I have recently started to learn about Lisp, and I have this simple code which defines a function(using defun) and it creates an array of four elements and then it assigns the value 7 to the first position of the array
(defun func(setf array (make-array '(4))))
(setf (aref array 0) 7)
but it prints FUNC in the output, why is that??
A function always needs a parameter list, even if it is empty
(defun func () ; <- here
(setf array (make-array '(4)))
(setf (aref array 0) 7))
Evaluating this form defines and registers the function named func and returns the symbol it is registered with. That's why you see FUNC then, and it is quite right and expected.
There are other problems:
When you do (setf array...) here, array is not defined. In principle, anything could happen, but in practice, a global variable will be created and set which might or might not be declaimed special (i. e. of dynamic extent).
You should create a local variable instead. This is usually done using let:
(defun func ()
(let ((array (make-array '(4))))
(setf (aref array 0) 7)))
This does not do much, since the return value is the last value assigned by setf, which is 7. You would most likely want to return the array:
(defun func ()
(let ((array (make-array '(4))))
(setf (aref array 0) 7))
array))
Note that elements 1, 2, and 3 are not initialized, so trying to read from them has undefined behaviour later (unless you set their values before, of course). If you want to treat the array as being filled only to the first element, you could use a fill-pointer:
(defun func ()
(let ((array (make-array 4 :fill-pointer 0)))
(vector-push 7 array)
array))
If you just need the exact structure of your array, you could copy a literal:
(defun func ()
(copy-seq #(7 0 0 0)))
Related
I've got a data structure that consists of two parts:
A hash table mapping symbols to indices
A vector of vectors containing data
For example:
(defparameter *h* (make-hash-table))
(setf (gethash 'a *h*) 0)
(setf (gethash 'b *h*) 1)
(setf (gethash 'c *h*) 2)
(defparameter *v-of-v* #(#(1 2 3 4) ;vector a
#(5 6 7 8) ;vector b
#(9 10 11 12))) ;vector c
I'd like to define a symbol macro to get at vector a without going through the hashmap. At the REPL:
(define-symbol-macro a (aref *v-of-v* 0))
works fine:
* a
#(1 2 3 4)
but there could be potentially many named vectors, and I don't know what the mappings will be ahead of time, so I need to automate this process:
(defun do-all-names ()
(maphash #'(lambda (key index)
(define-symbol-macro key (aref *v-of-v* index)))
*h*))
But that does nothing. And neither does any of the combinations I have tried of making do-all-names a macro, back-quote/comma templates, etc. I am beginning to wonder if this doesn't have something to do with the define-symbol-macro itself. It seems a little used feature, and On Lisp only mentions it twice. Not too many mentions here nor elsewhere either. In this case I'm using SBCL 2.1
Anyone have any ideas?
You need something like above to do it at runtime:
(defun do-all-names ()
(maphash #'(lambda (key index)
(eval `(define-symbol-macro ,key (aref *v-of-v* ,index)))
*h*))
DEFINE-SYMBOL-MACRO is a macro and does not evaluate all its arguments. So you need to generate a new macro form for each argument pair and evaluate it.
The other way to do it, usually at compile time, is to write a macro which generates these forms on the toplevel:
(progn
(define-symbol-macro a (aref *v-of-v* 0))
(define-symbol-macro b (aref *v-of-v* 1))
; ....
)
I'm not too sure on what you mean by "I don't know what the mappings will be ahead of time".
You could do something like:
(macrolet ((define-accessors ()
`(progn
,#(loop for key being the hash-keys of *h*
collect
`(define-symbol-macro ,key (aref *v-of-v* ,(gethash key *h*)))))))
(define-accessors))
If you know you do not require global access, then, you could do:
(defmacro with-named-vector-accessors (&body body) ; is that the name you want?
`(symbol-macrolet (,#(loop for key being the hash-keys of *h*
collect `(,key (aref *v-of-v* ,(gethash key *h*)))))
,#body))
;;; Example Usage:
(with-named-vector-accessors
(list a b c)) ;=> (#(1 2 3 4) #(5 6 7 8) #(9 10 11 12))
Also,
If you know *h* and the indices each symbol maps to at macroexpansion time, the above works.
