I've written a very simple program:
(define size 2)
(print size)
When I run this code, I get following error:
*** - EVAL: undefined function DEFINE
What does the error mean? How can I resolve it?
define is not a part of the ANSI Common Lisp language which is implemented by GNU CLISP (I think you are confusing CL with Scheme).
When defining a variable, use defvar,
for a function use defun.
You might want to get a book, e.g., ANSI Common Lisp.
define in the Scheme programming language means defining a variable or a function,
defvar or defparameter in the Common Lisp programming language (which is what CLISP implements, and which is different from Scheme) means defining a variable.
defun in Common Lisp means defining a function.
CL-USER 195 > (defparameter size 2)
SIZE
CL-USER 196 > (print size)
2
2
Related
This question already has answers here:
What makes Lisp macros so special?
(15 answers)
Closed 3 months ago.
I keep reading that Lisp macros are one of the most powerful features of the language. But reading over the specifications and manuals, they are just functions whose arguments are unevaluated.
Given any macro (defmacro example (arg1 ... argN) (body-forms)) I could just write (defun example (arg1 ... argN) ... (body-forms)) with the last body-form turned into a list and then call it like (eval (example 'arg1 ... 'argN)) to emulate the same behavior of the macro. If this were the case, then macros would just be syntactic sugar, but I doubt that syntactic sugar would be called a powerful language feature. What am I missing? Are there cases where I cannot carry out this procedure to emulate a macro?
I can't talk about powerful because it can be a little bit subjective, but macros are regular Lisp functions that work on Lisp data, so they are as expressive as other functions. This isn't the case with templates or generic functions in other languages that rely more on static types and are more restricted (on purpose).
In some way, yes macros are simple syntactic facilities, but you are focused in your emulation on the dynamic semantics of macros, ie. how you can run code that evaluates macros at runtime. However:
the code using eval is not equivalent to expanded code
the preprocessing/compile-time aspect of macros is not emulated
Lexical scope
Function, like +, do not inherit the lexical scope:
(let ((x 30))
(+ 3 4))
Inside the definition of +, you cannot access x. Being able to do so is what "dynamic scope" is about (more precisely, see dynamic extent, indefinite scope variables). But nowadays it is quite the exception to rely on dynamic scope. Most functions use lexical scope, and this is the case for eval too.
The eval function evaluates a form in the null lexical environment, and it never has access to the surrounding lexical bindings. As such, it behaves like any regular function.
So, in you example, calling eval on the transformed source code will not work, since arg1 to argnN will probably be unbound (it depends on what your macro does).
In order to have an equivalent form, you have to inject bindings in the transformed code, or expand at a higher level:
(defun expand-square (var)
(list '* var var))
;; instead of:
(defun foo (x) (eval (expand-square 'x))) ;; x unbound during eval
;; inject bindings
(defun foo (x) (eval `(let ((z ,x)) (expand-square z))))
;; or expand the top-level form
(eval `(defun foo (x) ,(expand-square 'x)))
Note that macros (in Common Lisp) also have access to the lexical environment through &environment parameters in their lambda-list. The use of this environment is implementation dependent, but can be used to access the declarations associated with a variable, for example.
Notice also how in the last example you evaluate the code when defining the function, and not when running it. This is the second thing about macro.
Expansion time
In order to emulate macros you could locally replace a call to a macro by a form that emulates it at runtime (using let to captures all the bindings you want to see inside the expanded code, which is tedious), but then you would miss the useful aspect of macros that is: generating code ahead of time.
The last example above shows how you can quote defun and wrap it in eval, and basically you would need to do that for all functions if you wanted to emulate the preprocessing work done by macros.
The macro system is a way to integrate this preprocessing step in the language in a way that is simple to use.
Conclusion
Macros themselves are a nice way to abstract things when functions can't. For example you can have a more human-friendly, stable syntax that hides implementation details. That's how you define pattern-matching abilities in Common Lisp that make it look like they are part of the language, without too much runtime penalty or verbosity.
