On Thu, 29 Sep 2011 15:23:18 -0400, Jonathan M Davis <jmdavisp...@gmx.com>
wrote:
On Thursday, September 29, 2011 15:05:56 Steven Schveighoffer wrote:
If all the data the calculated value depends on is immutable, then the
two
threads loading the value at the same time will be loading the same
value. If you're writing a 42 to an int from 2 threads, there is no
deadlock or race issue. Writing a 42 over a 42 does not cause any
problems.
An excellent point, but that's assuming that the data being used is all
immutable, and that particular stipulation was not given previously. But
if
that stipulation is there, then you're right. Otherwise, the locking is
still
needed.
Well, the object itself is immutable. All that is needed is to ensure any
static data used is also immutable.
Wait, we have that -- pure functions :)
So what if lazy initialization is allowed for immutable as long as the
function being assigned from is pure?
-Steve
