double get fractional part?

[Yin99]

If I have:

8.24
7.56
1.11

as doubles, how do I easily get just the ".24, .56, .11" ? Thanks,

Yin

[Peter Duniho]

On 2007-11-20 11:44:43 -0800, Yin99 <[e-mail]:

If I have:

8.24
7.56
1.11

as doubles, how do I easily get just the ".24, .56, .11" ? Thanks,

Well, there's no shortage of methods I'm sure, but the following would work:

double value, fraction;

value = 8.24;
fraction = value - (int) value;

Pete

[Fred Mellender]

assuming the number is positive:

double x = myNumber-Math.Floor(myNumber);

"Yin99" <[e-mail] in message
news:84**********************************@b15g2000 hsa.googlegroups.com...

If I have:

8.24
7.56
1.11

as doubles, how do I easily get just the ".24, .56, .11" ? Thanks,

Yin

[=?iso-8859-1?Q?Lo=EFc_Berthollet?=]

I have a doubt : what occurs when <doublevalue exceeds <intcapacities ?
Loïc

"Peter Duniho" <Np*********@NnOwSlPiAnMk.coma écrit dans le message de
groupe de discussion : 2007112011594443658-NpOeStPeAdM@NnOwSlPiAnMkcom...

On 2007-11-20 11:44:43 -0800, Yin99 <[e-mail]:

>If I have:

8.24
7.56
1.11

as doubles, how do I easily get just the ".24, .56, .11" ? Thanks,

Well, there's no shortage of methods I'm sure, but the following would
work:

double value, fraction;

value = 8.24;
fraction = value - (int) value;

Pete

[Peter Duniho]

On Sun, 30 Dec 2007 01:02:55 -0800, Loïc Berthollet
<lo*************@hotmail.comwrote:

I have a doubt : what occurs when <doublevalue exceeds <int>
capacities ?

Well, using the code you quoted you get an exception, of course.

If you have numbers that aren't within the range of int, you'll have to
use some other method obviously. For example, the Math.Truncate() method
can be used in place of casting to int.

Pete

[=?ISO-8859-1?Q?Arne_Vajh=F8j?=]

Loïc Berthollet wrote:

"Peter Duniho" <Np*********@NnOwSlPiAnMk.coma écrit dans le message de
groupe de discussion : 2007112011594443658-NpOeStPeAdM@NnOwSlPiAnMkcom...

>On 2007-11-20 11:44:43 -0800, Yin99 <[e-mail]:

>>If I have:

8.24
7.56
1.11

as doubles, how do I easily get just the ".24, .56, .11" ?

Well, there's no shortage of methods I'm sure, but the following would
work:

double value, fraction;

value = 8.24;
fraction = value - (int) value;

I have a doubt : what occurs when <doublevalue exceeds <intcapacities ?

You should use:

fraction = value - (long) value;

or nicer:

fraction = value - Math.Floor(value);

to avoid the problem.

Arne

[=?UTF-8?B?QXJuZSBWYWpow7hq?=]

Peter Duniho wrote:

On Sun, 30 Dec 2007 01:02:55 -0800, Loïc Berthollet
<lo*************@hotmail.comwrote:

>I have a doubt : what occurs when <doublevalue exceeds <int>
capacities ?

Well, using the code you quoted you get an exception, of course.

Only if compiled with /checked+.

Else the results will become very "weird looking".

If you have numbers that aren't within the range of int, you'll have to
use some other method obviously. For example, the Math.Truncate()
method can be used in place of casting to int.

That is probably better than my suggestion of Math.Floor - most
people will prefer -0.24 instead og 0.76 when applied to -8.24 !

Arne