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Use of # a.k.a. read-macro
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I'm using mapcar in Common Lisp, and I've seen examples that use #' and ' in front of the + but they both appear to do the same thing. Does it matter which is used?
CL-USER> (mapcar '+ '(1 2) '(2 3))
(3 5)
CL-USER> (mapcar #'+ '(1 2) '(2 3))
(3 5)
They are often interchangeable, but are not the same thing. mapcar takes a function designator – a symbol designating a symbol-function or a function designating itself – which will be coerced to a function object. #'foo is a reader macro that expands to (function foo), which returns a function object by looking up the functional value of foo in the lexical environment, i.e. the innermost flet, labels, or the global function definition. Thus, nothing needs to be done to coerce it. If you use a quoted symbol with mapcar, however, it is coerced to a function using symbol-function, which will not work with non-global function definitions, as there are no fboundp symbols involved in this case. For example:
CL-USER> (flet ((frob (x) (1+ x)))
(mapcar #'frob '(1 2 3)))
(2 3 4)
CL-USER> (flet ((frob (x) (1+ x)))
(mapcar 'frob '(1 2 3)))
; Evaluation aborted on #<CCL::UNDEFINED-FUNCTION-CALL #x18AB7F1E>.
Also, when exactly the coercion happens, is implementation dependent, which has consequences for self or mutualy redefining code.
In general, using #' seems to be widely preferred nowadays, probably because it is felt to be more consistent – using #' everywhere you want to pass a function is a simple rule. I rarely see quoted symbols used with functions taking function designators in CL. In other dialects however, especially older or dynamically scoped ones like Emacs Lisp, passing quoted symbols to these functions is quite common.
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.
By the end of chapter 10, on the advanced section there is this question:
10.9. Write a destructive function CHOP that shortens any non-NIL list to a list of one element. (CHOP '(FEE FIE FOE FUM)) should return
(FEE).
This is the answer-sheet solution:
(defun chop (x)
(if (consp x) (setf (cdr x) nil))
x)
I understand this solution. However, before checking out the official solution I tried:
(defun chop (xs)
(cond ((null xs) xs)
(t (setf xs (list (car xs))))))
Using this as a global variable for tests:
(defparameter teste-chop '(a b c d))
I tried on the REPL:
CL-USER> (chop teste-chop)
(A)
As you can see, the function returns the expected result.
Unfortunately, the side-effect to mutate the original list does not happen:
CL-USER> teste-chop
(A B C D)
Why it did not change?
Since I was setting the field (setf) of the whole list to be only its car wrapped as a new list, I was expecting the cdr of the original list to be vanished.
Apparently, the pointers are not automatically removed.
Since I have very strong gaps in low-level stuff (like pointers) I thought that some answer to this question could educate me on why this happens.
The point is about how parameters to functions are passed in Common Lisp. They are passed by value. This means that, when a function is called, all arguments are evaluated, and their values are assigned to new, local variables, the parameters of the function. So, consider your function:
(defun chop (xs)
(cond ((null xs) xs)
(t (setf xs (list (car xs))))))
When you call it with:
(chop teste-chop)
the value of teste-chop, that is the list (a b c d) is assigned to the function parameter xs. In the last line of the body of function, by using setf, you are assigning a new value, (list (car xs)) to xs, that is you are assigning the list (a) to this local variable.
Since this is the last expression of the function, such value is also returned by the function, so that the evaluation of (chop test-chop) returns the value (a).
In this process, as you can see, the special variable teste-chop is not concerned in any way, apart from calculating its value at the beginning of the evaluation of the function call. And for this reason its value is not changed.
Other forms of parameter passing are used in other languages, like for instance by name, so that the behaviour of function call could be different.
Note that instead, in the first function, with (setf (cdr x) nil) a data structure is modified, that is a part of a cons cell. Since the global variable is bound to that cell, also the global variable will appear modified (even if, in a certain sense, it is not modified, since it remains bound to the same cons cell).
