Java rounding up to an int using Math.ceil

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陌清茗
陌清茗 2020-11-29 17:42
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 se

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  •  温柔的废话
    2020-11-29 18:26

    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. To read more about imprecision of doubles see this question.

    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:
    Please note that this doesn't work for b < 0.

    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 Wikipedia: enter 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.

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