this post was submitted on 24 Nov 2023
1 points (100.0% liked)

Lisp

53 readers
3 users here now

founded 1 year ago
MODERATORS
rglullis
defmacro
1
Constructing a macro 'for' (alien.top)
submitted 1 year ago by Revolutionary_Utena@alien.top to c/lisp
19 comments fedilink hide all child comments
 

Hi. Need help with creating the 'for' macro. It must take a parameter, an initial value, an end value, and a loop step. Macro must return amount of iteration. Various operations should take place inside it. It should be designed without leaks.The preliminary version looks like this:

(defmacro for ((param-name start-value end-value &optional (step1 1)) &body body)
  (let* ((func-name (gensym))
(start (gensym))
(param-name (gensym))
(comparison (if (< step1 0) '< '>))
(end (gensym))
(step (gensym))
(k (gensym)))
'(labels ((,func-name (,param-name ,start ,end ,step ,k)
(let ((new-exprs (progn ,@body))
(newK (+ 1 ,k)))
(if (,comparison ,end ,param-name)
(,func-name (+ ,param-name ,step) ,start ,end ,step newK)
newK))))
(,func-name ,param-name ,start-value ,end-value ,step1 0))))

I understand that it looks terrible. I don't understand how you can access the parameter without violating the rule about leaks.

you are viewing a single comment's thread
view the rest of the comments
[–] Revolutionary_Utena@alien.top 1 points 1 year ago (1 children)

The teacher said to do it this way. As I understand it, when using a macro, you need to refer to a parameter, for example (let ((s 0))

(for (i 2 N)

((inch s (sin i)))

s) Here it is necessary to refer to the modified sh. Therefore, it is passed to the recursive param-name call.

[–] jacobb11@alien.top 1 points 1 year ago (1 children)

Here it is necessary to refer to the modified sh. Therefore, it is passed to the recursive param-name call.

You can bind a variable outside the function and still refer to it inside the function. Here is an example that does so with ",end":

(defmacro for ((param-name start-value end-value &optional (step1 1)) &body body)
  (let* ((func-name (gensym))
         (start (gensym))
         (param-name (gensym))
         (comparison (if (< step1 0) '< '>))
         (end (gensym))
         (step (gensym))
         (k (gensym)))
    `(let ((,end ,end-value))
      (labels ((,func-name (,param-name ,start ,step ,k)
                   (let ((new-exprs (progn ,@body))
                         (newK (+ 1 ,k)))
                     (if (,comparison ,end ,param-name)
                         (,func-name (+ ,param-name ,step) ,start ,step newK)
                         newK))))
         (,func-name ,param-name ,start-value ,step1 0)))))
[–] Revolutionary_Utena@alien.top 1 points 1 year ago 
(defmacro for ((param start-value end-value &optional (step1 1)) &body body)
  (let* ((func-name (gensym))
         (start (gensym))
         (param-name (gensym))
         (comparison (if (< step1 0) '< '>))
         (end (gensym))
         (step (gensym))
         (k (gensym)))
    `(let ((,end ,end-value)
           (,step ,step1)
           (,start ,start-value))
      (labels ((,func-name (,param-name ,k)
                   (let ((new-exprs (progn ,@body))
                         (newK (+ 1 ,k)))
                     (if (,comparison ,end ,param-name)
                         (,func-name (+ ,param-name ,step) newK)
                         newK))))
         (,func-name ,param 0)))))

(let ((s 0))
(for (i 2 5)
(incf s (sin i)))
s)

I made some changes but when I try to call macros with parameter (i 2 5) compiler give me error about unbound variable i.