Puzzled at storeAppend. :-(

Hi, pals!

I am puzzled at storeAppend. At the end of the function,
storeSwapOut(e) is called. I think storeSwapOut is used to swap
an object from memory to disk. Does that mean when data is appended
to a StoreEntry by storeAppend, the StoreEntry is always finally swapped
out?

I do not think so. I think storeSwapOutAble will judge whether the
StoreEntry will be swapped out. Am I correct?

But I do not understand how storeSwapOutAble judge whether the StoreEntry
will be written to disk.

Best regards,
George Ma
Received on Tue Apr 23 2002 - 05:52:58 MDT