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How to "round" a float to an int?

Peter

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

3 Replies

Know the answer? Post it — somebody with the same question will find it here.

Mamta M15y ago
Accepted 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.
Peter15y ago
Yes, but try (int)Math.Round(6.5f).
Blocked Account15y ago
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.