Foo *f = ::new((void*)0) Foo();

Is that guaranteed by the standard to just do nothing (except assign f
to (Foo*)0 of course)?

5.3.4s13 implies this because this placement new (18.4.1.3) has an
empty throw() exception specification but I want to make sure because
I've heard nobody mention this before (and I searched google a bit
before I posted).

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Me wrote:
> Foo *f = ::new((void*)0) Foo();
> Is that guaranteed by the standard to just do nothing (except assign f
> to (Foo*)0 of course)?

Yes, according to the standard that is true.  As you mentioned, 5.3.4
thus results in:
-the placement new allocation function returns null
-since it has a throw() clause the 0 is interpreted as a bad_alloc
-thus the initliazer is not called
-and the new operator returns 0


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