The object #f or #t is not applicable

Question

I'm trying to create a function that gets the sum of the square of the larger 2 of 3 numbers passed in. (Its exercise 1.3 in SICP)

When I run the following code I get the error ";The object #f is not applicable." If I switch the 3 and the 1 in my function call the error message will say #t instead of #f.

(define (sumOfSquareOfLargerTwoNumbers a b c) (
  cond (
    ( (and (> (+ a b) (+ a c) ) (> (+ a b) (+ b c) ) ) (+ (square a) (square b) ) ) 
    ( (and (> (+ a c) (+ a b) ) (> (+ a c) (+ b c) ) ) (+ (square a) (square c) ) )
    ( (and (> (+ b c) (+ a b) ) (> (+ b c) (+ a c) ) ) (+ (square b) (square c) ) )
  )
))

(sumOfSquareOfLargerTwoNumbers 1 2 3)

I was assuming the appropriate condition would return true and I'd get the square of the larger two numbers. Could someone please explain why I'm getting this error instead?

Brackets in Scheme are not like curly braces in other languages :) resist the temptation to open/close them in the places where you'd normally do so for curly braces, the conventions for indenting in Scheme are different. — Óscar López

Óscar López Óscar López · Accepted Answer · 2020-05-18T00:44:21

There are too many brackets in front of cond and that's causing the problem:

(cond (((and

The proper syntax for your solution should be:

(define (sumOfSquareOfLargerTwoNumbers a b c)
  (cond ((and (> (+ a b) (+ a c)) (> (+ a b) (+ b c)))
         (+ (square a) (square b)))
        ((and (> (+ a c) (+ a b)) (> (+ a c) (+ b c)))
         (+ (square a) (square c)))
        ((and (> (+ b c) (+ a b)) (> (+ b c) (+ a c)))
         (+ (square b) (square c)))))

What was happening is that the condition evaluated to a boolean, and the unexpected surrounding brackets made it look like a procedure application, so you ended up with something like this:

(#t 'something)

Which of course fails, because #t or #f are not procedures and cannot be applied. Just be careful with the brackets and use a good IDE with syntax coloring and code formatting, and you won't have this problem again.

The object #f or #t is not applicable

1 Answers