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 se

Answer 1

When dividing two integers, e.g.,

int c = (int) a / (int) b;

the result is an int, the value of which is a divided by b, rounded toward zero. Because the result is already rounded, ceil() doesn't do anything. Note that this rounding is not the same as floor(), which rounds towards negative infinity. So, 3/2 equals 1 (and floor(1.5) equals 1.0, but (-3)/2 equals -1 (but floor(-1.5) equals -2.0).

This is significant because if a/b were always the same as floor(a / (double) b), then you could just implement ceil() of a/b as -( (-a) / b).

The suggestion of getting ceil(a/b) from

int n = (a + b - 1) / b;, which is equivalent to a / b + (b - 1) / b, or (a - 1) / b + 1

works because ceil(a/b) is always one greater than floor(a/b), except when a/b is a whole number. So, you want to bump it to (or past) the next whole number, unless a/b is a whole number. Adding 1 - 1 / b will do this. For whole numbers, it won't quite push them up to the next whole number. For everything else, it will.

Yikes. Hopefully that makes sense. I'm sure there's a more mathematically elegant way to explain it.