C++ test for divisibility with double

Question

why doesn´t ((num / i) % 1 == 0) work in C++ when num is a double? and how would I instead write this code, that checks for factorials by checking if it leaves a remainder (etc 0.3333).

int getFactorials (double num)
{
    int total = 0;      // if (total / 2) is equal too 'num' it is a perfect number.

    for (int i = 1; i < num; i++)
    {
        if ((num / i) % 1 == 0)
        {
            cout << i << endl;
        }
    }
    return 0;
}

Do you mean [factorial](http://en.wikipedia.org/wiki/Factorial) or [factor](http://en.wikipedia.org/wiki/Integer_factorization)? — user93353, Nov 09 '12 at 09:33
@didierc: if i cast the double (num) to a integer, thus algorithm will not work — Tom Lilletveit, Nov 09 '12 at 09:41
This looks wrong anyway. " doing % 1" returns the remainder when dividing by one. Which will always be zero. — jcoder, Nov 09 '12 at 09:44
@J99: not if the number is a double then it will leave 0.4243 something if it does not divide even. That is why this code works only with a floating point value — Tom Lilletveit, Nov 09 '12 at 09:52
@J99 I´ve got that now, I come from Java, i think the modulus operator worked with floating point values over there, or I might have dreamt it in one of my exciting dreams. — Tom Lilletveit, Nov 09 '12 at 10:01
Ah ok, fair enough :) As others have said you probably want fmod then :) — jcoder, Nov 09 '12 at 10:03

score 7 · Answer 1 · answered Nov 09 '12 at 09:34

7

The % operator is only allowed on integral types (and user defined types which overload it). For floating point, you need the function fmod.

answered Nov 09 '12 at 09:34

James Kanze

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score 6 · Accepted Answer · answered Nov 09 '12 at 09:32

6

Actually what you want to do is check if n is divisible by i so all you have to change is

if ((num / i) % 1 == 0)

into

if (num % i == 0)

You should know that this is error-prone because you are using double as a type for num. You should use an int instead.

answered Nov 09 '12 at 09:32

alestanis

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Shouldn't `(num / i) % 1 == 0` work too? (if rounding errors didn't exist) – irrelephant Nov 09 '12 at 09:33
if (num % i == 0) would not work. for example if I divide 28 by 4 the answer is 7 not zero, but 4 is an divisor of 28 – Tom Lilletveit Nov 09 '12 at 09:38
2

@TomLilletveit The `%` operator gives you the **remainder** of the division of `n` by `i`, so if `i` divides `n` it will give you zero. – alestanis Nov 09 '12 at 09:50
Note that casting to `int` doesn't work unless the `double` is really supposed to be close to an integer value. And since floating point numbers don't represent integers exactly once the exponent part is large, they shouldn't be used as such. Use `long long` for large integers. – Potatoswatter Nov 09 '12 at 10:05

score 2 · Answer 3 · answered Nov 09 '12 at 09:37

The % operator is defined by C++11 §5.6/4:

if the quotient a/b is representable in the type of the result, (a/b)*b + a%b is equal to a.

This is meaningless for floating point types. The definition of modulus depends on division rounding to an integer.

As James Kanze says, use std::fmod instead.

score 0 · Answer 4 · answered Nov 09 '12 at 09:36

0

This code return you a double object

(num / i)

modulo operation is allowed only for int type. You can convert this to int type

( ((int) (num / i) ) % i)

answered Nov 09 '12 at 09:36

Jacek

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C++ test for divisibility with double

4 Answers4