Convert Ratio To Decimal

May 15, 2015

Here’s our solution, which was indeed written within fifteen minutes:

(define (decimal num denom digits) (let loop ((num (* (modulo num denom) 10)) (out (cons #\. (reverse (string->list (number->string (quotient num denom)))))) (digits digits)) (if (zero? digits) (list->string (reverse out)) (loop (* (modulo num denom) 10) (cons (integer->char (+ (quotient num denom) 48)) out) (- digits 1)))))

This is nothing more than the grade-school long-division algorithm, generating the output one digit at a time; out is initialized to the part of the quotient to the left of the decimal point, in reverse:

> (decimal 3227 557 30) "5.793536804308797127468581687612"

You can run the program at http://ideone.com/hiNpKQ.

Posted by programmingpraxis

Filed in Exercises

5 Comments »

5 Responses to “Convert Ratio To Decimal”

Paul said

May 15, 2015 at 3:28 PM

In Python.

from decimal import Decimal, getcontext

def round_frac1(num, den, digits):
    getcontext().prec = digits
    return Decimal(num) / Decimal(den)
    
def round_frac2(num, den, digits):
    res = ["" if num >= 0 else "-"]
    div, mod = divmod(abs(num), den)
    res += [str(div), "."]
    for d in range(digits):
        if mod == 0:
            break
        div, mod = divmod(mod * 10, den)
        res.append(str(div))
    return "".join(res)

print(round_frac1(3227, 557, 31)) # 5.793536804308797127468581687612
print(round_frac2(3227, 557, 30)) # 5.793536804308797127468581687612

Globules said

May 15, 2015 at 3:48 PM

A quick ‘n dirty Haskell version.

-- The result of a/b with n digits after the decimal point.
divide n a b = let (q, r) = a `quotRem` b
               in show q ++ "." ++ concatMap show (take n (r `fr` b))
  where a `fr` b = let (q, r) = (10*a) `quotRem` b in q : (r `fr` b)

A couple of examples in ghci:

λ> putStrLn $ divide 16 13 17
0.7647058823529411
λ> 13/17
0.7647058823529411
λ> putStrLn $ divide 30 3227 557
5.793536804308797127468581687612
λ>

Vaibhav Jain said
May 16, 2015 at 11:58 AM
In C++

#include

using namespace std;

int main()
{
int den,num,dig,temp1,temp2;
cout <> den;
cout <> num;
cout <> dig;
cout << num/den << "."; // Whole number part of decimal followed by decimal point
temp1=num;
for (int i=0;i<30;i++)
{
temp2=(temp1%den)*10;
cout << temp2/den;
temp1=temp2%den;
}
return 0;
}
Vaibhav Jain said
May 16, 2015 at 12:01 PM
In C++

Christophe said

May 27, 2015 at 3:23 PM

Clojure
I admit, it took me longer than 15 minutes..

(defn long-division
  [nom denom decimals]
  (loop [result   (str (quot nom denom) ".")
         nom          (* (mod nom denom) 10)
         decimals                   decimals]
    (if (>  decimals 0 )
      (recur (str result (quot nom denom))
             (* (mod nom denom) 10)
             (dec decimals))
      result)))



(println  (long-division 3227 557 30))

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