They rely on simple term-rewriting functions that are integrated in the language, but you can emulate their behavior either at compile-time or runtime yourself if you want. They can be used to perform different kinds of abstraction that are usually missing or more cumbersome to do in other languages, but are also limited: they don't "understand" code by themselves, they don't give access to all the facilities of the compiler (type propagation, etc.). If you want more you can use more advanced libraries or compiler tools (see deftransform), but macros at least are portable.
Macros are not just functions whose arguments are unevaluated. Macros are functions between programming languages. In other words a macro is a function whose argument is a fragment of source code of a programming language which includes the macro, and whose value is a fragment of source code of a language which does not include the macro (or which includes it in a simpler way).
In very ancient, very rudimentary, Lisps, before people really understood what macros were, you could simulate macros with things called FEXPRs combined with EVAL. A FEXPR was simply a function which did not evaluate its arguments. This worked in such Lisps only because they were completely dynamically scoped, and the cost of it working was that compilation of such things was not possible at all. Those are two enormous costs.
In any modern Lisp, this won't work at all. You can write a toy version of FEXPRs as a macro (this may be buggy):
(defmacro deffex (fx args &body body)
(assert (every (lambda (arg)
(and (symbolp arg)
(not (member arg lambda-list-keywords))))
args)
(args) "not a simple lambda list")
`(defmacro ,fx ,args
`(let ,(mapcar (lambda (argname argval)
`(,argname ',argval))
',args (list ,#args))
,#',body)))
So now we could try to write a trivial binding construct I'll call with using this thing:
(deffex with (var val form)
(eval `(let ((,var ,val)) ,form)))
And this seems to work:
> (with a 1 a)
1
Of course, we're paying the cost that no code which uses this construct can ever be compiled so all our programs will be extremely slow, but perhaps that is a cost we're willing to accept (it's not, but never mind).
Except, of course, it doesn't work, at all:
> (with a 1
(with b 2
(+ a b)))
Error: The variable a is unbound.
Oh dear.
Why doesn't it work? It doesn't work because Common Lisp is lexically scoped, and eval is a function: it can't see the lexical bindings.
So not only does this kind of approach prevent compilation in a modern Lisp, it doesn't work at all.
People often, at this point, suggest some kind of kludge solution which would allow eval to be able to see lexical bindings. The cost of such a solution is that all the lexical bindings need to exist in compiled code: no variable can ever be compiled away, not even its name. That's essentially saying that no good compilers can ever be used, even for the small part of your programs you can compile at all in a language which makes extensive use of macros like CL. For instance, if you ever use defun you're not going to be able to compile the code in its body. People do use defun occasionally, I think.
So this approach simply won't work: it worked by happenstance in very old Lisps but it can't work, even at the huge cost of preventing compilation, in any modern Lisp.
More to the point this approach obfuscates the understanding of what macros are: as I said at the start, macros are functions between programming languages, and understanding that is critical. When you are designing macros you are implementing a new programming language.
Just came across a quote from Peter Norvig in Paradigms in AI Programing:
In the past, eval was seen as the key to Lisp's flexibility. In
modern Lisps with lexical scoping, such as Common Lisp, eval is used
less often (in fact, in Scheme there is no eval at all). Instead,
programmers are expected to use lambda to create a new function, and
then apply or funcall the function.
There are considerations on why eval is not a good idea, such as this, however I am more interested in the interaction between eval and dynamic (special?) variables and how introduction of lexicals contributed to eval losing favor. What were some common eval idioms before the introduction of lexical variables?
Unfortunately, I don't have a citation handy, but from what I remember hearing, the pattern that that text might be referring to is a way to emulate lexical scoping. That is, in lexical Common Lisp, we can simply write:
(defun constant-adder (a)
(lambda (b) (+ a b)))
In a language which used dynamic scope for all variables (but with the same syntax otherwise), lexical scope could be emulated by constructing a lambda expression:
(defun constant-adder (a)
(eval `(lambda (b) (+ (quote ,a) b))))
That is, we are constructing at runtime an expression which has the value of a substituted in, rather than having the language capture a automatically.
