Daniel Lehman writes:
> I'll admit that the upper 32 bits being set was because of a quirk in
> our C++ code, but similar values passed to Windows worked
This would need some test cases.
--
Alexandre Julliard
julli...@winehq.org
> It can't, the code explicitly casts handle to unsigned before the
> shift.
It's cast to a 64-bit unsigned, so the upper 32 bits are kept for the shift
One of the pointers passed in was: 0xff26a8
after the shift, it became:
0xff26a8 => 0x3fc9aa
When assigned to the 32
Daniel Lehman wrote:
> 0x26a8 becomes 0xc9aa instead of just 0x9aa
It can't, the code explicitly casts handle to unsigned before the shift.
--
Dmitry.
