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Scheme The object is not applicable


Scheme The object is not applicable

By : user2952978
Date : November 19 2020, 03:59 PM
hope this fix your issue , You had min3 in the wrong place in the definition of sum:
code :
(define (sum x y z)
  (- (+ x y z) (min3 x y z)))


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Scheme Error Object Is Not Applicable

Scheme Error Object Is Not Applicable


By : user2549591
Date : March 29 2020, 07:55 AM
this one helps. I am writing a Scheme function that detects if a word is in a list of words. My code uses an if statement and memq to return either #t or #f. However, something is causing the first parameter to return the error that the object is not applicable. , Parentheses matter:
code :
(define in?                                                                     
  (lambda (y xs)                                                                
    (if (memq y xs) #t #f)))
(define in?                                                                     
  (lambda (y xs)                                                                
    (and (memq y xs) #t)))
Scheme object is not applicable - 2 different ways

Scheme object is not applicable - 2 different ways


By : Philip Clegg
Date : March 29 2020, 07:55 AM
To fix this issue In the first case, you're trying to apply (car roster) to addifnotcontains and (cdr roster):
code :
(cons ((car roster) addifnotcontains (cdr roster)))
      ^   <--    This is one expression   -->    ^ 
(cons (car roster) (addifnotcontains (cdr roster) item))
Scheme: The object () is not applicable

Scheme: The object () is not applicable


By : Mr.Roboto
Date : March 29 2020, 07:55 AM
I wish did fix the issue. The error, as usually happens in lisp languages, depends on the wrong use of the parentheses, in this case the extra parentheses that enclose the body of the function.
Remove it and the function should work:
code :
(define a (lambda (l i) 
            (cond ((null? l) l)
                  (else (cons (cons (car l) i) (a (cdr l) i))))))
Object not applicable in MIT Scheme (a different Ackermann's function)

Object not applicable in MIT Scheme (a different Ackermann's function)


By : Henry.z
Date : March 29 2020, 07:55 AM
it helps some times I found this version of Ackermann's function and tried to code it in MIT Scheme Lisp with no success: , There is an error (too many parentheses) in the last expression:
code :
(acker2 (m (- n 1)))
(acker2 m (- n 1))
Object not applicable in my scheme program

Object not applicable in my scheme program


By : kkxlynx
Date : March 29 2020, 07:55 AM
Hope that helps There are wrong parentheses all over the place, and the indentation can be improved. For instance, notice the errors in this snippet of code:
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