RE: [SDK]Double to long in c++ ?

["Kim Aldis" <XSI(at)kim-aldis.co.uk>]

 

(LONG)( 1.6 + 0.25) => 1                                                      

 

(LONG)(1.6 + 0.5) => 2

 

The correct round of 1.6 is 2, so adding 0.25 gives an incorrect round. Surely?

 

From: owner-xsi(at)Softimage.COM [mailto:owner-xsi(at)Softimage.COM] On Behalf Of Luc-Eric Rousseau
Sent: 07 June 2007 18:14
To: XSI(at)Softimage.COM
Subject: RE: [SDK]Double to long in c++ ?

 

I perfer to use 0.25, because if you're rendering in fields in XSI the time will be 1.0, 1.5, 2.0,

2.5.  but in fact any small espilon number is fine

 

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From: owner-xsi(at)Softimage.COM [mailto:owner-xsi(at)Softimage.COM] On Behalf Of Kim Aldis
Posted At: Thursday, June 07, 2007 1:07 PM
Posted To: xsi
Conversation: [SDK]Double to long in c++ ?
Subject: RE: [SDK]Double to long in c++ ?

Shouldn’t that be 0.5, not 0.25?

 

From: owner-xsi(at)Softimage.COM [mailto:owner-xsi(at)Softimage.COM] On Behalf Of Luc-Eric Rousseau
Sent: 07 June 2007 02:10
To: XSI(at)Softimage.COM
Subject: RE: [SDK]Double to long in c++ ?

 

these additionnal casts are not necessary because the entire _expression_ is being promoted to 'double' already, and you do need my +0.25 epsilon to control the truncation that hapens with the cast to 'long'

 

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From: owner-xsi(at)Softimage.COM [mailto:owner-xsi(at)Softimage.COM] On Behalf Of Wessam Al-Bahnasi
Posted At: Thursday, June 07, 2007 12:03 AM
Posted To: xsi
Conversation: [SDK]Double to long in c++ ?
Subject: RE: [SDK]Double to long in c++ ?

Or better yet:

 

double t = (double)T + (double)oOffset;
long id = (long)(t * 3.0 * (double)nbPoints);

 

Wessam Bahnassi

Microsoft DirectX MVP,

Programmer

Electronic Arts

--

'Talk is cheap because supply exceeds demand'

 

From: owner-xsi(at)Softimage.COM [mailto:owner-xsi(at)Softimage.COM] On Behalf Of Luc-Eric Rousseau
Sent: Wednesday, June 06, 2007 3:31 PM
To: XSI(at)Softimage.COM
Subject: RE: [SDK]Double to long in c++ ?

 

I don't really know the answer to your question, so I appologize if this is wrong, but the cast operator has higher precedence than the multiply. so it's going to take that 't', cast it to a long, and do the rest in longs.

I think you wanted to have another set of parenthesis like (long)(t*3*nbPoints)

 

in any case,

casting a double to long just like that is not a good idea, you should add an epsilon to compensate for any numerical innacuracies.

For example  :  long id = (long)(t*3*nbPoints + 0.25);

 

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From: owner-xsi(at)Softimage.COM [mailto:owner-xsi(at)Softimage.COM] On Behalf Of Ahmidou Lyazidi
Posted At: Wednesday, June 06, 2007 3:20 PM
Posted To: xsi
Conversation: [SDK]Double to long in c++ ?
Subject: [SDK]Double to long in c++ ?

Hi
In an Operator,we are trying to convert the datas passed by OperatorContex.GetTime to a long.The thing that is strange is when I log the value
it seems good, but when I mutiply it with a fixed value it seems to round it and give me 2 equal values for 2 differant frame
here is the code and the log:

long oOffset = oOp.GetParameterValue(L"TimeOffset");

CTime T = ctxt.GetTime();
Application().LogMessage(L"Current operator Time " + T.GetAsText());

double t = double(T) + oOffset;
long id = (long) t * 3 * nbPoints;