I got "scheme application not a procedure" in the last recursive calling of a function
Solution 1:
You intend to execute two expressions inside the consequent part of the if
, but if
only allows one expression in the consequent and one in the alternative.
Surrounding both expressions between parenthesis (as you did) won't work: the resulting expression will be evaluated as a function application of the first expression with the second expression as its argument, producing the error "application: not a procedure; expected a procedure that can be applied to arguments ..."
, because (time-prime-test n)
does not evaluate to a procedure, it evaluates to #<void>
.
You can fix the problem by either using a cond
:
(define (search-for-primes n m)
(cond ((< n m)
(time-prime-test n)
(search-for-primes (+ n 1) m))
(else
(display " calculating stopped. "))))
Or a begin
:
(define (search-for-primes n m)
(if (< n m)
(begin
(time-prime-test n)
(search-for-primes (+ n 1) m))
(display " calculating stopped. ")))
Solution 2:
((time-prime-test n)
(search-for-primes (+ n 1) m))
This will try to apply the result of time-prime-test
as a procedure. time-prime-test
doesn't return a procedure. Use begin
:
(begin
(time-prime-test n)
(search-for-primes (+ n 1) m))