Vista - Round DOWN of a number of double type

I want to get the integer part of a number declared as double type.
For instance, for 8.76, I want to get 8 instead of 9. In other words,
I want to cut off anything after the decimal point. Any advice on how
to do this?

"Curious" <fir5tsight@xxxxxx> wrote in message
news:aaee8f7e-68bb-42b8-b175-193acf98cc17@xxxxxxAs you don't tell us what language, in general terms: cast it to an integer
type. That will truncate the decimal portion of the double.

On Jul 16, 10:09*am, "PvdG42" <p...@xxxxxx> wrote:Thanks for the suggestion! FYI, I use C#.NET.

Curious wrote:What if it's negative?

Anyway, Math.Truncate does what you appear to want.

Andrew

On Jul 16, 10:48*am, "Andrew Morton" <a...@xxxxxx-press.co.uk.invalid>
wrote:I did a little experiment. a cast from double to int works fine for
both positive and negative numbers. It simply drops the residuals
after the decimal point.

However, if I use Math.Truncate, it gives me a runtime error, "Cannot
implicitly convert type 'double' to 'int'". FYI, the way I use it is
below:

double a = 8.9;
int correct1;
correct1 = Math.Truncate(a);

mLogSw.WriteLine(correct1.ToString());


double j = -4.3;
int correct2;
correct2 = Math.Truncate(j);

mLogSw.WriteLine(correct2.ToString());

double k = -8.7;
int correct3;
correct3 = Math.Truncate(k);
mLogSw.WriteLine(correct3.ToString());

"Curious" wroteI suggest you use the method you've found to work.

(You didn't mention in your original post that you wanted to get the integer
part into an integer variable.)

If you check the help for the return type of the Math.Truncate method you'll
see why you got the runtime error, however; although I know nothing of C#, I
suspect the resolution to getting the result of Math.Truncate to be an Int
would make using Math.Truncate redundant.

HTH,

Andrew