(However, your quote suggests that there was also no lambda in use. I have no plausible explanations for what would be happening in that case.)
eval
Evaluates form in the current dynamic environment and the null lexical environment.
This means that
(defun f (a) (eval '(1+ a)))
; warning: The variable A is defined but never used.
The compiler notices that, contrary to a naive expectation, a is not passed to eval.
Indeed:
(f 10)
; error: UNBOUND-VARIABLE: The variable A is unbound.
I'm currently reading the book Land of LISP, and I'm just working through the first chapter. In there, there is a little program written where the computer guesses numbers between 1 and 100. Its code is as follows:
(defparameter *small* 1)
(defparameter *big* 100)
(defun guess-my-number ()
(ash (+ *small* *big*) -1))
(defun smaller ()
(setf *big* (1- (guess-my-number)))
(guess-my-number))
(defun bigger ()
(setf *small* (1+ (guess-my-number)))
(guess-my-number))
(defun start-over ()
(defparameter *small* 1)
(defparameter *big* 100)
(guess-my-number))
So far, I understand what happens, and Using 'ash' in LISP to perform a binary search? helped me a lot in this. Nevertheless there's one thing left that puzzles me: As far as I have learned, you use setf to assign values to variables, and defparameter to initially define variables. I also have understood the difference between defparameter and defvar(at least I believe I do ;-)).
So now my question is: If I should use setf to assign a value to a variable once it had been initialized, why does the start-over function use defparameter and not setf? Is there a special reason for this, or is this just sloppiness?
The function is just:
(defun start-over ()
(setf *small* 1)
(setf *big* 100)
(guess-my-number))
It is already declared to be a special global variable. No need to do it inside the function again and again.
You CAN use DEFPARAMETER inside a function, but it is bad style.
DEFPARAMETER is for declaring global special variables and optional documentation for them. Once. If you need to do it several times, it's mostly done when a whole file or system gets reloaded. The file compiler also recognizes it in top-level position as a special declaration for a dynamically bound variable.
Example:
File 1:
(defparameter *foo* 10)
(defun foo ()
(let ((*foo* ...))
...))
File 2:
(defun foo-start ()
(defparameter *foo* 10))
(defun foo ()
(let ((*foo* ...))
...))
If Lisp compiles File 1 fresh with compile-file, the compiler recognizes the defparameter and in the following let we have a dynamic binding.
If Lisp compiles File 2 fresh with compile-file, the compiler doesn't recognize the defparameter and in the following let we have a lexical binding. If we compile it again, from this state, we have a dynamic binding.
So here version 1 is better, because it is easier to control and understand.
In your example DEFPARAMETER appears multiple times, which is not useful. I might ask, where is the variable defined and the answer would point to multiple source locations...
So: make sure that your program elements get mostly defined ONCE - unless you have a good reason not to do so.
So you have global variables. Those can be defined by defconstant (for really non-chainging stuff), defparameter (a constant that you can change) and defvar (a variable that does not overwrite if you load.
You use setf to alter the state of lexical as well and global variables. start-over could have used setf since the code doesn't really define it but change it. If you would have replaced defparameter with defvar start-over would stop working.
(defparameter *par* 5)
(setf *par* 6)
(defparameter *par* 5)
*par* ; ==> 5
(defvar *var* 5)
(setf *var* 6)
(defvar *var* 5)
*var* ; ==> 6
defconstant is like defparameter except once defined the CL implementation is free to inline it in code. Thus if you redefine a constant you need to redefine all functions that uses it or else it might use the old value.
(defconstant +c+ 5)
(defun test (x)
(+ x +c+))
(test 1) ; ==> 6
(defconstant +c+ 6)
(test 1) ; ==> 6 or 7
(defun test (x)
(+ x +c+))
(test 1) ; ==> 7
Normally, one would use defvar to initialy define global variables. The difference between defvar and defparameter is subtle, cf. the section in the CLHS and this plays a role here: defparameter (in contrast to defvar) assigns the value anew, whereas defvar would leave the old binding in place.
