Are symbols and names different? On Lisp by Paul Graham, which focuses on common lisp, has some discussions that seem to imply so, e.g.
Since lambda-expressions are also names of functions, they can also appear first in function calls:
((lambda (x) (* x 2) 3)
6
This makes it sound like symbols are names but names aren't symbols. But I don't understand what sort of Lisp "object" symbols are / could be.
This is also deriving from my question here on the sharp-quote (#') operator v. symbol-function. I'm suspecting the only reason these are different is because not all names are symbols but I don't have enough background to understand those answers yet (hence this question).
I'm also asking for clarification in elisp v. common lisp. I'm assuming this pertains to lexical form, which wasn't introduced in elisp until version 24 (24.1 I think).
Lambda Expressions are not names of functions. It's just that ((lambda (...) ...) ...) is allowed in Common Lisp, since it is defined in the standard as legal syntax.
The only allowed function names in Common Lisp are symbols and lists like (setf symbol).
For example one can write
(defun (setf foo) (...) ...)
Here (setf foo) is the function name.
Other function names don't exist in Common Lisp, only symbols and (setf symbol) names.
Common Lisp Hyperspec Glossary: Function Name.
function name n. 1. (in an environment) A symbol or a list (setf
symbol) that is the name of a function in that environment. 2. A
symbol or a list (setf symbol).
Note: the Common Lisp version of 1984 (as published in CLtL1) did only have symbols as function names. Thus the idea of a function name was not defined. The function to retrieve a function from a symbol was called SYMBOL-FUNCTION. In 1989 the ANSI CL standardization group decided to add setf lists as function names. It also introduced the function FDEFINITION, which is like SYMBOL-FUNCTION but also accepts other function names, besides symbols. See here: Issue FUNCTION-NAME.
I think Rainer's answer gets this right, but since the question appeared in a comment on my answer to another question, I'll include (an update to) my response from that comment.
A symbol is an actual object. You can inspect it, you can create them with make-symbol, etc. Importantly, symbols are one of the primary components of source code in Common Lisp. Function names, especially in the context that this question arose in (arguments to the function special operator) are either symbols or lists of the form (setf symbol) according to the glossary entry:
function name n. 1. (in an environment) A symbol or a list (setf symbol) that is the name of a function in that environment. 2. A
symbol or a list (setf symbol).
Function is a special operator that doesn't evaluate its arguments, so passing a symbol or setf list means something like:
(function car)
(function (setf car))
and not:
(function 'car)
(function '(setf car))
Now, lexical variables, e.g., x in (let ((x 42)) x), while represented by symbols in the source code, don't actually have any connection with the symbol at runtime. The compiled version of (let ((x 42)) x) doesn't need to know anything about the symbol x. Intuitively, this makes sense, because we'd expect the code (let ((y 42)) y) to compile to the same thing. However, when a variable is special, there is a connection with the symbol. The difference is clearest with:
(let ((x 42))
(symbol-value x))
;=> NIL
(let ((x 42))
(declare (special x)) ; or (defparameter x ...) or (defvar x ...) earlier
(symbol-value x))
;=> 42
We'd expect the same thing to be true of lexically scoped functions, e.g., the following code causes an error because there's no connection between the symbol x at runtime and the local function:
(flet ((x () 42))
(symbol-function 'x)) ; ERROR, no function value for symbol x
But even so, we can still do:
(flet ((x () 42))
(function x))
This is because function is special operator and can access the environment where is occurs. That means that (because it's special, and the implementation makes it work) it can know that x is defined as function here. It may be interesting to note, now, that since flet and labels are defined to take a function name, you can do:
(flet (((setf kar) (value kons)
...))
...)
Related
I am hoping someone can explain why tests 1-5 work but test 6 does not. I thought that quoting a lambda with ' and using #' in front of a lambda both returned pointers to the function with the only difference being that the #' will compile it first.
(defun test-1 (y)
(mapcar (lambda (x) (expt x 2))
'(1 2 3)))
(defun test-2 (y)
(mapcar (lambda (x) (expt x y))
'(1 2 3)))
(defun test-3 (y)
(mapcar #'(lambda (x) (expt x 2))
'(1 2 3)))
(defun test-4 (y)
(mapcar #'(lambda (x) (expt x y))
'(1 2 3)))
(defun test-5 (y)
(mapcar '(lambda (x) (expt x 2))
'(1 2 3)))
(defun test-6 (y)
(mapcar '(lambda (x) (expt x y))
'(1 2 3)))
I am using the free version of Franz Industries Allegro Common Lisp. The following are the outputs:
(test-1 2) ; --> (1 4 9)
(test-2 2) ; --> (1 4 9)
(test-3 2) ; --> (1 4 9)
(test-4 2) ; --> (1 4 9)
(test-5 2) ; --> (1 4 9)
(test-6 2) ; --> Error: Attempt to take the value of the unbound variable `Y'. [condition type: UNBOUND-VARIABLE]
For a start, you should be aware that your tests 1-4 are conforming Common Lisp, while your tests 5 and 6 are not. I believe Allegro is perfectly well allowed to do what it does for 5 and 6, but what it is doing is outside the standard. The bit of the standard that talks about this is the definition of functions like mapcar, which take function designators as argument, and the definition of a function designator:
function designator n. a designator for a function; that 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). The consequences are undefined if a symbol is used as a function designator but it does not have a global definition as a function, or it has a global definition as a macro or a special form. [...]
