how to always round up to the next integer [duplicate]

Answer 1

I think the easiest way is to divide two integers and increase by one :

int r = list.Count() / 10;
r += (list.Count() % 10 == 0 ? 0 : 1);

No need of libraries or functions.

edited with the right code.

Answer 2

You can use Math.Ceiling

http://msdn.microsoft.com/en-us/library/system.math.ceiling%28v=VS.100%29.aspx

Answer 3

Math.Ceiling((double)list.Count() / 10);

Answer 4

Check by using mod - if there is a remainder, simply increment the value by one.

Answer 5

This will also work:

c = (count - 1) / 10 + 1;

Answer 6

Xform to double (and back) for a simple ceil?

list.Count()/10 + (list.Count()%10 >0?1:0) - this bad, div + mod

edit 1st: on a 2n thought that's probably faster (depends on the optimization): div * mul (mul is faster than div and mod)

int c=list.Count()/10;
if (c*10<list.Count()) c++;

edit2 scarpe all. forgot the most natural (adding 9 ensures rounding up for integers)

(list.Count()+9)/10