Is there a build-in function in Common Lisp to coerce any thing into it's boolean value, like the bool function in Python does?
There's nothing specifically for this. You can't use boolean as the type argument to coerce. You can use:
(defun boolean-value (x)
(not (not x)))
This is analogous to the !!x idiom used in many other languages.
Sort of. (and form t) will return t if form is not one of the false values nil or ().
I.e. it is a macro, not a function, and it requires the additional argument t to do the trick.
Related
*Note: Despite having frequented StackOverflow for a long time, this is the first question that I have posted myself. Apologies if it's a bit verbose. Constructive criticism appreciated.
When I define a struct in Common Lisp using defstruct, a predicate function is automatically generated that tests whether its argument is of the type defined by the defstruct. Eg:
(defstruct book
title
author)
(let ((huck-finn (make-book :title "The Adventures of Huckleberry Finn" :author "Mark Twain")))
(book-p huck-finn))
=> True
However, when defining a class using defclass, such functions are seemingly not generated by default (is there a way to specify this?), so I'm trying to add this functionality myself, because I'd like a) for this syntax to be consistent between structs and classes, b) to have an abbreviation of (typep obj 'classname), which I need to write very often and is visually noisy,
and c) as a programming exercise, since I'm still relatively new to Lisp.
I could write a macro that defines a predicate function given the name of a class:
(defclass book ()
((title :initarg :title
:accessor title)
(author :initarg :author
:accessor author)))
;This...
(defmacro gen-predicate (classname)
...)
;...should expand to this...
(defun book-p (obj)
(typep obj 'book))
;...when called like this:
(gen-predicate 'book)
The name that I need to pass to defun must be of the form 'classname-p. Here's where I have difficulty. To create such a symbol, I could use the "symb" function from Paul Graham's On Lisp (p. 58). When it is run on the REPL:
(symb 'book '-p)
=> BOOK-P
My gen-predicate macro looks like this so far:
(defmacro gen-predicate (classname)
`(defun ,(symb classname '-p) (obj)
(typep obj ,classname)))
(macroexpand `(gen-predicate 'book))
=>
(PROGN
(EVAL-WHEN (:COMPILE-TOPLEVEL) (SB-C:%COMPILER-DEFUN '|'BOOK-P| 'NIL T))
(SB-IMPL::%DEFUN '|'BOOK-P|
(SB-INT:NAMED-LAMBDA |'BOOK-P|
(OBJ)
(BLOCK |'BOOK-P| (TYPEP OBJ 'BOOK)))
NIL 'NIL (SB-C:SOURCE-LOCATION)))
T
It would seem that the symbol created by (symb 'book '-p) is actually considered |'BOOK-P| by the implementation (SBCL), not BOOK-P. Sure enough, this now works:
(let ((huck-finn (make-instance 'book)))
(|'BOOK-P| huck-finn))
=> True
Why is the symbol created by symb interned as |'BOOK-P|? In On Lisp (same page as above) Graham says: "Any string can be the print-name of a symbol, even a string containing lowercase letters or macro characters like parentheses. When a symbol's name contains such oddities, it is printed within vertical bars." No such oddities exist in this case, do they? And am I correct in thinking that the "print-name" of a symbol is what is actually displayed on the standard output when the symbol is printed, and is, in the case of such oddities, distinct from the form of the symbol itself?
To be able to write function-defining macros like gen-predicate - whose defined functions are named based on the arguments passed to the macro - seems to me like something that Lisp hackers have probably been doing for ages. User Kaz says here (Merging symbols in common lisp) that the "mashing-up" of symbols can often be avoided, but that would defeat the purpose of this macro.
Finally, assuming I could get gen-predicate to work how I want, what would be the best way of ensuring that it be called for each new class as they are defined? Much in the same way as initialize-instance can be customized to perform certain actions upon instantiation of a class, is there a generic function called by defclass that can perform actions upon definition of a class?
Thank you.
That's a usual problem: what gets passed to a Macro?
Compare calls like this:
(symb 'book '-p)
and
(symb ''book '-p)
Your macro form is this:
(gen-predicate 'book)
GEN-PREDICATE is a macro. classname is a parameter for this macro.
Now what is the value of classname inside the macro during code expansion? Is it book or 'book?
Actually it is the latter, because you wrote (gen-predicate 'book). Remember: macros see source code and the argument source gets passed to the macro function - not the value. The argument is 'book. Thus this gets passed. (QUOTE BOOK) is the same, only printed differently. So it is a two element list. The first element is the symbol QUOTE and the second element is the symbol BOOK.
