I have the following structures:
(defstruct track
size
env
startpos
endpositions)
(defstruct state
pos
vel
action
cost
track
other)
I have a state and Im trying to access endpositions(list of lists)
(setq coluna_final (nth 1 (nth 0 (state-track-endpositions st))))
but I get the error: EVAL: undefined function STATE-TRACK-ENDPOSITIONS
What am I doing wrong?
The first defstruct defines (inter alia) function track-endpositions, and the second defines state-track. Lisp has no way to know that the latter returns a track (even if you declare the slot type, it will not define the function you want).
You can do it yourself:
(defun state-track-endpositions (st)
(track-endpositions (state-track st)))
Related
I want to create a lisp function which corresponds to a macro in C.
e.g., there is one HIWORD in win32 API, which is defined as a macro in the header file.
I tried to define it as below but was told that HIWORD is unresolved.
CL-USER 4 > (hiword #xFFFFFFFF)
Error: Foreign function HIWORD trying to call to unresolved external function "HIWORDW".
I just want to know how to create a wrapper for C macros like for C functions.
(fli:define-c-typedef DWORD (:unsigned :long))
(fli:define-c-typedef WORD (:unsigned :short))
(fli:define-foreign-function
(HIWORD "HIWORD" :dbcs)
((dwVal dword))
:result-type word :calling-convention :stdcall)
You cannot do this directly. C preprocessor macros are not preserved in the compilation process, i.e., there is simply no artifact in the generated object files, which would correspond to the C macro itself (though its expansion may be part of the object file multiple times). And since there is no artifact, there is nothing to bind to with FFI.
You can, however, provide a wrapper function
#define HIGHWORD(x) /* whatever */
int
highword_wrapper(int x)
{
return HIGHWORD(x);
}
and this one can be used with FFI.
No need to jump into another language. Shifting and masking in Lisp:
(defun hiword (val)
(logand (ash val -16) #xFFFF)
(defun loword (val) ;; ditto
(logand val #xFFFF))
Another way: using the ldb accessor with ranges expressed using byte syntax:
(defun hiword (val)
(ldb (byte 16 16) val) ;; get 16-bit-wide "byte" starting at bit 16.
(defun loword (val)
(ldb (byte 16 0) val) ;; get 16-bit-wide "byte" at position 0.
I'm trying to build a simple function that gets a number, checks if the number is more the zero and return the square root of the number:
#lang pl 03
(: sqrtt: Number -> Number)
(define (sqrtt root)
(cond [(null? root) error "no number ~s"]
[( < root 0) error "`sqrt' requires a non-negative input ~s"]
[else (sqrt root)]))
but the result I get when I'm trying to compile the function is:
type declaration: too many types after identifier in: (: sqrtt: Number
-> Number)
Why am I getting that error and how do I fix it?
Try this:
(define (sqrtt root)
(cond [(null? root) (error "no number ~s")]
[(< root 0) (error "`sqrt' requires a non-negative input ~s")]
[else (sqrt root)]))
You simply forgot the () around error. Remember that error is a procedure and, like all other procedures, to apply it you have to surround it with parentheses together with its arguments.
The error message you're getting tells you that you have too many types after an identifier in a : type declaration. Now in racket, sqrtt: counts as an identifier. What you probably meant was sqrtt :, with a space in between.
(: sqrtt : Number -> Number)
The difference is that type declarations of the form (: id : In ... -> Out) are treated specially, but those of the form (: id In ... -> Out) are not. And sqrtt: is counts as the id.
There's also the problem Oscar Lopez pointed out, where you're missing parens around the error calls. Whenever you call a function in racket, including error, you need to wrap the function call in parens.
Also, the (null? root) clause is useless, since root has the type Number and null? will always return false for numbers.
And another thing, depending on what the pl language does, if you get a type error from < afterwards, that's because < operates on only Real numbers, but the Number type can include complex numbers. So you might have to change the type to Real or something.
