Fold BIT_FIELD_REF of a reference

[Marc Glisse]

On Thu, 4 Apr 2013, Richard Biener wrote:

> On Thu, Apr 4, 2013 at 10:53 AM, Marc Glisse <marc.glisse@inria.fr> wrote:
>> On Thu, 4 Apr 2013, Richard Biener wrote:
>>
>>>>>> +      if ((handled_component_p (arg0) || TREE_CODE (arg0) == MEM_REF)
>>>>>
>>>>>
>>>>>
>>>>> This check means the optimization is not performed for
>>>>> BIT_FIELD_REF[a, *, CST] which I see no particularly good reason for.
>>>>
>>>>
>>>>
>>>> Er, are you trying to get rid of all BIT_FIELD_REFs? Why would you want
>>>> to
>>>> replace them with a MEM_REF? I actually think my patch already replaces
>>>> too
>>>> many.
>>>
>>>
>>> Yes, when I filed the bug I was working on bitfield lowering and the only
>>> BIT_FIELD_REFs that would survive would be bitfield extracts from
>>> registers.
>>
>>
>> Can't a vector (not in memory) count as a register?
>
> Sure.  A vector not in memory is a register.

So we need to have some test whether something is "a register" and not 
create a MEM_REF in that case.

>>> Thus, BIT_FIELD_REFs on memory would be lowered as
>>>
>>>  reg_2 = MEM[ ... ];
>>>  res_3 = BIT_FIELD_REF [reg_2, ...];
>>>
>>> with an appropriately aligned / bigger size memory MEM.
>>>
>>> As a first step I wanted to lower all BIT_FIELD_REFs that can be expressed
>>> directly as memory access (byte-aligned and byte-size) to regular memory
>>> accesses.
>>
>>
>> But the transformation on BIT_FIELD_REF[A,...] will take the address of A
>> even if A is not something that is ok with having its address taken.
>
> I'd like to see a case where this happens.

In the vector lowering pass, op0 is a SSA_NAME:

typedef double vec __attribute__((vector_size(64)));
vec f(vec x){
   return x+x;
}

>> I am probably missing something. Looking in tree-flow-inline.c, for
>> MEM_REF[a,...]:
>>
>>>        case MEM_REF:
>>>          {
>>>            tree base = TREE_OPERAND (exp, 0);
>>>            if (valueize
>>>                && TREE_CODE (base) == SSA_NAME)
>>>              base = (*valueize) (base);
>>
>>
>> valueize is 0.
>>
>>>            /* Hand back the decl for MEM[&decl, off].  */
>>>            if (TREE_CODE (base) == ADDR_EXPR)
>>
>>
>> not the case here.
>>
>>>              {
>>>                if (!integer_zerop (TREE_OPERAND (exp, 1)))
>>>                  {
>>>                    double_int off = mem_ref_offset (exp);
>>>                    gcc_assert (off.high == -1 || off.high == 0);
>>>                    byte_offset += off.to_shwi ();
>>>                  }
>>>                exp = TREE_OPERAND (base, 0);
>>>              }
>>>            goto done;
>>
>>
>> it returns a, which afaiu is an address.
>> For MEM_REF[&b] it does return b.
>
> No, it returns 'exp' which is still MEM_REF[&b].

(Actually, for MEM_REF[a] it returns the input unchanged, and for 
MEM_REF[&b] it returns b)
Thanks! I had completely misread that.

-- 
Marc Glisse