[llvm-dev] Selection DAG adding node question

[Ryan Taylor via llvm-dev]

So you want to create a new node, set new node's inputs the same as old
node's (along with the output?) and delete old node?

yes, this is possible.

On Wed, Oct 12, 2016 at 1:50 PM, SiliconValleyEmployee - <
hendrik.greving.samsung at gmail.com> wrote:

> Regardless of not being able to clone (which may or may not make sense),
> is selection DAG supporting (by the way, I am on llvm-3.6) what I am
> intending to do as it is right now?
>
> On Wed, Oct 12, 2016 at 10:29 AM, Ryan Taylor <ryta1203 at gmail.com> wrote:
>
>> It is my understanding that you cannot create a clone in the DAG. This is
>> due to common sub tree from what I understand.
>>
>> This is a problem for direct mem architectures, seems more and more
>> people are having issues with this but it can be worked around.
>>
>> LLVM assumes a LD/ST architecture.
>>
>> I'd love to hear otherwise.
>>
>> -Ryan
>>
>> On Wed, Oct 12, 2016 at 12:31 PM, SiliconValleyEmployee - via llvm-dev <
>> llvm-dev at lists.llvm.org> wrote:
>>
>>> I am having trouble adding a node to the selection DAG (e.g. during
>>> combine)
>>>
>>> E.g. node1 -> use1, use2
>>>
>>> Now if you add a node2, with node1 -> node2 with node2 number of output
>>> values equal node1 number of output values, then combine as well as e.g.
>>> promotion pass with replace all of node1's uses with node2, leaving node1
>>> dead. While this is kind of expected, does this mean
>>>
>>> a) always create new node (can you clone node, e.g. node1 above)?
>>> b) is this something that selection DAG is not intended for at all?
>>> c) sometimes it is not straight forward to create a new load, e.g. for
>>> loads, as DAG.getLoad will return "old" node if identical => a) can I clone
>>> a node?
>>>
>>> Thanks! Hendrik
>>>
>>> _______________________________________________
>>> LLVM Developers mailing list
>>> llvm-dev at lists.llvm.org
>>> http://lists.llvm.org/cgi-bin/mailman/listinfo/llvm-dev
>>>
>>>
>>
>
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