Re: What does MsgA(*) = MsgB(*) do?

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October 25, 2007 at 1:15 PM #18273
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    Yep, I think Tomas is correct. 

    The key is… when you do a transform inside an Orchestration the contaxt isn’t copied for you.  So the 2nd line (from above) would be needed to copy all the context data over if you need it inside the 2nd message.

    This is not the same behavior as mapping on the Receive Port.

    Stephen W. Thomas