bounce Posted November 1, 2007 Report Share Posted November 1, 2007 I apologize if any of my questions are answered in the documentation, but I couldn't find it anywhere. What is the behaviour supposed to be for the following C# code: ChartFX.WinForms.Chart graph = new ChartFX.WinForms.Chart(); for(int lcv = 0; lcv < 9; lcv++) { graph.Data.Y[0, lcv] = lcv; graph.Data.Y[1, lcv] = lcv; graph.Data.Y[2, lcv] = lcv; } graph.Data.Y[0, 4] = double.NaN; graph.Data.Y[1, 4] = double.PositiveInfinity; graph.Data.Y[2, 4] = double.NegativeInfinity; Should it act as if they were set to ChartFX.WinForms.DataValues.Hidden instead of the special double values? Speaking of which, I also have a few questions on ChartFX.WinForms.DataValues.Hidden: 1) Why define it as 1e108 instead of, say, NaN? 2) What is DataValues.InterpolateHidden initialized as? 3) If I'm setting both the X and Y value for a point, do I have to set both, either one, or a specific one to Hidden so that it won't be shown? Is there an advantage to doing it one way over the other? Thank you, Jerry Link to comment Share on other sites More sharing options...
Frank Posted November 1, 2007 Report Share Posted November 1, 2007 double.NaN. double.PositiveInfinity,double.NegativeInfinity can NOT be passed as values. > 1) Why define it as 1e108 instead of, say, NaN? Performance reasons > 2) What is DataValues.InterpolateHidden initialized as? false > 3) If I'm setting both the X and Y value for a point, do I have to set both, either one You should set them both hidden. However, if only the Y-Value is hidden the point will not show. Link to comment Share on other sites More sharing options...
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