As a final remark, in Common Lisp it is better not to modify constant data structures (like those obtained by evaluating '(a b c d)), since it could produce an undefined behaviour, depending on the implementation. So, if some structure should be modifiable, it should be built with the usual operators like cons or list (e.g. (list 'a 'b 'c 'd)).
I am pretty much of a newbie to common-lisp and only use it for fun. But I assume to know the difference between mapper and mapcan, as the documentation in the hyperspec and other places is pretty clear.
But what happens if the function mapcan calls on the list elements evaluates to an atom instead of a list? As mapcan uses nconc to append lists, I had expected that there would be an error if there is no list.
But if I try
(mapcan (lambda (x) (+ 2 x)) '(1 2 3 4))
it evaluates to '6' in sbcl and clisp. (There might not be a practical need for this example; I am just curious) I see the point that returning a value might be nicer than a simpler error but could find anything about mapcan returning the last value if there are no lists to nconc.
Is there a reason for this behavior?
According to the documentation for mapcan (mapcan (lambda (x) (+ 2 x)) '(1 2 3 4)) Should do the same as (apply #'nconc (mapcar (lambda (x) (+ 2 x)) '(1 2 3 4))) and it signals an error the error *** - NCONC: 5 is not a list in clisp.
The hyperspec only shows what nconc should be doing with proper lists and nil as the arguments before last. It does not have anything else described so what you are seeing is that sbcl and clisp perhaps share the algorithm from a lisp in public domain or that they have implemented it so similar they have the same implementation specific results.
You probably cannot assume other implementations will do the same so you should make sure the function passed to mapcan always return a fresh list or nil that can be nconc-ed within the specification.
For Racket programming language, why is lambda not considered a function?
For example, it can't be defined as a higher order function like this.
(define (my-lambda args body)
(lambda args body))
There's a key distinction that your question is missing:
lambda is syntax.
Procedures are values.
A lambda form is a form of expression whose value is a procedure. The question whether "lambda is a function" starts off with a type error, so to speak, because lambdas and procedures don't live in the same world.
But let's set that aside. The other way to look at this is by thinking of it in terms of evaluation rules. The default Scheme evaluation rule, for the application of a procedure to arguments, can be expressed in pseudo-code like this:
(define (eval-application expr env)
(let ((values
;; Evaluate each subexpression in the same environment as the
;; enclosing expression, and collect the result values.
(map (lambda (subexpr) (eval subexpr env))
expr)))
;; Apply the first value (which must be a procedure) to the
;; other ones in the results.
(apply (car values) (cdr values))))
In English:
Evaluate all of the subexpressions in the same environment as the "parent".
apply the first result (which must have evaluated to a procedure) to the list of the rest.
And now, another reason lambda can't be a procedure is that this evaluation rule doesn't work for lambda expressions. In particular, the point of lambda is to not evaluate its body right away! This, in particular, is what afflicts your my-lambda—if you try to use it this way:
(my-lambda (x) (+ x x))
...the (x) in the middle must be immediately evaluated as an invocation of a procedure named x in the environment where the whole expression appears. The (+ x x) must also be immediately evaluated.
So lambda requires its own evaluation rule. As Basile's answer points out, this is normally implemented as a primitive in the Scheme system implementation, but we can sketch it in pseudocode with something like this:
;;;
;;; Evaluate an expression of this form, returning a procedure:
;;;
;;; (lambda <formals> <body> ...)
;;;
(define (eval-lambda expr env)
(let ((formals (second expr))
(body (cddr expr)))
;; We don't evaluate `body` right away, we return a procedure.
(lambda args
;; `formals` is never evaluated, since it's not really an
;; expression on its own, but rather a subpart that cannot
;; be severed from its enclosing `lambda`. Or if we want to
;; say it all fancy, the `formals` is *syncategorematic*...
