error handling - How to remove an element from a list without using remove function in LISP - 


  (defun my_remove(e list1 list2)   (if (null list1)    nil   ((setf x car(list1))     (if (!= x e)     ((my_append  e list2 )     (setf y cdr(list1))     (my_remove(e y list2)))      ((setf y cdr(list1))     (my_remove(e y list2))      )))))

i trying write function remove element list getting error "it should lambada function" , don't know code correct or wrong.

problems code

there few problems code. first, let's @ standard indentation

(defun my_remove(e list1 list2)   (if (null list1)       nil       ((setf x car(list1))                   ; (i)        (if (!= x e)            ((my_append  e list2 )            ; (ii)             (setf y cdr(list1))              ; (iii)             (my_remove(e y list2)))          ; (iv)            ((setf y cdr(list1))              ; (v)             (my_remove(e y list2)))))))      ; (vi)

each of marked lines has problem. syntax function call in lisp is

(function argument…)

that means in line (i), you're trying call function named (setf x car(list1)) argument (if (!= x e) …). of course, that's not name of function, , suspect if was, didn't want call argument (if (!= x e) …). 

(setf x car (list1))

is trying to set x value of value of variable car (and there isn't one), , assign new value place (list1). since syntax function call (function argument…), want instead:

(setf x (car list1))

if you're trying sequence forms, might consider using cond, in can provide multiple forms, or progn (see in common lisp, why multi-expression bodies of (if) statements require (progn)?).

for instance, instead of 

           ((my_append  e list2 )            ; (ii)             (setf y cdr(list1))              ; (iii)             (my_remove(e y list2)))          ; (iv)

you want

           (progn             (my_append  e list2 )            ; (ii)             (setf y cdr(list1))              ; (iii)             (my_remove(e y list2)))          ; (iv)

you'll have problems that, too, though. in (iii) , (iv), you'll need use (cdr list1) , (my_remove e y list2), discussed above, have problem you're evaluating (ii) , (iii), you're discarding value.

a simplified approach

i think might benefit if think simple definition of my-remove. in general, list either empty list () or cons cell car first element of list , cdr rest of list. can use definition this, then:

  • remove(x,list)
    • if list empty, return list (it's empty, doesn't contain x)
    • if list not empty,
      • if first element of list x, return remove(x,rest(list))
      • else, first element of list not x, return new list first element first element of list, , rest remove(x,rest(list)).

in code, looks like:

(defun my-remove (element list)   (if (endp list)       list       (if (eql element (first list))           (my-remove element (rest list))           (list* (first list)                  (my-remove element (rest list)))))) 
cl-user> (my-remove 1 '(1 2 3 1 2 3)) (2 3 2 3)

those nested ifs bit ugly, , might want use cond here, though don't need multiple expression bodies permits:

(defun my-remove (element list)   (cond     ((endp list)      list)     ((eql element (first list))      (my-remove element (rest list)))     (t      (list* (first list) (my-remove element (rest list))))))

since cond clause no body test form evaluates true returns value of test form, can make last clause bit simpler:

(defun my-remove (element list)   (cond     ((endp list) list)     ((eql element (first list)) (my-remove element (rest list)))     ((list* (first list) (my-remove element (rest list))))))

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