Java math.random query, rounding (concept understanding)

Question

I'm trying to understand the logic of the math.random or more specifically what is occurring and or why.

I understand essentially with the code below for example due to java being a 0 based language we start at 0 for indexing, etc.

the below code prints out a value between 0 and 10 inclusive. 11 is exclusive as understood.

int result = (int) (Math.random () * 11);

My question is why for example if we were to get 0.99 * 10 then = 10.99 Why does the number not round up to 11?

Is this simply by design java will always round down to 0? or am I missing a concept?

Thank you for all your help.

Casting to an `int` will result in a `floor`. This means that your number will always get rounded to 0, so 10.99 will become 10 after casting to `int` — David Buzatu, Jul 22 '20 at 07:00
Does this answer your question? Could I submit the answer or do you have other questions? — David Buzatu, Jul 22 '20 at 07:06
I wasn't aware of the floor occurring, Thank-you David I understand now. — Artiems, Jul 22 '20 at 07:06

score 2 · Answer 1 · answered Jul 22 '20 at 07:09

2

As I mentioned in the comments: Casting to an int will result in a floor. This means that your number will always get rounded to 0, so 10.99 will become 10 after casting to int.

You can look at this reference as well

answered Jul 22 '20 at 07:09

David Buzatu

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Java math.random query, rounding (concept understanding)

1 Answers1