From this it is clear that a list like (lambda (...) ...) is not a function designator: it's just a list whose car happens to be lambda. What Allegro is doing is noticing that this list is in fact something that can be turned into a function and doing that.
Well, let's just write a version of mapcar which does what Allegro's does:
(defun mapcar/coercing (maybe-f &rest lists)
(apply #'mapcar (coerce maybe-f 'function) lists))
This just uses coerce which is a function which knows how to turn lists like this into functions, among other things. If its argument is already a function, coerce just returns it.
Now we can write the two tests using this function:
(defun test-5/coercing (y)
(mapcar/coercing '(lambda (x) (expt x 2))
'(1 2 3)))
(defun test-6/coercing (y)
(mapcar/coercing '(lambda (x) (expt x y))
'(1 2 3)))
So, after that preamble, why can't test-6/explicit work? Well the answer is that Common Lisp is (except for for special variables) lexically scoped. Lexical scope is just a fancy way of saying that the bindings (variables) that are available are exactly and only the bindings you can see by looking at the source of the program. (Except, in the case of CL for special bindings, which I'll ignore, since there are none here.)
So, given this, think about test-6/coercing, and in particular the call to mapcar/coercing: in that call, coerce has to turn the list (lambda (x) (expt z y)) into a function. So it does that. But the function it returns doesn't bind y and there is no binding for y visible in it: the function uses y 'free'.
The only way that this could work is if the function that coerce constructs for us were to dynamically look for a binding for y. Well, that's what dynamically-scoped languages do, but CL is not dynamically-scoped.
Perhaps a way of making this even clearer is to realise that we can lift the function creation right out of the function:
(defun test-7 (y f)
(mapcar f '(1 2 3)))
> (test-7 1 (coerce '(lambda (x) (expt x y)) 'function))
It's clear that this can't work in a lexically-scoped language.
So, then, how do tests 1-4 work?
Well, firstly there are only actually two tests here. In CL, lambda is a macro and (lambda (...) ...) is entirely equivalent to (function (lambda (...) ...)). And of course #'(lambda (...) ...) is also the same as (function (lambda (...) ...)): it's just a read-macro for it.
And (function ...) is a magic thing (a special form) which says 'this is a function'. The important thing about function is that it's not a function: it's a deeply magic thing which tells the evaluator (or the compiler) that its argument is the description of a function in the current lexical context, so, for instance in
(let ((x 1))
(function (lambda (y) (+ x y))))
The x referred to by the function this creates is the x bound by let. So in your tests 2 and 4 (which are the same):
(defun test-4 (y)
(mapcar (function (lambda (x) (expt x y)))
'(1 2 3)))
The binding of y which the function created refers to is the binding of y which is lexically visible, which is the argument of test-4 itself.
Let's add a y parameter to avoid closing over variables and see what kind of values we are manipulating:
USER> (type-of #'(lambda (x y) (expt x y)))
FUNCTION
USER> (type-of (lambda (x y) (expt x y)))
FUNCTION
USER> (type-of '(lambda (x y) (expt x y)))
CONS
As you can see, the two first lambda-like forms are evaluated as functions, while the third is evaluated as a cons-cell. As far as Lisp is concerned, the third argument is just a tree of symbols with no meaning.
Reader macros
I thought that quoting a lambda with ' and using #' in front of a lambda both returned pointers to the function with the only difference being that the #' will compile it first.
Let's go back to the definitions, ' and #' are reader macros, respectively Single-Quote and Sharpsign Single-Quote. They are found in front of other forms, for example 'f is read as (quote f) and #'f is read as (function f). At read-time, f and the resulting forms are just unevaluated data.
We will see below how both special operators are interpreted, but what matters really is the lexical scope, so let's open a parenthesis.
Lexical environment
Lexical environments are the set of bindings in effect at some point of your code. When you evaluate a let or an flet it enriches the current environment with new bindings. When you call EVAL on an expression, you start evaluating from a null lexical environment, even if the call to eval itself is in a non-null environment.
Here x is just unbound during eval:
(let ((x 3)) (eval '(list x))) ;; ERROR
Here we build a let to be evaluated by eval:
(eval '(let ((x 3)) (list x)))
=> (3)
That's all for the crash course on lexical environments.
