ODP.net Oracle Decimal Number precision problem when filling a dataset. Exception: Arithmetic operation resulted in an overflow

Question

I am working in c# .net 2 (Visual Studio 2005 SP1) attempting to fill a dataset with the results from a select * from table from an Oracle10g database. The .net framework, IDE and database cannot be changed at this client site.

I'm connecting using the ODP.net provider the dll version is 2.102.2.20

When I run the fill command I get an Exception:

Arithmetic operation resulted in an overflow

Also if I attempt to view the offending column in the Visual Studio designer (Show Table Data) I get for every row for this column in the table. The code works perfectly if my query selects other columns with integers for example omitting this column.

The column in question looks fine when I view it in the database from Toad, data looks like:

919.742866695572

I need the precision as it's required for a monte carlo simulation.

If instead of using a data adapter to fill the datatable I use a datareader and call dataReader.getValue(columnIndex) I get the same error but if I call dataReader.GetOracleDecimal(columnIndex) then I get the result I am looking for, no error.

I would rather use data adapter and filling a dataset (note these are untyped datasets as I couldn't get autogenerated strongly typed datasets to work from an oracle db). I don't want to use datareader and walk through the results (pick out the column values) as I am trying to write this as a generic method to work for many scenarios regardless of number of columns, index of decimal columns that would require specific get calls by datatype.

Can anyone help? Can I use new versions of the ODP.net dlls to connect to the older Oracle10g database? am wondering if this will help.

Thanks

Answer 1

The problem is that the precision of the result value is too high to convert to a System.Decimal without some data loss. I forget the exact number of digits allowed, but it is around 18 or so. Is it acceptable to round() the result value to that many digits? In the example you gave, a round(MyColumn, 15) or so should be sufficient...