(let ((bindings (make-bindings formals args)))
;; When the procedure we return is called, *then* we evaluate
;; the `body`--but in an extended environment that binds its
;; formal parameters to the arguments supplied in that call.
(eval `(begin ,#body) (extend-environment env bindings))))))
;;;
;;; "Tie" each formal parameter of the procedure to the corresponding
;;; argument values supplied in a given call. Returns the bindings
;;; as an association list.
;;;
(define (make-bindings formals args)
(cond ((symbol? formals)
`((,formals . args)))
((pair? formals)
`((,(car formals) . ,(car args))
,#(make-bindings (cdr formals) (cdr args))))))
To understand this pseudocode, the time-tested thing is to study one of the many Scheme books that demonstrate how to build a meta-circular interpreter (a Scheme interpreter written in Scheme). See for example this section of Structure and Interpretation of Computer programs.
lambda needs to be a core language feature (like if, let, define are) in Scheme because it is constructing a closure so needs to manage the set of closed or free variables (and somehow put their binding in the closure).
For example:
(define (translate d) (lambda (x) (+ d x)))
When you invoke or evaluate (translate 3) the d is 3 so the dynamically constructed closure should remember that d is bound to 3. BTW, you generally want the result of (translate 3) and of (translate 7) be two different closures sharing some common code (but having different bindings for d).
Read also about λ-calculus.
Explaining that all in details requires an entire book. Fortunately, C. Queinnec has written it, so read his Lisp In Small Pieces book.
(If you read French, you could read the latest French version of that book)
See also the Kernel programming language.
Read also wikipage about evaluation strategy.
PS. You could, and some Lisp implementations (notably MELT and probably SBCL) do that, define lambda as some macro -e.g. which would expand to building some closure in an implementation specific way (but lambda cannot be defined as a function).
A function call (e0 e1 e2) is evaluated like this
e0 is evaluated, the result is (hopefully) a function f
e1 is evaluated, the result is a value v1
e2 is evaluated, the result is a value v2
The function body of f is evaluated in an environment in which
the formal parameters are bound to the values v1 and v2.
Note that all expressions e0, e1, and, e2 are evaluated before the body of the function is activated.
This means that a function call like (foo #t 2 (/ 3 0)) will result in an error when (/ 3 0) is evaluated - before control is handed over to the body of foo.
Now consider the special form lambda. In (lambda (x) (+ x 1)) this creates a function of one variable x which when called with a value v will compute (+ v 1).
If in contrast lambda were a function, then the expressions (x) and (+ x 1) are evaluated before the body of lambda is activated. And now (x) will most likely produce an error - since (x) means call the function x with no arguments.
In short: Function calls will always evaluate all arguments, before the control is passed to the function body. If some expressions are not to be evaluated a special form is needed.
Here lambda is a form, that don't evaluate all subforms - so lambda needs to be a special form.
In Scheme lingo we use the term procedure instead of function throughout the standard report. Thus since this is about scheme dialects I'll use the term procedure.
In eager languages like standard #!racket and #!r6rs procedures get their arguments evaluated before the body is evaluated with the new lexical environment. Thus since if and lambda has special evaluation rules than for procedures special forms and macros are the way to introduce new syntax.
In a lazy language like #!lazy racket evaluation is by need and thus many forms that are implemented as macros/special forms in an eager language can be implemented as procedure. eg. you can make if as a procedure using cond but you cannot make cond using if because the terms themselves would be evaluated as forms on access and eg (cond (#t 'true-value)) would fail since #t is not a procedure. lambda has similar issue with the argument list.
I understand that, because there are separate namespaces in Common Lisp for functions and variables, you can do this:
((lambda (x) (* 2 x)) 3)
and you can also do this:
(funcall #'(lambda (x) (* 2 x)) 3)
When should we use #' as opposed to not using it? I read in another StackOverflow question that #' was only kept around for historic reasons and shouldn't be used anymore. Is this true? My question is not a duplicate, I am asking about when I would use these in my code.
