Scheme Error Object Is Not Applicable

问题

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.

(define in?                                                                     
  (lambda (y xs)                                                                
    ((if (memq( y xs )) #t #f)))) 

回答1:

Parentheses matter:

(define in?                                                                     
  (lambda (y xs)                                                                
    (if (memq y xs) #t #f)))

so

• you have double parentheses before if
• you put memq parameters between parentheses

BTW, you can also express this as

(define in?                                                                     
  (lambda (y xs)                                                                
    (and (memq y xs) #t)))

来源：https://stackoverflow.com/questions/26287081/scheme-error-object-is-not-applicable