If you know *h* at macroexpansion but the indices each symbol maps to will change after macroexpansion, you will want to collect (,key (aref *v-of-v* (gethash ,key *h*))).
PS: If you find loop ugly for hash-tables, you could use the iterate library with the syntax:
(iter (for (key value) in-hashtable *h*)
(collect `(,key (aref *v-of-v* ,value))))
I have a code which takes a list and returns all possible permutations by the parameter result.
But when I compile I have an error which says *** - =: (1+ INDEX) is not a number.
Is this message true or I messed up the code generally?
I am new to lisp I can looking for a fix and also open to suggestions from fucntional programmers.
;; Creates permutatiions of a given list and returns it via parameter
(defun create-permuations (source)
(setf result (list))
(create-permuations-helper source 0 '() result)
result)
(defmacro create-permuations-helper (source index cur result)
(if (= (list-length cur) index)
(cons cur result)
(loop for i from 0 to (list-length cur) do
(create-permuations-helper source (1+ index)
(append cur (list (nth i source))) result))))
99% of times when a compiler reports an error you can trust it to be true. Here Index is the list (1+ index), literally the 1+ symbol followed by the index symbol. This is so because you are using a macro, and macros operate on code.
In your macro, you do not return a form to be evaluated, you execute code during macro-expansion that depends on itself. That alone is an undefined behaviour. For example:
(defmacro a (x)
(if (plusp x)
(a (- x 1))
nil))
In the body of a, you want to expand code using a recursive call to itself. But the macro is not yet fully known and cannot be until the whole macro is defined.
Maybe the particular lisp implementation binds a to the macro function in body of the macro, which is a strange thing to do, or you evaluated the definition twice. The first time the compiler assumes a is an unknown function, then binds a to a macro, and the second time it tries to expand the macro.
Anyway macro are not supposed to be recursive.
In the example, since the macro does not evaluate its argument, the nested call to the macro is given the literal expression (- x 1), and not its actual value, which cannot be known anyway since x is unknown. You are crossing a level of abstraction here by trying to evaluate things at macroexpansion time.
But, macros can expand into code that refers to themselves.
(defmacro a (x)
(if (plusp x)
`(b (a ,(- x 1)))
nil))
Now, (a 2) expands into (b (a 1)), which itself macroexpands into (b (b (a 0))), and finally reaches a fixpoint which is (b (b nil)).
The difference is that the macro produces a piece of code and returns, which the compiler macroexpands again, whereas in the first example, the macro must already be expanded in the body of its own definition.
Possible implementation
One way to solve your problem is to define a local function that has access to a variable defined in your main function. Then, the local function can set it, and you do not need to pass a variable by reference (which is not possible to do):
(defun permut (list)
(let (result)
(labels ((recurse (stack list)
(if list
(dolist (x list)
(recurse (cons x stack)
(remove x list :count 1)))
(push stack result))))
(recurse nil list))
result))
Alternatively, you can split the process in two; first, define permut-helper, which is a higher-order function that takes a callback function; it generates permutations and calls the callback for each one:
(defun permut-helper (stack list callback)
(if list
(dolist (x list)
(permut-helper (cons x stack)
(remove x list :count 1)
callback))
(funcall callback stack)))
You call it with a function that pushes results into a list of permutations:
(defun permut (list)
(let (result)
(flet ((add-result (permutation)
(push permutation result)))
(permut-helper nil list #'add-result))
result))
I am wondering how one can achieve the following. Suppose I have a list of variables that are bound by some let above. I would like to turn this list into a list of the values to which those variables are bound.
That is, suppose we have
(define make-plist-from-variables (variables)
(let ((keys variables)
(values (mapcar #'identity variables)))
(if (eq (length keys) (length values))
(make-plist keys values)
nil))))
What can I use in place of #'identity to unpack those values properly?
At the moment, the following call produces the following output.
CL-USER> (let ((a 2) (b 3)) (make-plist-from-variables '(a b)))
(A A B B)
I would like it to be (A 2 B 3)
It needs to be a macro because there is no way to fetch a variable's lexical value based on its symbol.
