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))
tuo
Updated on June 05, 2022Comments
-
tuo almost 2 years
so here is the code:
(define (time-prime-test n) (newline) (display n) (start-prime-test n (runtime))) (define (start-prime-test n start-time) (if (prime? n) (report-prime (- (runtime) start-time)))) (define (report-prime elapsed-time) (display " *** ") (display elapsed-time)) (define (search-for-primes n m) (if (< n m) ((time-prime-test n) (search-for-primes (+ n 1) m)) (display " calculating stopped. "))) (search-for-primes 100000 100020)
and i got this error after "calculating stopped." has been displayed. like below:
100017 100018 100019 * 54 calculating stopped. . . application: not a procedure; expected a procedure that can be applied to arguments
given: #<void>
arguments...:
#<void>