Java to C# - intValue() equivalent in c#

I am trying to re-write a piece of code which was in Java to C#, and I ran into a problem, basically I am creating a method which returns a string, but the string returned from c# and java is not the same, therefore the code is fauly. One problem is the following code,

I have this Java Code:

Double localDouble1 = new Double(d1 / 100.0D);
int l = localDouble1.intValue();

And I want to re-write it in C#, I have tried

Double localDouble1 = d1 / 100.0D;
int l = Convert.ToInt32(localDouble1);

It compiles and works, but the result is different, in my particular scenario, the Java int l variable contains 0, and the c# one retur开发者_开发知识库ns 1.

Is there a better method to achieve what I need to do, the same as in Java.

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Convert.ToInt32(double) rounds to the nearest integer. intValue() doesn't - it truncates, just as a cast does (as documented).

Just cast instead:

int l = (int) localDouble1;

(Also try to avoid names like l which look like 1 :)

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Try this.

Double localDouble1 = d1 / 100.0D;

int l=Convert.ToInt32(Math.Floor(localDouble1));

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Because Convert.ToInt32 If value is halfway between two whole numbers, the even number is returned; that is, 5.5 is converted to 6, and 6.5 is converted to 7. Explicit numeric conversions table

  cast int l = (int) localDouble1;