Special operators
FUNCTION
Special operator FUNCTION takes an argument that is either the name of a function (symbol or setf), or a lambda expression; in particular:
The value of function is the functional value of name in the current lexical environment.
Here the lambda expression is evaluated in the current lexical environment, which means it can refer to variable outside the lambda expression. That's the definition of closures, they capture the surrounding bindings.
NB. you do not need to prefix lambda with #', because there is a macro named (lambda ...) that expands into (function (lambda ...)). It looks like this could expand recursively forever, but this is not the case: at first the macro is expanded so that (lambda ...) becomes (function (lambda ...)), then the special operator function knows how to evaluate the lambda expression itself.
This means that (lambda ...) and #'(lambda ...) are equivalent. Note in particular that there is nothing about whether one form is compiled or not at this point, the compiler will see the same expression after macroexpansion.
QUOTE
Special operator QUOTE evaluates (quote f) as f, where f itself is unevaluated. In test-5 and test-6, there is no function, just an unevaluated structured expression that can be interpreted as code.
Type coercion
Now, certain functions like MAPCAR are used to apply functions. Notice how the specification says that the function parameter is a function designator:
function --- a designator for a function that must take as many arguments as there are lists.
A designator for a type is not necessarily a value of that type, but can be a value that can be coerced to that type. Sometimes a user wants to specify a pathname, and enters a string, but a string is not a value of type pathname: the system has to converts the string into a pathname.
Common Lisp defines a COERCE function with rules regarding how values can be converted to other values. In you case, mapcar first does (coerce (lambda ...) 'function). This is defined as follows:
If the result-type is function, and object is a lambda expression, then the result is a closure of object in the null lexical environment.
The value is thus evaluated in a null lexical environment, so it does not have access to the surrounding bindings; y is a free variable in your lambda expression, and since it is evaluated in a null environment, it is unbound. That's why test-5 pass but test-6 fails.
Name resolution, compilers and late binding
There is a difference whether you write #'f or 'f when referring to a function f where f is a symbol: in the first case, the expression evaluated to an object of type function, and in the second case, you only evaluate a symbol.
Name resolution for this function can change depending and how the compiler works. With a symbol as a function designator, the function does not even need to be defined, the name is resolved when the symbols has to be coerced as a function.
When you write #'f, some compilers may remove one level of indirection and directly make your code jump to the code associated with the function, without having to resolve the name at runtime.
However, this also means that with such compilers (e.g. SBCL), you need to recompile some call sites on function redefinition, as-if the function was declared inline, otherwise some old code will still reference the previous definition of #'f. This is something that is not necessarily important to consider at the beginning, but it can be a source of confusion to keep in mind when you are live coding.
In this post, I ask tangentially why when I declare in SBCL
(defun a (&rest x)
x)
and then check what the function cell holds
(describe 'a)
COMMON-LISP-USER::A
[symbol]
A names a compiled function:
Lambda-list: (&REST X)
Derived type: (FUNCTION * (VALUES LIST &OPTIONAL))
Source form:
(LAMBDA (&REST X) (BLOCK A X))
I see this particular breakdown of the original function. Could someone explain what this output means? I'm especially confused by the last line
Source form:
(LAMBDA (&REST X) (BLOCK A X))
This is mysterious because for some reason not clear to me Lisp has transformed the original function into a lambda expression. It would also be nice to know the details of how a function broken down like this is then called. This example is SBCL. In Elisp
(symbol-function 'a)
gives
(lambda (&rest x) x)
again, bizarre. As I said in the other post, this is easier to understand in Scheme -- but that created confusion in the answers. So once more I ask, Why has Lisp taken a normal function declaration and seemingly stored it as a lambda expression?
I'm still a bit unclear what you are confused about, but here is an attempt to explain it. I will stick to CL (and mostly to ANSI CL), because elisp has a lot of historical oddities which just make things hard to understand (there is an appendix on elisp). Pre-ANSI CL was also a lot less clear on various things.
I'll try to explain things by writing a macro which is a simple version of defun: I'll call this defun/simple, and an example of its use will be
(defun/simple foo (x)
(+ x x))
So what I need to do is to work out what the expansion of this macro should be, so that it does something broadly equivalent (but simpler than) defun.
The function namespace & fdefinition
First of all I assume you are comfortable with the idea that, in CL (and elisp) the namespace of functions is different than the namespace of variable bindings: both languages are lisp-2s. So in a form like (f x), f is looked up in the namespace of function bindings, while x is looked up in the namespace of variable bindings. This means that forms like
(let ((sin 0.0))
(sin sin))
are fine in CL or elisp, while in Scheme they would be an error, as 0.0 is not a function, because Scheme is a lisp-1.