Here are two simple functions that use push on a variable passed in:
(defun push-rest (var) (push 99 (rest var)))
and
(defun just-push (something) (push 5 something))
The first one will permanently mutate the var passed. The second does not. This is quite confusing for someone who is learning the scoping behavior of this language:
CL-USER> (defparameter something (list 1 2))
SOMETHING
CL-USER> something
(1 2)
CL-USER> (just-push something)
(5 1 2)
CL-USER> something
(1 2)
CL-USER> (push-rest something)
(99 2)
CL-USER> something
(1 99 2)
In push-rest why isn't the var's scope local to the function like in just-push, when they are both using the same function, push?
According to Peter Siebel's Practical Common Lisp, Chapter 6. Variables: This might help you a lot:
As with all Common Lisp variables, function parameters hold object references. Thus, you can assign a new value to a function parameter within the body of the function, and it will not affect the bindings created for another call to the same function. But if the object passed to a function is mutable and you change it in the function, the changes will be visible to the caller since both the caller and the callee will be referencing the same object.
And a footnote:
In compiler-writer terms Common Lisp functions are "pass-by-value." However, the values that are passed are references to objects.
(Pass by value also essentially means copy; but we aren't copying the object; we are copying the reference/pointer to the object.)
As I noted in another comment:
Lisp doesn't pass objects. Lisp passes copies of object references to functions. Or you could think of them as pointers. setf assigns a new pointer created by the function to something else. The previous pointer/binding is not touched. But if the function instead operates on this pointer, rather than setting it, then it operates on the original object the pointer points too. if you are a C++ guy, this might make much more sense for you.
You can't push on a variable passed. Lisp does not pass variables.
Lisp passes objects.
You need to understand evaluation.
(just-push something)
Lisp sees that just-push is a function.
Now it evaluates something. The value of something is a list (1 2).
Then it calls just-push with the single argument (1 2).
just-push will never see the variable, it does not care. All it gets are objects.
(defun push-rest (some-list) (push 99 (rest some-list)))
Above pushes 99 onto the rest, a cons, of the list passed. Since that cons is visible outside, the change is visible outside.
(defun just-push (something) (push 5 something))
Above pushes 5 to the list pointed to by something. Since something is not visible outside and no other change has made, that change is not visible outside.
push works differently when it's passed a symbol or list as it's second argument. Pehaps you might understand it better if you do macroexpand on the two different.
(macroexpand '(push 99 (rest var)))
;;==>
(let* ((#:g5374 99))
(let* ((#:temp-5373 var))
(let* ((#:g5375 (rest #:temp-5373)))
(system::%rplacd #:temp-5373 (cons #:g5374 #:g5375)))))
Now most of this is to not evaluate the arguments more than once so we can in this case rewrite it to:
(rplacd var (cons 99 (rest var)))
Now this mutates the cdr of var such that every binding to the same value or lists that has the same object in it's structure gets altered. Now lets try the other one:
(macroexpand '(push 5 something))
; ==>
(setq something (cons 5 something))
Here is creates a new list starting with 5 and alters the local functions binding something to that value, that in the beginning pointed to the original structure. If you have the original structure in a variable lst it won't get changed since it's a completely different binding than something. You can fix your problem with a macro:
(defmacro just-push (lst)
(if (symbolp lst)
`(push 5 ,lst)
(error "macro-argument-not-symbol")))
This only accepts variables as argument and mutates it to a new list having 5 as it's first element and the original list as it's tail. (just-push x) is just an abbreviation for (push 5 x).