To address what to use: In general, defvar and friends are used as top-level forms, not inside some function (closures being the most notable exception in the context of defun). I would use setf, not defparameter.
For a beginner symbols, variables, etc. can be a bit surprising. Symbols are surprisingly featureful. Just to mention a few things you can ask a symbol for it's symbol-value, symbol-package, symbol-name, symbol-function etc. In addition symbols can have a varying amount of information declared about them, for example a type, that provides advice the compile might leverage to create better code. This is true of all symbols, for example *, the symbol you use for multiplication has a symbol-function that does that multiplication. It also has a symbol-value, i.e. the last value returned in the current REPL.
One critical bit of declarative information about symbols is if they are "special." (Yeah, it's a dumb name.) For good reasons it's good practice to declare all global symbols to be special. defvar, defparameter, and defconstant do that for you. We have a convention that all special variables are spelled with * on the front and back, *standard-output* for example. This convention is so common that some compilers will warn you if you neglect to follow it.
One benefit of declaring a symbol as special is that it will suppress the warning you get when you misspell a variable in your functions. For example (defun faster (how-much) (incf *speed* hw-much)) will generate a warning about hw-much.
The coolest feature of a symbol that is special is that it is managed with what we call dynamic scoping, in contrast to lexical scope. Recall how * has the value of the last result in the REPL. Well in some implementations you can have multiple REPLs (each running in it's own thread) each will want to have it's own *, it's own *standard-output*, etc. etc. This is easy in Common Lisp; the threads establish a "dynamic extent" and bind the specials that should be local to that REPL.
So yes, you could just setf *small* in you example; but if you never declare it to be special then you are going to get warnings about how the compiler thinks you misspelled it.
I'm programming on Ubuntu using GCL. From the documentation on Common Lisp from various sources, I understand that let creates local variables, and setq sets the values of existing variables. In cases below, I need to create two variables and sum their values.
Using setq
(defun add_using_setq ()
(setq a 3) ; a never existed before , but still I'm able to assign value, what is its scope?
(setq b 4) ; b never existed before, but still I'm able to assign value, what is its scope?
(+ a b))
Using let
(defun add_using_let ( )
(let ((x 3) (y 4)) ; creating variables x and y
(+ x y)))
In both the cases I seem to achieve the same result; what is the difference between using setq and let here? Why can't I use setq (since it is syntactically easy) in all the places where I need to use let?
setq assigns a value to a variable, whereas let introduces new variables/bindings. E.g., look what happens in
(let ((x 3))
(print x) ; a
(let ((x 89))
(print x) ; b
(setq x 73)
(print x)) ; c
(print x)) ; d
3 ; a
89 ; b
73 ; c
3 ; d
The outer let creates a local variable x, and the inner let creates another local variable shadowing the inner one. Notice that using let to shadow the variable doesn't affect the shadowed variable's value; the x in line d is the x introduced by the outer let, and its value hasn't changed. setq only affects the variable that it is called with. This example shows setq used with local variables, but it can also be with special variables (meaning, dynamically scoped, and usually defined with defparameter or defvar:
CL-USER> (defparameter *foo* 34)
*FOO*
CL-USER> (setq *foo* 93)
93
CL-USER> *foo*
93
Note that setq doesn't (portably) create variables, whereas let, defvar, defparameter, &c. do. The behavior of setq when called with an argument that isn't a variable (yet) isn't defined, and it's up to an implementation to decide what to do. For instance, SBCL complains loudly:
CL-USER> (setq new-x 89)
; in: SETQ NEW-X
; (SETQ NEW-X 89)
;
; caught WARNING:
; undefined variable: NEW-X
;
; compilation unit finished
; Undefined variable:
; NEW-X
; caught 1 WARNING condition
89
Of course, the best ways to get a better understanding of these concepts are to read and write more Lisp code (which comes with time) and to read the entries in the HyperSpec and follow the cross references, especially the glossary entries. E.g., the short descriptions from the HyperSpec for setq and let include:
SETQ
Assigns values to variables.
LET
let and let* create new variable bindings and execute a series
of forms that use these bindings.
