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Re: Fold BIT_FIELD_REF of a reference


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


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