So we need some way of accessing that namespace, and in CL the most general way of doing that is fdefinition: (fdefinition <function name>) gets the function definition of <function name>, where <function name> is something which names a function, which for our purposes will be a symbol.
fdefinition is what CL calls an accessor: this means that the setf macro knows what to do with it, so that we can mutate the function binding of a symbol by (setf (fdefinition ...) ...). (This is not true: what we can access and mutate with fdefinition is the top-level function binding of a symbol, we can't access or mutate lexical function bindings, and CL provides no way to do this, but this does not matter here.)
So this tells us what our macro expansion needs to look like: we want to set the (top-level) definition of the name to some function object. The expansion of the macro should be like this:
(defun/simple foo (x)
x)
should expand to something involving
(setf (fdefinition 'foo) <form which makes a function>)
So we can write this bit of the macro now:
(defmacro defun/simple (name arglist &body forms)
`(progn
(setf (fdefinition ',name)
,(make-function-form name arglist forms))
',name))
This is the complete definition of this macro. It uses progn in its expansion so that the result of expanding it is the name of the function being defined, which is the same as defun: the expansion does all its real work by side-effect.
But defun/simple relies on a helper function, called make-function-form, which I haven't defined yet, so you can't actually use it yet.
Function forms
So now we need to write make-function-form. This function is called at macroexpansion time: it's job is not to make a function: it's to return a bit of source code which will make a function, which I'm calling a 'function form'.
So, what do function forms look like in CL? Well, there's really only one such form in portable CL (this might be wrong, but I think it is true), which is a form constructed using the special operator function. So we're going to need to return some form which looks like (function ...). Well, what can ... be? There are two cases for function.
(function <name>) denotes the function named by <name> in the current lexical environment. So (function car) is the function we call when we say (car x).
(function (lambda ...)) denotes a function specified by (lambda ...): a lambda expression.
The second of these is the only (caveats as above) way we can construct a form which denotes a new function. So make-function-form is going to need to return this second variety of function form.
So we can write an initial version of make-function-form:
(defun make-function-form (name arglist forms)
(declare (ignore name))
`(function (lambda ,arglist ,#forms)))
And this is enough for defun/simple to work:
> (defun/simple plus/2 (a b)
(+ a b))
plus/2
> (plus/2 1 2)
3
But it's not quite right yet: one of the things that functions defined by defun can do is return from themselves: they know their own name and can use return-from to return from it:
> (defun silly (x)
(return-from silly 3)
(explode-the-world x))
silly
> (silly 'yes)
3
defun/simple can't do this, yet. To do this, make-function-form needs to insert a suitable block around the body of the function:
(defun make-function-form (name arglist forms)
`(function (lambda ,arglist
(block ,name
,#forms))))
And now:
> (defun/simple silly (x)
(return-from silly 3)
(explode-the-world x))
silly
> (silly 'yes)
3
And all is well.
This is the final definition of defun/simple and its auxiliary function.
Looking at the expansion of defun/simple
We can do this with macroexpand in the usual way:
> (macroexpand '(defun/simple foo (x) x))
(progn
(setf (fdefinition 'foo)
#'(lambda (x)
(block foo
x)))
'foo)
t
The only thing that's confusing here is that, because (function ...) is common in source code, there's syntactic sugar for it which is #'...: this is the same reason that quote has special syntax.
It's worth looking at the macroexpansion of real defun forms: they usually have a bunch of implementation-specific stuff in them, but you can find the same thing there. Here's an example from LW:
> (macroexpand '(defun foo (x) x))
(compiler-let ((dspec::*location* '(:inside (defun foo) :listener)))
(compiler::top-level-form-name (defun foo)
(dspec:install-defun 'foo
(dspec:location)
#'(lambda (x)
(declare (system::source-level
#<eq Hash Table{0} 42101FCD5B>))
(declare (lambda-name foo))
x))))
t
Well, there's a lot of extra stuff in here, and LW obviously has some trick around this (declare (lambda-name ...)) form which lets return-from work without an explicit block. But you can see that basically the same thing is going on.
Conclusion: how you make functions
In conclusion: a macro like defun, or any other function-defining form, needs to expand to a form which, when evaluated, will construct a function. CL offers exactly one such form: (function (lambda ...)): that's how you make functions in CL. So something like defun necessarily has to expand to something like this. (To be precise: any portable version of defun: implementations are somewhat free to do implementation-magic & may do so. However they are not free to add a new special operator.)
What you are seeing when you call describe is that, after SBCL has compiled your function, it's remembered what the source form was, and the source form was exactly the one you would have got from the defun/simple macro given here.
