Mathematics simple example

Question

I am trying to do simple math. a=5, b=a/2 &(round up), which has to be 5/2 = 3 when I round up , c=b-a, 3-5=2. my question when I divide 5/2 and round up I got 2 instead of 3 for b. can some one help me how I can round up and get 3? Thanks

int a=5;

int b=(int)Math.ceil(a/2);

int c=b-a;


System.out.println(b);

System.out.println(c);

Read up on integer arithmetic in Java. – PM 77-1 Apr 01 '16 at 13:47 — PM 77-1, Apr 01 '16 at 13:47
@haylu Check this: http://stackoverflow.com/tour – user987339 Apr 01 '16 at 15:05 — user987339, Apr 01 '16 at 15:05

Cootri · Answer 1 · 2016-04-01T14:13:13.860

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you should use floating-point value of 2 instead of integer in this line:

int b = (int) Math.ceil(a / 2f);//also you can use "2.0" or "2d" instead of "2f"

your version uses int division (which produces int value of 2) and 2f constant forces to use floating-point division and promote a value to float (after it division produces float value of 2.5)

It is possible because of JLS spec for Multiplicative Operators:

Binary numeric promotion is performed on the operands (§5.6.2)

And that is how numeric promotion works for int and float division operands in our case:

Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules: If either operand is of type double, the other is converted to double. Otherwise, if either operand is of type float, the other is converted to float.

edited Apr 01 '16 at 14:13

answered Apr 01 '16 at 13:42

Cootri

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`(a/2d)` should also work, to enforce double division . – Arnaud Apr 01 '16 at 13:46
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Nit: it's not `float` division, it is floating-point division (since `2.0` is a `double`). Similarly, `2.5` is a `double`, not a `float`. – Andy Turner Apr 01 '16 at 13:55
thanks for the reply, you're right – Cootri Apr 01 '16 at 14:03
it works.Thanks @Cootri – Apr 01 '16 at 14:11
glad it helped! Welcome to the Stackoverflow and not forget to mark an answer as accepted to close this question :) – Cootri Apr 01 '16 at 14:17

user987339 · Answer 2 · 2016-04-01T14:08:54.973

3

You can use a little trick to get things faster:

int b =(a + 1) / 2;

Since you are dividing with 2, in case the number is odd with + 1, you'll get exact value for upper rounding. And if number is even with + 1 rounding will give the half of it as expected.

EDIT

Or in general, if you want to calculate Math.ceil(a / b) you can do it with following formula : (a + b - 1) / b, assuming a and b being positive.

edited Apr 01 '16 at 14:08

answered Apr 01 '16 at 13:49

user987339

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More generally, to round up `a / b`, you add `b - 1` to the numerator: `(a + b - 1) / b` (provided `a` and `b` are positive). – Andy Turner Apr 01 '16 at 13:51
Yes, that generalizes this issue. Thank you for pointing this out. – user987339 Apr 01 '16 at 13:53

Mathematics simple example

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