Simple way to copy or clone a DataRow?

I'm looking for a simple way to make a clone of a DataRow. Kind of like taking a snapshot of that Row and saving it. The values of original Row are then free to change but we still have another saved copy which doesn't change. Is this the correct way to do it?

DataRow Source, Destination;
// Assume we create some columns and fill them with values
Destination.ItemArray = Source.ItemArray;

Will this just set Snapshot's ItemArray reference to point to the one in Source or does it actually make a separate copy? Should I do this instead?

Destination.ItemArray = Source.ItemArray.Clone();

EDIT: I don't think the second code snippet actually compiles.

Solution 1:

You can use ImportRow method to copy Row from DataTable to DataTable with the same schema:

var row = SourceTable.Rows[RowNum];
DestinationTable.ImportRow(row);

Update:

With your new Edit, I believe:

var desRow = dataTable.NewRow();
var sourceRow = dataTable.Rows[rowNum];
desRow.ItemArray = sourceRow.ItemArray.Clone() as object[];

will work

Solution 2:

^{Note: cuongle's helfpul answer has all the ingredients, but the solution can be streamlined (no need for .ItemArray) and can be reframed to better match the question as asked.}

To create an (isolated) clone of a given System.Data.DataRow instance, you can do the following:

// Assume that variable `table` contains the source data table.

// Create an auxiliary, empty, column-structure-only clone of the source data table.
var tableAux = table.Clone();
// Note: .Copy(), by contrast, would clone the data rows also.

// Select the data row to clone, e.g. the 2nd one:
var row = table.Rows[1];

// Import the data row of interest into the aux. table.
// This creates a *shallow clone* of it.
// Note: If you'll be *reusing* the aux. table for single-row cloning later, call
//       tableAux.Clear() first.
tableAux.ImportRow(row);

// Extract the cloned row from the aux. table:
var rowClone = tableAux.Rows[0];

Note: Shallow cloning is performed, which works as-is with column values that are value type instances, but more work would be needed to also create independent copies of column values containing reference type instances (and creating such independent copies isn't always possible).