Java rounding up to an int using Math.ceil

Question:

int total = (int) Math.ceil(157/32);

Why does it still return 4? 157/32 = 4.90625, I need to round up, I’ve looked around and this seems to be the right method.

I tried total as double type, but get 4.0.

What am I doing wrong?

 

Answer:

You are doing 157/32 which is dividing two integers with each other, which always result in a rounded down integer. Therefore the (int) Math.ceil(...) isn’t doing anything. There are three possible solutions to achieve what you want. I recommend using either option 1 or option 2. Please do NOT use option 0.

Option 0

Convert a and b to a double, and you can use the division and Math.ceil as you wanted it to work. However I strongly discourage the use of this approach, because double division can be imprecise.

int n = (int) Math.ceil((double) a / b));

Option 1

int n = a / b + (a % b == 0) ? 0 : 1; 

You do a / b with always floor if a and b are both integers. Then you have an inline if-statement witch checks whether or not you should ceil instead of floor. So +1 or +0, if there is a remainder with the division you need +1. a % b == 0 checks for the remainder.

Option 2

This option is very short, but maybe for some less intuitive. I think this less intuitive approach would be faster than the double division and comparison approach:

int n = (a + b - 1) / b;

To reduce the chance of overflow you could use the following. However please note that it doesn’t work for a = 0 and b > 1.

int n = (a - 1) / b + 1;

Explanation behind the “less intuitive approach”

Since dividing two integer in Java (and most other programming languages) will always floor the result. So:

int a, b;
int result = a/b (is the same as floor(a/b) )

But we don’t want floor(a/b), but ceil(a/b), and using the definitions and plots from Wikipediaenter image description here

With these plots of the floor and ceil function you can see the relationship.

Floor function Ceil function

You can see that floor(x) <= ceil(x). We need floor(x + s) = ceil(x). So we need to find s. If we take 1/2 <= s < 1 it will be just right (try some numbers and you will see it does, I find it hard myself to prove this). And `1/2 <= (b-1) / b < 1, so

ceil(a/b) = floor(a/b + s)
          = floor(a/b + (b-1)/b)
          = floor( (a+b-1)/b) )

This is not a real proof, but I hope your are satisfied with it. If someone can explain it better I would appreciate it too. Maybe ask it on MathOverflow.

shareimprove this answer

https://stackoverflow.com/questions/7139382/java-rounding-up-to-an-int-using-math-ceil

 

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