Notes
lambda as a macro
In ANSI CL, lambda is defined as a macro whose expansion is a suitable (function (lambda ...)) form:
> (macroexpand '(lambda (x) x))
#'(lambda (x) x)
t
> (car (macroexpand '(lambda (x) x)))
function
This means that you don't have to write (function (lambda ...)) yourself: you can rely on the macro definition of lambda doing it for you. Historically, lambda wasn't always a macro in CL: I can't find my copy of CLtL1, but I'm pretty certain it was not defined as one there. I'm reasonably sure that the macro definition of lambda arrived so that it was possible to write ISLisp-compatible programs on top of CL. It has to be in the language because lambda is in the CL package and so users can't portably define macros for it (although quite often they did define such a macro, or at least I did). I have not relied on this macro definition above.
defun/simple does not purport to be a proper clone of defun: its only purpose is to show how such a macro can be written. In particular it doesn't deal with declarations properly, I think: they need to be lifted out of the block & are not.
Elisp
Elisp is much more horrible than CL. In particular, in CL there is a well-defined function type, which is disjoint from lists:
> (typep '(lambda ()) 'function)
nil
> (typep '(lambda ()) 'list)
t
> (typep (function (lambda ())) 'function)
t
> (typep (function (lambda ())) 'list)
nil
(Note in particular that (function (lambda ())) is a function, not a list: function is doing its job of making a function.)
In elisp, however, an interpreted function is just a list whose car is lambda (caveat: if lexical binding is on this is not the case: it's then a list whose car is closure). So in elisp (without lexical binding):
ELISP> (function (lambda (x) x))
(lambda (x)
x)
And
ELISP> (defun foo (x) x)
foo
ELISP> (symbol-function 'foo)
(lambda (x)
x)
The elisp intepreter then just interprets this list, in just the way you could yourself. function in elisp is almost the same thing as quote.
But function isn't quite the same as quote in elisp: the byte-compiler knows that, when it comes across a form like (function (lambda ...)) that this is a function form, and it should byte-compile the body. So, we can look at the expansion of defun in elisp:
ELISP> (macroexpand '(defun foo (x) x))
(defalias 'foo
#'(lambda (x)
x))
(It turns out that defalias is the primitive thing now.)
But if I put this definition in a file, which I byte compile and load, then:
ELISP> (symbol-function 'foo)
#[(x)
"\207"
[x]
1]
And you can explore this a bit further: if you put this in a file:
(fset 'foo '(lambda (x) x))
and then byte compile and load that, then
ELISP> (symbol-function 'foo)
(lambda (x)
x)
So the byte compiler didn't do anything with foo because it didn't get the hint that it should. But foo is still a fine function:
ELISP> (foo 1)
1 (#o1, #x1, ?\C-a)
It just isn't compiled. This is also why, if writing elisp code with anonymous functions in it, you should use function (or equivalently #'). (And finally, of course, (function ...) does the right thing if lexical scoping is on.)
Other ways of making functions in CL
Finally, I've said above that function & specifically (function (lambda ...)) is the only primitive way to make new functions in CL. I'm not completely sure that's true, especially given CLOS (almost any CLOS will have some kind of class instances of which are functions but which can be subclassed). But it does not matter: it is a way and that's sufficient.
DEFUN is a defining macro. Macros transform code.
In Common Lisp:
(defun foo (a)
(+ a 42))
Above is a definition form, but it will be transformed by DEFUN into some other code.
The effect is similar to
(setf (symbol-function 'foo)
(lambda (a)
(block foo
(+ a 42))))
Above sets the function cell of the symbol FOO to a function. The BLOCK construct is added by SBCL, since in Common Lisp named functions defined by DEFUN create a BLOCK with the same name as the function name. This block name can then be used by RETURN-FROM to enable a non-local return from a specific function.
Additionally DEFUN does implementation specific things. Implementations also record development information: the source code, the location of the definition, etc.
Scheme has DEFINE:
(define (foo a)
(+ a 10))
This will set FOO to a function object.
Reading this question got me thinking about what constitutes a valid car of an expression. Obviously, symbols and lambdas can be "called" using the usual syntax. According to the hyperspec,
function name n. 1. (in an environment) A symbol or a list (setf symbol) that is the name of a function in that environment. 2. A symbol or a list (setf symbol).
So, theoretically, (setf some-name) is a function name. I decided to give it a try.
(defun (setf try-this) ()
(format t "Don't name your functions like this, kids :)"))
((setf try-this))
(funcall '(setf try-this))
(setf (try-this))
GNU CLISP, SBCL, and ABCL will all let me define this function. However, SBCL and ABCL won't let me call it using any of the syntaxes shown in the snippet. CLISP, on the other hand, will run the first two but still errs on the third.
I'm curious about which compiler is behaving correctly. Since SBCL and ABCL agree, I would hazard a guess that a correct implementation should reject that code. As a second question, how would I call my incredibly-contrived not-useful function from the code snippet, since the things I tried above aren't working portably. Or, perhaps more usefully,
A SETF function has to take at least one argument, which is the new value to be stored in the place. It can take additional arguments as well, these will be filled in from arguments in the place expression in the call to SETF.
