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Tagged pointers are a common optimization when implementing dynamic languages: take advantage of alignment requirements that mean the low two or three bits of a pointer will always be zero, and use them to store type information.

Suppose you're using the Boehm garbage collector, which basically works by looking at active data for things that look like pointers. Tagged pointers don't look like pointers, in the sense that their low bits are nonzero.

Is this a showstopper, i.e. do you have to ditch tagged pointers if you're using Boehm? Or does it have a way around this problem?

rwallace
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  • Why use Boehm for a language implementation? If you're advanced enough to think about saving a few bits of type information, you can surely write a hand-tailored GC that doesn't fall to this and is way smarter than Boehm for your language? Right? –  Jan 10 '11 at 19:43
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    In principle perhaps, but in the short term it's nontrivial to do better than Boehm (note that e.g. Mono used it until recently) and in the long term I'm aiming for interoperability with C (in which scenario Boehm is arguably a near optimal solution). – rwallace Jan 10 '11 at 19:48

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AFAIK Boehm can handle this with the right options. It is capable, at a small price, of detecting interior pointers. It is also possible to write your own scanning code. Basically there are probably enough hooks to handle just about anything.

I have written my own collector, it is precise on the heap and conservative on the stack. It does not touch C made pointers. For some applications it will be faster because it knows a lot about my language allocated objects and doesn't care about other stuff which is managed, say, using traditional C++ destructors.

However it isn't incremental or generational, and it doesn't handle threads as well (it's not smart enough to stop threads with signals). On the plus side, however, it doesn't require magic linkage techniques which Boehm does (to capture mallocs, etc). On the seriously minus side you can't put managed objects into unmanaged ones.

Yttrill
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