you look at your macro, the main problem is that nested usage of it could cause problems. Both lexically or dynamically.
lexically it could be the same variable, which is rebound.
dynamically, if it is a special variable it could also be rebound
Using a generated symbol at macro expansion time would make sure that different and expanded code would not share bindings.
I just discovered Racket a few days ago, and I'm trying to get more comfortable with it by writing a little script that generates images to represent source code using #lang slideshow.
I know that when programming in a functional paradigm it's good practice to create almost all your variables with let, but I find that it introduces too many levels of nesting and that Racket's let has an overcomplicated API which requires superfluous parentheses. I'm sure this is to remove ambiguity when using let in more powerful ways, but for my purposes it's just an annoyance. Consequently, I'm creating all my variables with define, and writing blocks with begin if I need to (such as in the body of an if statement).
The problem is that I've repeatedly been getting what seem to be very mysterious errors. I'm sure I'm just making some silly beginner's mistake, being new to the language, but I really can't seem to find the source of the complaint.
Here's the offending code:
(define sub-code (foldr ht-append (rectangle 0 0) (map internal-style (rest code))))
although what we're defining sub-code to seems pretty irrelevant. If I replace it with
(define sub-code '())
I receive the same error. DrRacket is saying that define is being used in an expression context. I understand what this error would normally mean - IE that it would raise when you write code like (print (define x 10)), but I can't see what would trigger it here.
If it helps, this define is at the beginning of a begin block, inside an if statement
(if (list? code)
(begin
(define sub-code '())
; a few more define statements and finally an expression ))
The specific error message DrRacket is printing is
define: not allowed in an expression context in: (define sub-code (quote ()))
I thought maybe define isn't allowed in begin blocks, but I checked the docs and one of the examples for begin is
(begin
(define x 10)
x)
So I don't really know what to do. Thanks in advance!
Definitions are allowed in a 'body' context, like in lambda and let among others. The consequent and alternate clauses of if are not body contexts; they are expression contexts and therefore definitions are not allowed.
begin is special - begin in a body context allows definitions, but begin in an expression contexts forbids definitions. Your case falls in to the later.
For example:
(define (foo . args) #| body context #|)
(define foo (lambda args #| body context |#))
(define (foo . args)
(let (...)
#| body context |#))
Syntactic keywords that requires expressions: if, cond, case, and, or, when, unless, do, begin. Check out the formal syntax in any Scheme report (r{4,5,6,7}rs); look for <body>, <sequence>, <command>, and <expression>.
Also, if you need a body context in an expression, just wrap a let syntactic form, as such:
(if test
(let ()
(define foo 'foo)
(list foo foo))
alternate)
As GoZoner explained, you can't use define in an expression context.
What could you do instead?
Use let:
(if (list? code)
(let ([x '()])
x)
...
Or it would work with an "empty" let and define:
(if (list? code)
(let ()
(define x '())
x)
...
But that's a bit silly.
Or use cond and define:
(cond [(list? code)
(define x '())
x]
...
This last way -- using cond and define -- is closest to what the current Racket style guide recommends.
Here's more details, from the Racket docs.
The different contexts are required because macros must expand differently, depending on which language forms are allowed.
As others have said, definitions are not allowed in expression contexts ("expr ..." in the docs), but are ok in other contexts.
In other doc entries, "body ..." indicates an internal-definition context (guide, reference), for example in lambda bodies, and "form ..." indicates all non-expression contexts, like in the docs for begin.
Or you could wrap the expressions in (begin)
e.g.(begin
(define x 10)
(define y 100)
(define z 1000))