How to round a float to an int?

Question

Hi how do I round a float to an int such that if the decimal part is .5 or more then the value is rounded up otherwise the value is rounded down? For example 3.2f => 3; 3.5f => 4; 3.7f => 4; 4.2f => 4; 4.5f => 5; 4.7f => 5 If I use (int)Math.Round(3.5f) the result is 4 (ie what I want); whereas with (int)Math.Round(4.5f) the result is 4 (ie not what I want). Is there a Math method which does rounding the way I want it? Thanks Peter

Mamta M · Answer

Peter What you need is (int)Math.Round(4.5f MidpointRounding.AwayFromZero) This will return you 5 as you desire. More on the MidPointRounding can be found here http://msdn.microsoft.com/en-us/library/system.midpointrounding.aspx Hope that helps.

Peter · Answer

Yes but try (int)Math.Round(6.5f).

Manish Dwivedi · Answer

Hi Peter Math.Round is working like that as you want. I checked that. it gives ( int ) Math .Round( 7.5f ) result = 8 and when i put ( int ) Math .Round( 7.4f ) result= 7 I think this is you want. And It works perfect.

How to "round" a float to an int?