Round Up a double to int

Question

I have a number ("double") from int/int (such as 10/3).

What's the best way to Approximation by Excess and convert it to int on C#?

Answer 1

Are you asking about System.Math.Ceiling?

Math.Ceiling(0.2) == 1
Math.Ceiling(0.8) == 1
Math.Ceiling(2.6) == 3
Math.Ceiling(-1.4) == -1

Answer 2

int scaled = (int)Math.Ceiling( (double) 10 / 3 ) ;

Answer 3

By "Approximation by Excess", I assume you're trying to "round up" the number of type double. So, @Doug McClean's "ceiling" method works just fine.

Here is a note: If you start with double x = 0.8; and you do the type conversion by (int)x; you get 0. Or, if you do (int)Math.Round(x); you get 1. If you start with double y = 0.4; and you do the type conversion by (int)y; you get 0. Or, if you do (int)Math.Round(y); you get 0.

Answer 4

Consider 2.42 , you can say it's 242/100 btw you can simplify it to 121/50 .