> On Mar 16, 2016, at 1:42 PM, Brent Royal-Gordon via swift-dev
> <[email protected]> wrote:
>
> This is damned clever.
Yes, I agree!
>> The MSB bit also becomes set if you increment or decrement a retain count
>> too far. That means we can implement the RR fast path with a single
>> conditional branch after the increment or decrement:
>
> You don't talk much about it, but I think you're suggesting that the strong
> and unowned refcounts would still be packed into the same fields. It's worth
> noting that your clever high-bit-indicates-slow-path trick would *not* work
> on the unowned refcount, though that matters less since it doesn't get
> checked as often.
My understanding is that there would not be an unowned refcount here (inline in
the object). The first time you make a weak reference to the object it gets
converted to have the out-of-band refcount structure, and the strong and
unowned refcounts both get stored there (in an undefined-by-this-proposal way).
>> // out-of-object refcount (MSB bits 0b10x)
>> // or refcount has overflowed (MSB bits 0b111)
>> // or refcount is constant (MSB bits 0b110)
>
> Is this going to work on 32-bit? Even if we assume that valid pointers don't
> start with 0b1, can we assume that they don't start with 1b01?
The out-of-object refcount flag here is only 2 bits (0b10 in the high bits), so
presumably you store the other 2 bits of the pointer in the low bits which
would always be zero, assuming word-aligned pointers. So in 32-bit you can just
left shift by 2 and still have the whole normal pointer length available.
- Greg
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