On Thu, 4 Apr 2013, Richard Biener wrote:
On Thu, Apr 4, 2013 at 10:53 AM, Marc Glisse <marc.gli...@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