Just to be clear. In an Algol dialect the equivalent code would be something like:
public class Node
{
private int value;
private Node next;
public Node(int value, Node next)
{
this.value = value;
this.next = next;
}
public static void pushRest(Node var)
{
Node n = new Node(99, var.next); // a new node with 99 and chained with the rest of var
var.next = n; // argument gets mutated to have new node as next
}
public static void justPush(Node var)
{
var = new Node(5, var); // overwrite var
System.out.print("var in justPush is: ");
var.print();
}
public void print()
{
System.out.print(String.valueOf(value) + " ");
if ( next == null )
System.out.println();
else
next.print();
}
public static void main (String[] args)
{
Node n = new Node( 10, new Node(20, null));
n.print(); // displays "10 20"
pushRest(n); // mutates list
n.print(); // displays "10 99 20"
justPush(n); // displays "var in justPush is :5 10 99 20"
n.print(); // displays "10 99 20"
}
}
(push item place)
It work as follows　when the form is used to instruct the place where this is referred to in the setf:
(setf place (cons item place))
Basedon your profile, it looks like you have familiarity with C-like languages. push is a macro, and is the following equivalence is roughly true (except for the fact that this would cause x to be evaluated twice, whereas push won't):
(push x y) === (setf x (list* x y))
That's almost a C macro. Consider a similar incf macro (CL actually defines an incf, but that's not important now):
(incf x) === (setf x (+ 1 x))
In C, if you do something like
void bar( int *xs ) {
xs[0] = xs[0] + 1; /* INCF( xs[0] ) */
}
void foo( int x ) {
x = x + 1; /* INCF( x ) */
}
and have calls like
bar(zs); /* where zs[0] is 10 */
printf( "%d", zs[0] ); /* 11, not 10 */
foo(z); /* where z is 10 */
printf( "%d", z ); /* 10, not 11 */
The same thing is happening in the Lisp code. In your first code example, you're modifying contents of some structure. In your second code example, you're modifying the value of lexical variable. The first you'll see across function calls, because the structure is preserved across function calls. The second you won't see, because the lexical variable only has lexical scope.
Sometimes I wonder if Lisp aficionados (myself included) promote the idea that Lisp is different so much that we confuse people into thinking that nothing's the same.
Well, the title is a mouthful, so I will expand on it. I have the following code (it is incomplete, mostly just for illustration):
(use '[clojure.zip :only [up down right node])
(defn in-zip? [form]
(contains? (-> 'clojure.zip ns-publics vals set) (first form)))
(defn do-something-to-zip-form [fx loc rest]
;; this is where I would do the transform, but for now, I will just
;; return the actual form
form)
(defn transform-zip [form]
(if (in-zip? form)
(do-something-to-zip-form form)
form))
(defmacro gozip [body]
(clojure.walk/postwalk transform-zip body))
The purpose of in-zip? is to take a form and determine whether the evaluated form calls to a function in clojure.zip . So, something like (is-zip? '(clojure.zip/down loc) or (is-zip? '(up loc)) should return true, any form that isn't calling a function within clojure.zip should return false.
I want to be able to call gozip with a form, and have every call to a function in clojure.zip be replaced by my do-something-to-zip-form transformation. Some examples:
(gozip (-> loc down right right (clojure.zip/update 3))
In the above expression, I would like it to run the transform in 5 places (loc,down,right,right, and update) because those are all functions within clojure.zip.
(gozip (let [d (down loc)] (node loc)))
In the above expression, I would like to run transform in 2 places (down, node).
Sorry about being so pedantic about explaining what I am interested in, it is just that I am having trouble explaining exactly what I want, seems easier through examples. I am looking to use gozip in clojure and clojurescript code.
The macro, transform!, as defined below seems to work for => (transform! ["foo" 1 2 3]). The purpose is to take in a list, with the first element being a string that represents a function in the namespace. Then wrapping everything into swap!.
The problem is that transform! doesn't work for => (transform! coll), where (def coll ["foo" 1 2 3]). I am getting this mystery exception:
#<UnsupportedOperationException java.lang.UnsupportedOperationException: nth not supported on this type: Symbol>
The function:
(defmacro transform!
" Takes string input and update data with corresponding command function.
"
[[f & args]] ;; note double brackets
`(swap! *image* ~(ns-resolve *ns* (symbol f)) ~#args))
I find it strange that it works for one case and not the other.
Macros work at compile-time and operate on code, not on runtime data. In the case of (transform! coll), the macro is being passed a single, unevaluated argument: the symbol coll.
You don't actually need a macro; a regular function will suffice:
(defn transform! [[f & args]]
(apply swap! *image* (resolve (symbol f)) args)))
Resolving vars at runtime could be considered a code smell, so think about whether you really need to do it.
You're passing a symbol to the macro, namely coll. It will try to pull that symbol apart according to the destructuring statement [f & args], which won't be possible of course.
You can also use (resolve symbol) instead of (ns-resolve *ns* symbol).