I can't seem to find the answer I'm looking for regarding a simple question: how do I round up any number to the nearest int?
For example, whenever the number is 0.2, 0.7, 0.2222, 0.4324, 0.99999 I would want the outcome to be 1.
So far I have
int b = (int) Math.ceil(a / 100);It doesn't seem to be doing the job, though.
67 Answers
Math.ceil() is the correct function to call. I'm guessing a is an int, which would make a / 100 perform integer arithmetic. Try Math.ceil(a / 100.0) instead.
int a = 142;
System.out.println(a / 100);
System.out.println(Math.ceil(a / 100));
System.out.println(a / 100.0);
System.out.println(Math.ceil(a / 100.0));
System.out.println((int) Math.ceil(a / 100.0));Outputs:
1
1.0
1.42
2.0
2See
3I don't know why you are dividing by 100 but here my assumption int a;
int b = (int) Math.ceil( ((double)a) / 100);or
int b = (int) Math.ceil( a / 100.0);int RoundedUp = (int) Math.ceil(RandomReal);This seemed to do the perfect job. Worked everytime.
10 years later but that problem still caught me.
So this is the answer to those that are too late as me.
This does not work
int b = (int) Math.ceil(a / 100);Cause the result a / 100 turns out to be an integer and it's rounded so Math.ceil
can't do anything about it.
You have to avoid the rounded operation with this
int b = (int) Math.ceil((float) a / 100);Now it works.
The easiest way to do this is just:
You will receive a float or double and want it to convert it to the closest round up then just do System.out.println((int)Math.ceil(yourfloat)); it'll work perfectly
Just another option. Use basics of math:
Math.ceil(p / K) is same as ((p-1) // K) + 1
Assuming a as double and we need a rounded number with no decimal place . Use Math.round() function.
This goes as my solution .
double a = 0.99999;
int rounded_a = (int)Math.round(a);
System.out.println("a:"+rounded_a );
Output :
a:1