Thus the macro now calls the function SYMB with the argument value (QUOTE BOOK) or, shorter, 'BOOK.
If you want to generate the predicate without the quote character, you need to write:
(gen-predicate book)
Alternatively you can also change the macro:
(symb classname '-p)
would be:
(symbol (if (and (consp classname)
(eq (first classname) 'quote))
(second classname)
classname))
Compare
We write
(defun foo () 'bar)
and not
(defun 'foo () 'bar) ; note the quoted FOO
DEFUN is a macro and the first argument is the function name. It's a similar problem then...
Second part of the question
I don't really know any good answer to that. I can't remember any easy way to run code (for example to define a function) after a class definition.
Maybe use the MOP, but that's ugly.
write a custom macro DEFINE-CLASS which does what you want: expands into DEFCLASS and the DEFUN.
iterate over all symbols in a package, find the classes and define the corresponding predicates
To address the second part of the question, classes are themselves objects, thanks to the MOP, so it might be possible to write an :after method on initialize-instance specialized on STANDARD-CLASS. But you should check the MOP to see whether defining such a method is allowed or not.
If it's possible, then yes, you can run code in response to the creation of a class; however, since you don't know the name of the class being created until runtime, you cannot spell it textually in the source, so you cannot use your macro (unless you use eval). You'd rather use something like
(let ((classname (class-name class)))
(compile (generate-my-predicate-symbol classname)
(lambda (x) (typep x classname))))
I think Rainer's suggestion to write your own DEFINE-CLASS macro is the way to go, I mean, the way a seasoned Lisper most likely would do it, if there aren't any other considerations at play. But I'm not really a seasoned Lisper, so I might be wrong ;)
I am still interested in the question which has been answered.
continuation in common lisp by macros — regarding an implemetation in OnLisp
What will happen if Paul Graham's assumption is correct especially when change from (A 5) to (B 1)? What is cont bound to here?
And one more confusion when the text says
=bind, is intended to be used in the same way as multiple-value-bind. It takes a list of parameters, an expression, and a body of code: the parameters are bound to the values returned by the expression, and the code body is evaluated with those bindings.
I cannot see the binding directly from the macro definition of =bind which looks like
(defmacro =bind (parms expr &body body)
`(let ((*cont* #'(lambda ,parms ,#body))) ,expr))
Does the binding happens only when =values comes in later?
The macro sets the continuation, *cont*, to be a lambda which takes all of your variables as arguments, and then evaluates the expression expr. The expression is expected to call the continuation with its final value, which can be done indirectly by calling the =values function, or directly with funcall. Unlike Scheme, where the continuation is implicitly called with the return value of any expression, you must explicitly write your code in continuation-passing style by calling *cont* or using =values instead of returning from any function.
I'm trying to learn and understand the Lisp programming language to a deep level. The function + evaluates its arguments in applicative order:
(+ 1 (+ 1 2))
(+ 1 2) will be evaluated and then (+ 1 3) will be evaluated, but the if function works differently:
(if (> 1 2) (not-defined 1 2) 1)
As the form (not-defined 1 2) isn't evaluated, the program doesn't break.
How can the same syntax lead to different argument evaluation? How is the if function defined so that its arguments aren't evaluated?
if is a special operator, not an ordinary function.
This means that the normal rule that the rest elements in the compound form are evaluated before the function associated with the first element is invoked is not applicable (in that it is similar to macro forms).
The way this is implemented in a compiler and/or an interpreter is that one looks at the compound form and decides what to do with it based on its first element:
if it is a special operator, it does its special thing;
if it is a macro, its macro-function gets the whole form;
otherwise it is treated as a function - even if no function is defined.
Note that some special forms can be defined as macros expanding to other special forms, but some special forms must actually be present.
E.g., one can define if in terms of cond:
(defmacro my-if (condition yes no)
`(cond (,condition ,yes)
(t ,no)))
and vice versa (much more complicated - actually, cond is a macro, usually expanding into a sequence of ifs).
PS. Note that the distinction between system-supplied macros and special operators, while technically crisp and clear (see special-operator-p and macro-function), is ideologically blurred because
An implementation is free to implement a Common Lisp special operator
as a macro. An implementation is free to implement any macro operator
as a special operator, but only if an equivalent definition of the
macro is also provided.
sds's answer answers this question well, but there are a few more general aspects that I think are worth mentioning. As that answer and others have pointed out, if, is built into the language as a special operator, because it really is a kind of primitive. Most importantly, if is not a function.