When you use SETF, it has to have an even number of arguments: every place you're assigning to needs a value to be assigned.
So it should be:
(defun (setf try-this) (new-value)
(format t "You tried to store ~S~%" new-value))
(setf (try-this) 3)
(funcall #'(setf try-this) 'foo)
You can't use
((setf try-this) 'bar)
because the car of a form does not contain a function name. It can only be a symbol or a lambda expression (although implementations may allow other formats as extensions).
The Lisp forum thread Define macro alias? has an example of creating function alias using a form such as
(setf (symbol-function 'zero?) #'zerop)
This works fine, making zero? a valid predicate. Is it possible to parametrize this form without resorting to macros? I'd like to be able to call the following and have it create function?:
(define-predicate-alias 'functionp)`
My take was approximately:
(defun defalias (old new)
(setf (symbol-function (make-symbol new))
(symbol-function old)))
(defun define-predicate-alias (predicate-function-name)
(let ((alias (format nil "~A?" (string-right-trim "-pP" predicate-function-name))))
(defalias predicate-function-name alias)))
(define-predicate-alias 'zerop)
(zero? '())
This fails when trying to call zero? saying
The function COMMON-LISP-USER::ZERO? is undefined.
make-symbol creates an uninterned symbol. That's why zero? is undefined.
Replace your (make-symbol new) with e.g. (intern new *package*). (Or you may want to think more carefully in which package to intern your new symbol.)
Your code makes a symbol, via MAKE-SYMBOL, but you don't put it into a package.
Use the function INTERN to add a symbol to a package.
To expand on Lars' answer, choose the right package. In this case the default might be to use the same package from the aliased function:
About style:
Anything that begins with DEF should actually be a macro. If you have a function, don't use a name beginning with "DEF". If you look at the Common Lisp language, all those are macro. For example: With those defining forms, one would typically expect that they have a side-effect during compilation of files: the compiler gets informed about them. A function can't.
If I put something like this in a file
(define-predicate-alias zerop)
(zero? '())
and then compile the file, I would expect to not see any warnings about an undefined ZERO?. Thus a macro needs to expand (define-predicate-alias 'zerop) into something which makes the new ZERO? known into the compile-time environment.
I would also make the new name the first argument.
Thus use something like MAKE-PREDICATE-ALIAS instead of DEFINE-PREDICATE-ALIAS, for the function.
There are already some answers that explain how you can do this, but I'd point out:
Naming conventions, P, and -P
Common Lisp has a naming convention that is mostly adhered to (there are exceptions, even in the standard library), that if a type name is multiple words (contains a -), then its predicate is named with -P suffix, whereas if it doesn't, the suffix is just P. So we'd have keyboardp and lcd-monitor-p. It's good then, that you're using (string-right-trim "-pP" predicate-function-name)), but since the …P and …-P names in the standard, and those generated by, e.g., defstruct, will be using P, not p, you might just use (string-right-trim "-P" predicate-function-name)). Of course, even this has the possible issues with some names (e.g., pop), but I guess that just comes with the territory.
Symbol names, format, and *print-case*
More importantly, using format to create symbol names for subsequent interning is dangerous, because format doesn't always print a symbol's name with the characters in the same case that they actually appear in its name. E.g.,
(let ((*print-case* :downcase))
(list (intern (symbol-name 'foo))
(intern (format nil "~A" 'foo))))
;=> (FOO |foo|) ; first symbol has name "FOO", second has name "foo"
You may be better off using string concatenation and extracting symbol names directly. This means you could write code like (this is slightly different use case, since the other questions already explain how you can do what you're trying to do):
(defmacro defpredicate (symbol)
(flet ((predicate-name (symbol)
(let* ((name (symbol-name symbol))
(suffix (if (find #\- name) "-P" "P")))
(intern (concatenate 'string name suffix)))))
`(defun ,(predicate-name symbol) (x)
(typep x ',symbol)))) ; however you're checking the type
(macroexpand-1 '(defpredicate zero))
;=> (DEFUN ZEROP (X) (TYPEP X 'ZERO))
(macroexpand-1 '(defpredicate lcd-monitor))
;=> (DEFUN LCD-MONITOR-P (X) (TYPEP X 'LCD-MONITOR))
I've written an ad hoc parser generator that creates code to convert an old and little known 7-bit character set into unicode. The call to the parser generator expands into a bunch of defuns enclosed in a progn, which then get compiled. I only want to expose one of the generated defuns--the top-level one--to the rest of the system; all the others are internal to the parser and only get called from within the dynamic scope of the top-level one. Therefore, the other defuns generated have uninterned names (created with gensym). This strategy works fine with SBCL, but I recently tested it for the first time with CLISP, and I get errors like:
*** - FUNCALL: undefined function #:G16985
It seems that CLISP can't handle functions with uninterned names. (Interestingly enough, the system compiled without a problem.) EDIT: It seems that it can handle functions with uninterned names in most cases. See the answer by Rörd below.