(defmacro make-plist-from-variables (&rest variables)
(loop :for binding :in variables
:collect `',binding :into result
:collect binding :into result
:finally (return `(list ,#result))))
(macroexpand-1 '(make-plist-from-variables a b))
; ==> (list 'a a 'b b)
(let ((a 2) (b 3))
(make-plist-from-variables a b))
; ==> (a 2 b 3)
EDIT
Implementation without loop using mapcan:
(defmacro make-plist-from-variables (&rest variables)
`(list ,#(mapcan (lambda (v) `(',v ,v)) variables))
Functions don't have access to the lexical environment of their callers.
More precisely, during evaluation you cannot access the values of lexical variables knowing only their symbols. Only macros have access to environment objects.
Special variables
You can use dynamic binding:
(defun foo ()
(declare (special a))
(symbol-value 'a))
(let ((a 3))
(declare (special a))
(foo))
=> 3
In your case, you would collect the symbol along its value, by using SYMBOL-vaLUE on all your symbols.
Related to your question is how to dynamically bind variables to values where the variable names and/or values are known at evaluation time; see special operator PROGV.
Macros
You could obtain e.g. an association list by writing the following code:
(acons 'a a (acons 'b b nil))
Depending on the use case behind your question, you may want to have a macro that expands into such code, that references the variables you want to evaluate.
I think I just use setq (or setf, I'm not really sure the difference), but I don't understand how to reference the [i][j]-th element in an array in lisp.
My start condition is this:
? (setq x (make-array '(3 3)))
#2A((0 0 0) (0 0 0) (0 0 0))
I want to alter, say, the 2nd item of the 3rd "row" to give this:
? ;;; What Lisp code goes here?!
#2A((0 0 0) (0 0 0) (0 "blue" 0))
The following, which I would have thought close, gives an error:
(setq (nth 1 (nth 2 x)) "blue")
So what's the correct syntax?
Thanks!
I think proper way is to use setf with aref like this:
(setf (aref x 2 1) "blue")
For more details see reference.
You can find a dictionary of the ARRAY operations in the Common Lisp HyperSpec (the web version of the ANSI Common Lisp standard:
http://www.lispworks.com/documentation/lw50/CLHS/Body/c_arrays.htm
AREF and (SETF AREF) are documented here:
http://www.lispworks.com/documentation/lw50/CLHS/Body/f_aref.htm
The syntax to set an array element is: (setf (aref array &rest subscripts) new-element).
Basically if you want to set something in Common Lisp, you just need to know how to get it:
(aref my-array 4 5 2) ; access the contents of an array at 4,5,2.
Then the set operation is schematically:
(setf <accessor code> new-content)
This means here:
(setf (aref my-array 4 5 2) 'foobar) ; set the content of the array at 4,5,2 to
; the symbol FOOBAR
The correct invocation is
(setf (aref x 2 1) "blue")
setq is used when you're assigning to a variable. Only setf knows how to "reach into" compound objects as with setting a value in your array. Of course, setf also knows how to assign to variables, so if you stick with setf you'll always be okay.
I would like to define a type specifier that describes a list of things of the same type. So I would like to have (list-of integer) similar to (array integer) (which is built-in). I am able to create it for a specific type, like this:
(defun elements-are-integer (seq)
(every #'(lambda (x) (typep x 'integer)) seq))
(deftype list-of-integer ()
'(and list (satisfies elements-are-integer)))
However, this means I have to do this for every possible type. How can I change this code so that the type would take another type as an argument, and construct the satisfies predicate on the fly? The problem is that the satisfies requires a global symbol, and I don't know how to define the predicate function in proper context (I guess I need to gensym it somehow, but how?). Also, the solution should work so that the type could be created inside another package.
Try this:
(defun elements-are-of-type (seq type)
(every #'(lambda (x) (typep x type)) seq))
(deftype list-of-type (type)
(let ((predicate (gensym)))
(setf (symbol-function predicate)
#'(lambda (seq) (elements-are-of-type seq type)) )
`(and list (satisfies ,predicate)) ))
(typep '(1 2 3) '(list-of-type integer))
; -> T
(typep '(1 2 a) '(list-of-type integer))
; -> NIL
(typep '(a b c) '(list-of-type symbol))
; -> T