That said, the functionality of if can be achieved using just functions and normal function calling where all the arguments are evaluated. Thus, conditionals can be implemented in the lambda calculus, on which languages in the family are somewhat based, but which doesn't have a conditional operator.
In the lambda calculus, one can define true and false as functions of two arguments. The arguments are presumed to be functions, and true calls the first of its arguments, and false calls the second. (This is a slight variation of Church booleans which simply return their first or second argument.)
true = λ[x y].(x)
false = λ[x y].(y)
(This is obviously a departure from boolean values in Common Lisp, where nil is false and anything else is true.) The benefit of this, though, is that we can use a boolean value to call one of two functions, depending on whether the boolean is true or false. Consider the Common Lisp form:
(if some-condition
then-part
else-part)
If were were using the booleans as defined above, then evaluating some-condition will produce either true or false, and if we were to call that result with the arguments
(lambda () then-part)
(lambda () else-part)
then only one of those would be called, so only one of then-part and else-part would actually be evaluated. In general, wrapping some forms up in a lambda is a good way to be able delay the evaluation of those forms.
The power of the Common Lisp macro system means that we could actually define an if macro using the types of booleans described above:
(defconstant true
(lambda (x y)
(declare (ignore y))
(funcall x)))
(defconstant false
(lambda (x y)
(declare (ignore x))
(funcall y)))
(defmacro new-if (test then &optional else)
`(funcall ,test
(lambda () ,then)
(lambda () ,else)))
With these definitions, some code like this:
(new-if (member 'a '(1 2 3))
(print "it's a member")
(print "it's not a member"))))
expands to this:
(FUNCALL (MEMBER 'A '(1 2 3)) ; assuming MEMBER were rewritten
(LAMBDA () (PRINT "it's a member")) ; to return `true` or `false`
(LAMBDA () (PRINT "it's not a member")))
In general, if there is some form and some of the arguments aren't getting evaluated, then the (car of the) form is either a Common Lisp special operator or a macro. If you need to write a function where the arguments will be evaluated, but you want some forms not to be evaluated, you can wrap them up in lambda expressions and have your function call those anonymous functions conditionally.
This is a possible way to implement if, if you didn't already have it in the language. Of course, modern computer hardware isn't based on a lambda calculus interpreter, but rather on CPUs that have test and jump instructions, so it's more efficient for the language to provide if a primitive and to compile down to the appropriate machine instructions.
Lisp syntax is regular, much more regular than other languages, but it's still not completely regular: for example in
(let ((x 0))
x)
let is not the name of a function and ((x 0)) is not a bad form in which a list that is not a lambda form has been used in the first position.
There are quite a few "special cases" (still a lot less than other languages, of course) where the general rule of each list being a function call is not followed, and if is one of them. Common Lisp has quite a few "special forms" (because absolute minimality was not the point) but you can get away for example in a scheme dialect with just five of them: if, progn, quote, lambda and set! (or six if you want macros).
While the syntax of Lisp is not totally uniform the underlying representation of code is however quite uniform (just lists and atoms) and the uniformity and simplicity of representation is what facilitates metaprogramming (macros).
"Lisp has no syntax" is a statement with some truth in it, but so it's the statement "Lisp has two syntaxes": one syntax is what uses the reader to convert from character streams to s-expressions, another syntax is what uses the compiler/evaluator to convert from s-expressions to executable code.
It's also true that Lisp has no syntax because neither of those two levels is fixed. Differently from other programming languages you can customize both the first step (using reader macros) and the second step (using macros).
It would not make any sense to do so. Example: (if (ask-user-should-i-quit) (quit) (continue)). Should that quit, even though the user does not want to?
IF is not a function in Lisp. It is a special built-in operator. Lisp a several built-in special operators. See: Special Forms. Those are not functions.
The arguments are not evaluated as for functions, because if is a special operator. Special operators can be evaluated in any arbitrary way, that's why they're called special.
Consider e.g.
(if (not (= x 0))
(/ y x))
If the division was always evaluated, there could be a division by zero error which obviously was not intended.
If isn't a function, it's a special form. If you wanted to implement similar functionality yourself, you could do so by defining a macro rather than a function.
This answer applies to Common Lisp, but it'll probably the same for most other Lisps (though in some if may be a macro rather than a special form).
