Getting the median of 3 values using scheme's - scheme

Question

The problem this time is to get the median of three values (easy)
I did this:
(define (med x y z) (car(cdr(x y z)))
and it was accepted but when testing it:
(med 3 4 5)
I get this error:
Error: attempt to call a non-procedure
(2 3 4)
And when entering letters instead of number i get:
(md x y z)
Error: undefined varia
y
(package user)
Using something besides x y z I get:
(md d l m)
Error: undefined variable
d
(package user)
the question was deleted dont know how anyway
write a function that return the median of 3 values
Sorry for editing the question I got that I should put the values in order first not just a sill car and cdr thing so I did so
33> (define (med x y z)
(if(and(
(<x y) (<y z) y
if(and(
(<y x) (<x z) x z)))))
Warning: invalid expression
(if (and< (<x y) (<y z) y if (and ((<y x) (<x z) x z))))
but as u see Im getting a warning so what is wronge ?

Answer 1

You probably want to create a list, like this:
(define (med x y z) (car(cdr(list x y z)))
However, it seems like a waste to bundle up the values into a list just to undo them again. This would have the same effect:
(define (med x y z) y)

Answer 2

Note that as defined, (med . rest) is equivalent to (cadr rest) (except that med only takes three values). Personally, I would expect a function that's supposed to return the median of values to return, well, the median, regardless of list order. For example, (med 4 2 5) would return 4 and (3 0 9 6 5) would return 5.
As for the syntax error (which doesn't matter so much for writing med, since there is a better way using sort, length and list-ref), you don't have your parentheses in the right spots. Here's another way of writing what you have now, lining up terms with their siblings and to the right of their ancestors:
(if (and (
(<x y)
(<y z)
y
if
(and (
(<y x)
(<x z)
x
z
) ) ) ) )
The format for if is:
(if test-expr true-expr false-expr)
All of your terms are sub-terms of the conditional, and there's no true-expr or false-expr. You'd want to write your code as:
(if (and ...)
y
(if (...) ; we know that (not (and (< x y) (< y z))
x
z))
Note that you might be ably to simplfy the later tests, since you know the earlier tests are false.
You could also use a cond, which is clearer than a nested sequence of ifs:
(cond (test result)
(test result)
(test result)
... )
For your code, it would be:
(cond ((and ...) y)
((...) x)
(else z))
There's nothing too special about else (it's a syntactical construct rather than an expression, not that it matters much). You can use any value that evaluates to true (e.g. #t) instead.
Remember, parentheses surround both the function and arguments in Lisp ((foo 1 2 3), not foo(1, 2 ,3)), unlike mathematical notation where parentheses surround just the arguments.

Answer 3

while outis did a pretty good job of explaining the issue, there's one other important bit you need to know
(<x y)
calls the function named <x with the parameter y. you need to make sure you add a space inbetween the < and the x, like so:
(< x y)