Round Up a double to int

2019-02-21 08:07发布

问题:

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#?

回答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


回答2:

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


回答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.



回答4:

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