How to calculate the sum of a digits of a number in Scheme?
Solution 1
An alternative method would be to loop over the digits by using modulo. I'm not as used to scheme syntax, but here's a function that works in Lisp for nonnegative integers (and with a little work could encompass decimals and negative values):
(defun sumofdigits(x)
(if (= x 0) 0
(+ (mod x 10)
(sumofdigits (/ ( x (mod x 10)) 10)))))
Solution 2
Something like this can do your digits thing arithmetically rather than string style:
(define (digits n)
(if (zero? n)
'()
(cons (remainder n 10) (digits2 (quotient n 10))))
Anyway, idk if its what you're doing but this question makes me think Project Euler. And if so, you're going to appreciate both of these functions in future problems.
Above is the hard part, this is the rest:
(foldr + (digits 12345) 0)
OR
(apply + (digits 1234))
EDIT  I got rid of intLength
above, but in case you still want it.
(define (intLength x)
(define (intLengthP x c)
(if (zero? x)
c
(intLengthP (quotient x 10) (+ c 1))
)
)
(intLengthP x 0))
Solution 3
Those #\1, #\2 things are characters. I hate to RTFM you, but the Racket docs are really good here. If you highlight string>list in DrRacket and hit F1, you should get a browser window with a bunch of useful information.
So as not to keep you in the dark; I think I'd probably use the "string" function as the missing step in your solution:
(map string (list #\a #\b))
... produces
(list "a" "b")
bearzk
Updated on June 04, 2022Comments

bearzk 7 months
I want to calculate the sum of digits of a number in Scheme. It should work like this:
>(sumofdigits 123) 6
My idea is to transform the number
123
to string"123"
and then transform it to a list'(1 2 3)
and then use(apply + '(1 2 3))
to get6
.but it's unfortunately not working like I imagined.
>(string>list(number>string 123)) '(#\1 #\2 #\3)
Apparently
'(#\1 #\2 #\3)
is not same as'(1 2 3)
... because I'm using languageracket
under DrRacket, so I can not use the function likechar>digit
.Can anyone help me fix this?