[LLVMdev] Getting the DSNode from a Pool Descriptor?

[Patrick Simmons]

Thanks for all your help so far.

My problem is that what I have are the pool descriptors, which I by 
traversing the uses of poolinit and accessing the first argument of each 
call.  I need to find the DSNode (in the original function) to which 
this pool descriptor corresponds.  The rub is that this pool descriptor 
of course does not exist except in the clone.

If I call getFuncInfo(), I get a NULL pointer.  getFuncInfoOrClone() 
returns the original function's FuncInfo structure, but this function's 
DSNode <-> PoolDescriptor mapping will not have my pool descriptor in it 
because my pool descriptor only existed in the clone.  I'm fine with the 
DSNode being the original function's DSNode -- in fact, I need that -- 
but I'm really at a loss as to how to get out of this catch-22.

--Patrick

On 03/19/10 04:10, Harmen van der Spek wrote:
> Hi Patrick,
>
> That's right. DSNodes are coupled to the original function. For function clones, you first need
> to get the original function, and then use the DSNode from that function. FuncInfo
> contains the information if a function is a clone and what the original function is.
>
> If you want to find the corresponding DSNode for some instruction, you must call
>
> PA::FuncInfo::MapValueToOriginal( value )
>
> Then you can get the DSNode from the DSGraph:
> 	
>      llvm::DSNodeHandle handle = _dsg->getNodeForValue(origVal);
>      DSNode *node = handle.getNode();
>
> origVal is obtained by calling MapValueToOriginal on a cloned Value.
>
> All those mappings are quite confusing. I've been thinking about splitting Pool allocation in two phases,
> one in which the clones are generated (which should be internal functions) and then, instead of maintaining all these mappings, just
> rerun Top-Down DSA on that result. In that case, it would be much easier to find DSNodes. Anyway, this was just
> a thought, I've not really tried anything like that.
>
> Harmen
>
>
> On Mar 19, 2010, at 9:14 AM, Patrick Alexander Simmons wrote:
>
>    
>> 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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