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Re: Fold BIT_FIELD_REF of a reference


On Thu, Apr 4, 2013 at 12:11 PM, Marc Glisse <marc.glisse@inria.fr> wrote:
> 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.

Ah, of course.  Easy on the gimple side - just check is_gimple_reg ().

>
>>>> 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


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