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From: Alexander Terekhov (terekhov_at_[hidden])
Date: 2003-05-23 09:41:38

Larry Evans wrote:
> weighted reference counts eliminate around half the synchronization.
> This is because, AFAICT, the only time synchronization is needed is
> when a smart weighted pointer releases it's pointee, but not
> when it acquires a new pointee (because the reference count is
> gotten from the other, or source, smart weighted pointer).

I don't understand. Please elaborate (explain with more details...
or simply drop a link ;-) ).


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