On Monday, 6 May 2013 at 18:46:56 UTC, Andrei Alexandrescu wrote:
Any concurrent operation (in this case read from one thread and
write from another) requires a handshake between threads, most
often in the form of an release write coupled with an acquire
read. Whenever the handshake is absent but concurrent
operations on shared memory do occur, the code is broken. The
beauty of the TLS-based pattern is that in the steady state
there's no need for a shared read and handshake.
Andrei
Hmm, are you referring to the same lack of a barrier that the
others are also referring to?
As far as I can see, there shouldn't be a need for any other
handshake in this example.
As long as the object is fully initialized before _static is
written to (easy enough with just a memory barrier), there is no
penalty for subsequent reads whatsoever.
Right?