My questions is: Is this a problem with CLISP, or is it a limitation of Common Lisp that certain implementations (e.g. SBCL) happen to overcome?
EDIT:
For example, the macro expansion of the top-level generated function (called parse) has an expression like this:
(PRINC (#:G75735 #:G75731 #:G75733 #:G75734) #:G75732)
Evaluating this expression (by calling parse) causes an error like the one above, even though the function is definitely defined within the very same macro expansion:
(DEFUN #:G75735 (#:G75742 #:G75743 #:G75744) (DECLARE (OPTIMIZE (DEBUG 2)))
(DECLARE (LEXER #:G75742) (CONS #:G75743 #:G75744))
(MULTIPLE-VALUE-BIND (#:G75745 #:G75746) (POP-TOKEN #:G75742)
...
The two instances of #:G75735 are definitely the same symbol--not two different symbols with the same name. As I said, this works with SBCL, but not with CLISP.
EDIT:
SO user Joshua Taylor has pointed out that this is due to a long standing CLISP bug.
You don't show one of the lines that give you the error, so I can only guess, but the only thing that could cause this problem as far as I can see is that you are referring to the name of the symbol instead of the symbol itself when trying to call it.
If you were referring to the symbol itself, all your lisp implementation would have to do is lookup that symbol's symbol-function. Whether it's interned or not couldn't possibly matter.
May I ask why you haven't considered another way to hide the functions, i.e. a labels statement or defining the functions within a new package that exports only the one external function?
EDIT: The following example is copied literally from an interaction with the CLISP prompt.
As you can see, calling the function named by a gensym is working as expected.
[1]> (defmacro test ()
(let ((name (gensym)))
`(progn
(defun ,name () (format t "Hello!"))
(,name))))
TEST
[2]> (test)
Hello!
NIL
Maybe your code that's trying to call the function gets evaluated before the defun? If there's any code in the macro expansion besides the various defuns, it may be implementation-dependent what gets evaluated first, and so the behaviour of SBCL and CLISP may differ without any of them violating the standard.
EDIT 2: Some further investigation shows that CLISP's behaviour varies depending upon whether the code is interpreted directly or whether it's first compiled and then interpreted. You can see the difference by either directly loading a Lisp file in CLISP or by first calling compile-file on it and then loading the FASL.
You can see what's going on by looking at the first restart that CLISP offers. It says something like "Input a value to be used instead of (FDEFINITION '#:G3219)." So for compiled code, CLISP quotes the symbol and refers to it by name.
It seems though that this behaviour is standard-conforming. The following definition can be found in the HyperSpec:
function designator n. a designator for a function; that 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). The consequences are undefined if a symbol is used as a function designator but it does not have a global definition as a function, or it has a global definition as a macro or a special form. See also extended function designator.
I think an uninterned symbol matches the "a symbol is used as a function designator but it does not have a global definition as a function" case for unspecified consequences.
EDIT 3: (I can agree that I'm not sure whether CLISP's behaviour is a bug or not. Someone more experienced with details of the standard's terminology should judge this. It comes down to whether the function cell of an uninterned symbol - i.e. a symbol that cannot be referred to by name, only by having a direct hold on the symbol object - would be considered a "global definition" or not)
Anyway, here's an example solution that solves the problem in CLISP by interning the symbols in a throwaway package, avoiding the matter of uninterned symbols:
(defmacro test ()
(let* ((pkg (make-package (gensym)))
(name (intern (symbol-name (gensym)) pkg)))
`(progn
(defun ,name () (format t "Hello!"))
(,name))))
(test)
EDIT 4: As Joshua Taylor notes in a comment to the question, this seems to be a case of the (10 year old) CLISP bug #180.
I've tested both workarounds suggested in that bug report and found that replacing the progn with locally actually doesn't help, but replacing it with let () does.
You can most certainly define functions whose names are uninterned symbols. For instance:
CL-USER> (defun #:foo (x)
(list x))
#:FOO
CL-USER> (defparameter *name-of-function* *)
*NAME-OF-FUNCTION*
CL-USER> *name-of-function*
#:FOO
CL-USER> (funcall *name-of-function* 3)
(3)
However, the sharpsign colon syntax introduces a new symbol each time such a form is read read:
#: introduces an uninterned symbol whose name is symbol-name. Every time this syntax is encountered, a distinct uninterned symbol is created. The symbol-name must have the syntax of a symbol with no package prefix.
