weird Math.round
Author
Hey,
I just needed to use the Math.round(..) func from Java. But I get wrong results.
If I do: Math.round(0.5), I get 1 ... all good so far. But with Math.round(-0.5) I get 0.0 ... why?!! shouldnt it be -1 ?
thanks
Quote:Posted by Instruo (next post)
its probably roudning .0 to .5 (non-inclusive) down and .5 (inclusive) to 1.0 (non inclusive) up. in which case, -0.5 would round to 0.0
If you want to round like that you could use some function like this pseudo code (sorry, I don't know Java and I'm not sure how close this is to a Java function):
int myrounding(float number){ if(number >= 0) return Math.round(number); number *= -1; int roundedNumber = Math.round(number); return (roundedNumber *= -1);}Something like that will make your negative numbers positive, round them, and make them negative again. This way when you put -0.5 in, it'll take it, turn it into 0.5, round up to 1 and then return -1.
I wouldn't think so. its probably roudning .0 to .5 (non-inclusive) down and .5 (inclusive) to 1.0 (non inclusive) up. in which case, -0.5 would round to 0.0
Quote:Original post by Instruo
I wouldn't think so. its probably roudning .0 to .5 (non-inclusive) down and .5 (inclusive) to 1.0 (non inclusive) up. in which case, -0.5 would round to 0.0
That's probably more true than my post. I'll go edit that.
Either way the proposed function in my post should still get it to do what you want it to do.
-0.5 would be rounded UP to the nearest whole number. -1 would mean rounding DOWN since -1 <
Let:
- f be you float
- N the greatest integer such as N <= f
- M the smallest integer such as f <= M
(so: N <= f <= M)
then round(f) = N if and only if (f-N) < (M-f) (strictly)
otherwise round(f) = M
Think about your case:
f = -0.5
N = -1, M = 0
f - N = 0.5
M - f = 0.5
is (f-N) < (M-f) ? No. Then round(-0.5) = M = 0
I hope this mathematical view clears things up.
- f be you float
- N the greatest integer such as N <= f
- M the smallest integer such as f <= M
(so: N <= f <= M)
then round(f) = N if and only if (f-N) < (M-f) (strictly)
otherwise round(f) = M
Think about your case:
f = -0.5
N = -1, M = 0
f - N = 0.5
M - f = 0.5
is (f-N) < (M-f) ? No. Then round(-0.5) = M = 0
I hope this mathematical view clears things up.
Quote:Original post by FreJa
Hey,
I just needed to use the Math.round(..) func from Java. But I get wrong results.
If I do: Math.round(0.5), I get 1 ... all good so far. But with Math.round(-0.5) I get 0.0 ... why?!! shouldnt it be -1 ?
No, it shouldn't. From the documentation:
Quote:Java docs for Math.round(float)
Returns the closest int to the argument. The result is rounded to an integer by adding 1/2, taking the floor of the result, and casting the result to type int.
Adding 1/2 to -0.5 produces 0; taking the floor of 0 yields 0; casting 0 to an int yields 0. Thus the answer is 0. If you'd used -0.6, you would have gotten -1, since adding 1/2 to -0.6 produces -0.1; taking the floor of this yields -1; then casting to an int yields -1, so the the result is -1.
To get the behavior you want, use the absolute value of the argument in round() instead. Then multiply the result by arg/|arg| (-1 for negative numbers, +1 for positive numbers). For instance, suppose the argument is arg = -0.5, as in your example. Then take |-0.5| = 0.5; round as usual, giving 1. Now multiply by arg/|arg| == -0.5/0.5 == -1, giving 1 * -1, which is -1, the desired value.
hope that helps,
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