Quote:
Originally Posted by Inverness
The original object is still executing the code, not the object in the with () block, so its not circumventing it.
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thisor.var = thisr.var; or thisr.var = thisor.var; in a with block
would be circumventing it.
e: I think you may have misunderstood Stefan, too.
The last post wasn't meant to single you out, but after I posted, I realized how much of a pain it would have been to get my point understood.