This means that even though something like
CL-USER> (list '#:foo '#:foo)
;=> (#:FOO #:FOO)
shows the same printed representation, you actually have two different symbols, as the following demonstrates:
CL-USER> (eq '#:foo '#:foo)
NIL
This means that if you try to call such a function by typing #: and then the name of the symbol naming the function, you're going to have trouble:
CL-USER> (#:foo 3)
; undefined function #:foo error
So, while you can call the function using something like the first example I gave, you can't do this last one. This can be kind of confusing, because the printed representation makes it look like this is what's happening. For instance, you could write such a factorial function like this:
(defun #1=#:fact (n &optional (acc 1))
(if (zerop n) acc
(#1# (1- n) (* acc n))))
using the special reader notation #1=#:fact and #1# to later refer to the same symbol. However, look what happens when you print that same form:
CL-USER> (pprint '(defun #1=#:fact (n &optional (acc 1))
(if (zerop n) acc
(#1# (1- n) (* acc n)))))
(DEFUN #:FACT (N &OPTIONAL (ACC 1))
(IF (ZEROP N)
ACC
(#:FACT (1- N) (* ACC N))))
If you take that printed output, and try to copy and paste it as a definition, the reader creates two symbols named "FACT" when it comes to the two occurrences of #:FACT, and the function won't work (and you might even get undefined function warnings):
CL-USER> (DEFUN #:FACT (N &OPTIONAL (ACC 1))
(IF (ZEROP N)
ACC
(#:FACT (1- N) (* ACC N))))
; in: DEFUN #:FACT
; (#:FACT (1- N) (* ACC N))
;
; caught STYLE-WARNING:
; undefined function: #:FACT
;
; compilation unit finished
; Undefined function:
; #:FACT
; caught 1 STYLE-WARNING condition
I hope I get the issue right. For me it works in CLISP.
I tried it like this: using a macro for creating a function with a GENSYM-ed name.
(defmacro test ()
(let ((name (gensym)))
`(progn
(defun ,name (x) (* x x))
',name)))
Now I can get the name (setf x (test)) and call it (funcall x 2).
Yes, it is perfectly fine defining functions that have names that are unintenred symbols. The problem is that you cannot then call them "by name", since you can't fetch the uninterned symbol by name (that is what "uninterned" means, essentially).
You would need to store the uninterned symbol in some sort of data structure, to then be able to fetch the symbol. Alternatively, store the defined function in some sort of data structure.
Surprisingly, CLISP bug 180 isn't actually an ANSI CL conformance bug. Not only that, but evidently, ANSI Common Lisp is itself so broken in this regard that even the progn based workaround is a courtesy of the implementation.
Common Lisp is a language intended for compilation, and compilation produces issues regarding the identity of objects which are placed into compiled files and later loaded ("externalized" objects). ANSI Common Lisp requires that literal objects reproduced from compiled files are only similar to the original objects. (CLHS 3.2.4 Literal Objects in Compiled Files).
Firstly, according to the definition similarity (3.2.4.2.2 Definition of Similarity), the rules for uninterned symbols is that similarity is name based. If we compile code with a literal that contains an uninterned symbol, then when we load the compiled file, we get a symbol which is similar and not (necessarily) the same object: a symbol which has the same name.
What if the same uninterned symbol is inserted into two different top-level forms which are then compiled as a file? When the file is loaded, are those two similar to each other at least? No, there is no such requirement.
But it gets worse: there is also no requirement that two occurrences of the same uninterned symbol in the same form will be externalized in such a way that their relative identity is preserved: that the re-loaded version of that object will have the same symbol object in all the places where the original was. In fact, the definition of similarity contains no provision for preserving the circular structure and substructure sharing. If we have a literal like '#1=(a b . #1#), as a literal in a compiled file, there appears to be no requirement that this be reproduced as a circular object with the same graph structure as the original (a graph isomorphism). The similarity rule for conses is given as naive recursion: two conses are similar if their respective cars and cdrs are similar. (The rule can't even be evaluated for circular objects; it doesn't terminate).
That the above works is because of implementations going beyond what is required in the spec; they are providing an extension consistent with (3.2.4.3 Extensions to Similarity Rules).
Thus, purely according to ANSI CL, we cannot expect to use macros with gensyms in compiled files, at least in some ways. The expectation expressed in code like the following runs afoul of the spec:
(defmacro foo (arg)
(let ((g (gensym))
(literal '(blah ,g ,g ,arg)))
...))
(defun bar ()
(foo 42))
The bar function contains a literal with two insertions of a gensym, which according to the similarity rules for conses and symbols need not reproduce as a list containing two occurrences of the same object in the second and third positions.
If the above works as expected, it's due to "extensions to the similarity rules".
So the answer to the "Why can't CLISP ..." question is that although CLISP does provide an extension for similarity which preserves the graph structure of literal forms, it doesn't do it across the entire compiled file, only within individual top level items within that file. (It uses *print-circle* to emit the individual items.) The bug is that CLISP doesn't conform to the best possible behavior users can imagine, or at least to a better behavior exhibited by other implementations.