I'm having a little trouble creating a recursive function in Scheme. I need to create a function called foo(x) that recursively does the addition of all the powers. For example foo(5) would be 5^4 + 4^3 + 3^2 + 2^1 + 1^0 = 701. The stopping condition is if x = 0 then return zero. Else then return x^x-1 + foo(x-1) Here's what I have so far for my function:
(define (foo x)
(cond ((zero? x) 0)
(else (+(expt(x (- x 1)))foo(- x 1)))))
You just have to be more careful with the parentheses, in particular notice that the correct way to call a procedure is like this: (foo x), instead of this: foo(x). This should work:
(define (foo x)
(cond ((zero? x) 0)
(else (+ (expt x (- x 1))
(foo (- x 1))))))
(foo 5)
=> 701
Allow me to ident the code. I just pasted it in DrRacket and hit CTRL+I then put the arguments to + on one line each:
(define (foo x)
(cond ((zero? x) 0)
(else (+ (expt (x (- x 1)))
foo
(- x 1)))))
So the base case is ok, but your default case looks very off. x is treated as a procedure since it has parentheses around it and - also uses x as if it's a number. It can't be both.
foo is not applied since it doesn't have parentheses around it so it evaluates to a procedure value, while + would expect all its arguments to be numeric.
The rules of Scheme are that parentheses matters. x and (x) are two totally different things. The first x can be any value, but (x) is an application so x have to evaluate to a procedure. Some exceptions are for special forms you need to know by heart like cond, and define but rather than that it's very important to know you change the meaning of a program by adding parentheses.
The correct definition of your procedure might be:
(define (foo x)
(if (zero? x)
0
(+ (expt x (- x 1))
(foo (- x 1)))))
(foo 5) ; ==> 701
Here I've changed cond to if since none of conds features were used. Seeing cond I expect either side effects or more than one predicate.