Why do we have to use funcall to call higher order functions in Common Lisp? For example, why do we have to use:
(defun foo (test-func args)
(funcall test-func args))
instead of the simpler:
(defun bar (test-func args)
(test-func args))
Coming from a procedural background, I'm a bit surprised by that since the languages I'm more used to (e.g. Python, C#) don't need the distinction. In particular, on the source level at least, the C# compiler transforms it to something like func.invoke().
The only problem I see is that this would mean we couldn't call a global function test-func anymore because it'd be shadowed, but that's hardly a problem.
Strictly speaking, funcall would not be needed, but there are some lisps (lisp-2 variants, such as Common Lisp) that separate the variable name space of the function name space. Lisp-1 variants (e.g. Scheme) do not make this distinction.
More specifically, in your case, test-func is in the variable name space.
(defun foo (test-func args)
(funcall test-func args))
Therefore you need a construct that actually searches the function object associated with this variable in the variable name space. In Common Lisp this construct is funcall.
See also this answer.
The majority of Lisps have two namespaces (functions and variables). A name is looked up in the function namespace when it appears as the first element in an S-expression, and in the variable namespace otherwise. This allows you to name your variables without worrying about whether they shadow functions: so you can name your variable list instead of having to mangle it into lst.
However, this means that when you store a function in a variable, you can't call it normally:
(setq list #'+) ; updates list in the variable namespace
(list 1 2 3) => (1 2 3) ; looks up list in the function namespace
Hence the need for funcall and apply:
(funcall list 1 2 3) => 6 ; looks up list in the variable namespace
(Not all Lisps have two namespaces: Scheme is an example of a Lisp with just one namespace.)
In Common Lisp, each symbol can be associated with its symbol-function and its symbol-value, among other things. When reading a list, by default, Common Lisp interprets:
arg1 as a function and so retrieves test-func's symbol-function, which is undefined -- thus function bar doesn't work
arg2 as something to be evaled -- thus function foo retrieves test-func's symbol-value, which, in your case, happens to be a function
I'm trying to pass a function as an argument and call that function within another function.
A piece of my code looks like this:
(defun getmove(strategy player board printflag)
(setq move (funcall strategy player board))
(if printflag
(printboard board))
strategy is passed as a symbol represented in a two dimensional list as something such as 'randomstrategy
I keep getting the error:
"FUNCALL: 'RANDOMSTRATEGY is not a function name; try using a symbol instead...
When I replace strategy with 'randomstrategy it works fine.
I can also call randomstrategy independently.
What is the problem?
The problem is that the variable strategy does not contain the symbol randomstrategy but rather the list (!) 'randomstrategy (which is a shorthand notation for (quote randomstrategy)).
Now, you could, of course, extract the symbol from the list by way of the function second, but that would only cover the real problem up, which is probably somewhere up the call chain. Try to determine why the argument that is passed to function getmove is 'randomstrategy, not randomstrategy as it should be. (Maybe you erroneously used a quote inside of a quoted list?)
Oh, and don't let yourself be confused by the fact that (funcall 'randomstrategy ...) works: the expression 'randomstrategy does not, after all, evaluate to itself, but to the symbol randomstrategy.
Is strategy a variable with a functional value? If not, then use the #' syntax macro before it, i.e. #'strategy, or just (if the function is global) 'strategy.
WHY? Because arguments of a funcall call are evaluated. And your strategy symbol is just a variable name in this case. Variable this value 'RANDOMSTRATEGY. But you should give to funcall a function. How to access function if we have a symbol?
Three cases:
Symbol may denote a variable with functional value.
Symbol may denote a global function (symbol-function is the accessor in this case.
Symbol may denote a local function (flet, labels and so on).
It looks like you forgot to define RANDOMSTRATEGY function.
(defun RANDOMSTRATEGY …)
Hmm
FUNCALL: 'RANDOMSTRATEGY
Maybe you have (setq strategy ''RANDOMSTRATEGY)?
Then strategy will evaluate to 'RANDOMSTRATEGY.
Did you notice ' before the symbol name? 'RANDOMSTRATEGY <=> (quote RANDOMSTRATEGY); it is not a proper function name.
Have you set strategy anywhere? It looks like a scoping issue.
Try this
(setq strategy 'randomstrategy)
(setq move (funcall strategy player board))
Not seeing the code, I'm imagining you're doing something like this:
(defun randomstrategy (a b c) ...)
and then doing this:
(getmove 'randomstrategy x y z)
What you want to do is pass the function "randomstrategy" to getmove using #':
(getmove #'randomstrategy x y z)
In CommonLisp, #' yields the function bound to the symbol, which is
what you want to pass to getmove.