You may want to read more about variables and bindings. let and let* also have some special behavior with dynamic variables and special declarations (but you probably won't need to know about that for a while), and in certain cases (that you probably won't need to know about for a while) when a variable isn't actually a variable, setq is actually equivalent to setf. The HyperSpec has more details.
There are some not-quite duplicate questions on Stack Overflow that may, nonetheless, help in understanding the use of the various variable definition and assignment operators available in Common Lisp:
setq and defvar in lisp
What's difference between defvar, defparameter, setf and setq
Assigning variables with setf, defvar, let and scope
In Lisp, how do I fix "Warning: Assumed Special?" (re: using setq on undefined variables)
Difference between let* and set? in Common Lisp
Let should almost always be the way you bind variables inside of a function definition -- except in the rare case where you want the value to be available to other functions in the same scope.
I like the description in the emacs lisp manual:
let is used to attach or bind a symbol to a value in such a way that the Lisp interpreter will not confuse the variable with a variable of the same name that is not part of the function.
To understand why the let special form is necessary, consider the situation in which you own a home that you generally refer to as ‘the house,’ as in the sentence, “The house needs painting.” If you are visiting a friend and your host refers to ‘the house,’ he is likely to be referring to his house, not yours, that is, to a different house.
If your friend is referring to his house and you think he is referring to your house, you may be in for some confusion. The same thing could happen in Lisp if a variable that is used inside of one function has the same name as a variable that is used inside of another function, and the two are not intended to refer to the same value. The let special form prevents this kind of confusion.
-- http://www.gnu.org/software/emacs/manual/html_node/eintr/let.html
(setq x y) assigns a new value y to the variable designated by the symbol x, optionally defining a new package-level variable 1. This means that after you called add_using_setq you will have two new package-level variables in the current package.
(add_using_setq)
(format t "~&~s, ~s" a b)
will print 3 4 - unlikely the desired outcome.
To contrast that, when you use let, you only assign new values to variables designated by symbols for the duration of the function, so this code will result in an error:
(add_using_let)
(format t "~&~s, ~s" a b)
Think about let as being equivalent to the following code:
(defun add-using-lambda ()
(funcall (lambda (a b) (+ a b)) 3 4))
As an aside, you really want to look into code written by other programmers to get an idea of how to name or format things. Beside being traditional it also has some typographic properties you don't really want to loose.
1 This behaviour is non-standard, but this is what happens in many popular implementations. Regardless of it being fairly predictable, it is considered a bad practice for other reasons, mostly all the same concerns that would discourage you from using global variables.
SETQ
you can get the value of the symbol out of the scope, as long as Lisp is still running. (it assign the value to the symbol)
LET
you can't get the value of the symbol defined with LET after Lisp has finished evaluating the form. (it bind the value to the symbol and create new binding to symbol)
consider example below:
;; with setq
CL-USER> (setq a 10)
CL-USER> a
10
;; with let
CL-USER> (let ((b 20))
(print b))
CL-USER> 20
CL-USER> b ; it will fail
; Evaluation aborted on #<UNBOUND-VARIABLE B {1003AC1563}>.
CL-USER>
The Racket Reference sections 11.9 Expanding Top-Level Forms and 13.2 Evaluation and Compilation use the term "top-level-form" and the descriptions of the functions (eval top-level-form [nm]) and (expand top-level-form) in the reference manual have "top-level-form" as their function argument, but I'm unclear about the definition of "top-level-form". What is the meaning of the term "top-level-form" within the Racket language?
The intuition here is that these functions all deal with "top-level forms" as opposed to a form that depends on a lexical environment. As a semi-obvious example, eval can only deal with top-level forms, which is why this:
(let ([x 10])
(eval '(* 3 x)))
doesn't work. The usual use of just "forms" is talking about any forms, such as inputs to macros -- which of course can have such references.
I have now found the precise definition of top-level-form in a kind of Backus-Naur format in the Racket Reference manual section 1.2.3.1
http://docs.racket-lang.org/reference/syntax-model.html#(part._fully-expanded)