It's not an issue of lisp-2 versus lisp-1. With lambda expressions in a position where a function value is needed, it's simply a stylistic choice. Some people like the visual marker of #' and some don't. The lambda macro already expands into the function form for which #' provides an abbreviation:
Macro LAMBDA
(lambda lambda-list [[declaration* | documentation]] form*)
== (function (lambda lambda-list [[declaration* | documentation]] form*))
== #'(lambda lambda-list [[declaration* | documentation]] form*)
#'x is just syntactic sugar for (function x), and the function special operator provides "the functional value of name in the current lexical environment." T
Special Operator FUNCTION
The value of function is the functional value of name in the current
lexical environment.
If name is a function name, the functional definition of that name is
that established by the innermost lexically enclosing flet, labels, or
macrolet form, if there is one. Otherwise the global functional
definition of the function name is returned.
While (lambda ...) is the name of a function, it's not a name that could ever be established by a flet, label, or macrolet form, so you're always getting "the global definition of the function name", which is just the lambda function. Since (lambda ...) expands to (function (lambda ...)), there's no difference. It's just a matter of style.
However, it is important to note that in the first case that you talked about,
((lambda (x) (* x 2)) 3)
you could not do:
(#'(lambda (x) (* x 2)) 3) ; or
((function (lambda (x) (* x 2))) 3)
The support for ((lambda ...) ...) is part of the language, unrelated to the fact that there's a definition of lambda as a macro. It's a particular type of compound form, namely a lambda form, which is described in the HyperSpec:
3.1.2.1.2.4 Lambda Forms
A lambda form is similar to a function form, except that the function
name is replaced by a lambda expression.
A lambda form is equivalent to using funcall of a lexical closure of
the lambda expression on the given arguments. (In practice, some
compilers are more likely to produce inline code for a lambda form
than for an arbitrary named function that has been declared inline;
however, such a difference is not semantic.)
For further information, see Section 3.1.3 (Lambda Expressions).
In common lisp there is map, which lets you do this kind of thing:
(map (lambda (x y) (/ x y)) (list 2 4 6 8 10 12) (list 1 2 3 4 5 6))
returning (2 2 2 2 2 2)
However now I am working at ACL2 and there is no such a thing as map.
So in my mind the only choice left I have is doing recursion to calculate what I want, unless there is another simpler and/or more efficient way of doing it.
... Which is exactly my question. Is there a better way of doing it than to create a recursive function called something like divide-two-lists? It just feels like something that a lisp-based language should naturally do instead of having you to create another function specifically just for it, hence why I am asking.
You could pretty easily write your own map. From the GNU Emacs guide:
(defun mapcar* (function &rest args)
"Apply FUNCTION to successive cars of all ARGS.
Return the list of results."
;; If no list is exhausted,
(if (not (memq nil args))
;; apply function to cars.
(cons (apply function (mapcar 'car args))
(apply 'mapcar* function
;; Recurse for rest of elements.
(mapcar 'cdr args)))))
(mapcar* 'cons '(a b c) '(1 2 3 4))
⇒ ((a . 1) (b . 2) (c . 3))
I'm unfamiliar with acl2, so you might have to change some functions (e.g. memq), or deal differently with how apply or &rest arguments work, but this is the meat of the code.
ACL2 is based on first order logic. In first order logic, statements like
(define (P R A) (R A))
are not allowed because R is being used as both a parameter and a function.
It is theoretically possible to get around this limitation by literally defining your own language within first order logic that includes the constructs for higher order logic. Otherwise, you are correct, your best option is to define something like divide-two-lists every single time you want to use a map function.
That's tedious, but it is how ACL2 was meant to be used.
This isn't exactly suitable to your question, but it's related, and so I mention it in case it helps someone else who is looking at your question.
Consider the book "std/util/defprojection", which provides a macro that lets you map a function across a list.