[LLVMdev] Getting the DSNode from a Pool Descriptor?

[Patrick Alexander Simmons]

Harmen, your suggestion of inverting the mapping almost worked (and 
Andrew was correct that the function I need is the same as the one in 
which poolinit appears).  Unfortunately, it appears that this mapping 
only considers the original function and not any of its clones.  Since 
the pool descriptor in question may very well only exist in a clone, I 
can't use this.  Is there another way?

Thanks,
--Patrick

Harmen van der Spek wrote:
> You might want to have a look at PoolAllocate.h.
>
> Per function, a PA::FuncInfo structure keeps track of all DSNodes that 
> should be pool allocated. ArgNodes contains pool arguments, NodesToPA
> contains nodes that are locally pool allocated and thus initialized 
> using poolinit.
>
> PoolDescriptors contains a mapping from DSNodes to pool descriptors, and
> you could easily invert this mapping.
>
> Finding a corresponding DSNode which is complete is not uniquely 
> determined. For example, if a function F uses a pool, but its DSNode
> is incomplete, it might be called from two different function G and H,
> which both have a complete DSNode that maps to the DSNode in F.
>
> You can assume that if a function is cloned, so that its DSNodes
> are pool allocated, those DSNodes originate from a complete DSNode
> somewhere higher in the call chain.
>
> Per function, a pool-allocated version can be generated. After that,
> function calls are rewritten to call the pool allocated version. This is 
> done in TransformFunctionBody.cpp. by calling TransformBody from 
> PoolAllocate.cpp.
>
>
>
> Harmen
>
>
>
> Patrick Alexander Simmons wrote:
>   
>> I figure (hopefully correctly) that I can iterate over all pool 
>> descriptors in a program by iterating over all users of poolinit and 
>> looking at the first argument.  However, once I have a pool descriptor, 
>> I need to get its corresponding DSNode in the function in which it is 
>> complete (or in the global graph if it is a global).  How do I do this?
>>
>> Thanks,
>> --Patrick
>>     
>
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