On Mon, Mar 25, 2013 at 6:33 PM, Wei Mi <w...@google.com> wrote:
>> I am trying to figure out a way not to lose the opportunity when shift
>> truncation is not combined in a bit test pattern. Can we keep the
>> explicit truncation in RTL, but generate truncation code in assembly?
>> Then only shift truncation which not combined in a bit test
>> pattershift truncationn will happen.
>>
>> (define_insn "*<shift_insn_and><mode>"
>> [(set (match_operand:SWI48 0 "nonimmediate_operand" "=rm")
>> (any_shiftrt:SWI48
>> (match_operand:SWI48 1 "nonimmediate_operand" "0")
>> (subreg:QI
>> (and:SI
>> (match_operand:SI 2 "nonimmediate_operand" "c")
>> (match_operand:SI 3 "const_int_operand" "n")) 0)))
>> (clobber (reg:CC FLAGS_REG))]
>> "ix86_binary_operator_ok (<CODE>, <MODE>mode, operands)"
>> {
>> if ((INTVAL (operands[3]) & (GET_MODE_BITSIZE (<MODE>mode)-1))
>> == GET_MODE_BITSIZE (<MODE>mode)-1)
>> return "and\t{%3, %2|%2, %3}\n\r shift\t{%b2, %0|%0, %b2}";
>> else
>> "shift\t{%2, %0|%0, %2}";
>> }
>
> Sorry, rectify a mistake:
>
> {
> if ((INTVAL (operands[3]) & (GET_MODE_BITSIZE (<MODE>mode)-1))
> == GET_MODE_BITSIZE (<MODE>mode)-1)
> return "shift\t{%2, %0|%0, %2}";
> else
> return "and\t{%3, %2|%2, %3}\n\r shift\t{%b2, %0|%0, %b2}";
> }
I'm not sure the existing patterns are wrong because SHIFT_COUNT_TRUNCATED
is false for x86 AFAIK, exactly because of the bit-test vs. shift instruction
differences. So there is no inconsistency. The i386 backend seems to
try to follow my suggestion as if SHIFT_COUNT_TRUNCATED didn't exist
(well, it's false, so it technically doesn't exist for i386) and recognizes
the shift with truncate with the *<shift_insn><mode>3_mask splitter.
But I'm not sure why it bothers to do it with a splitter instead of just
with a define_insn? Because the split code,
[(parallel [(set (match_dup 0)
(any_shiftrt:SWI48 (match_dup 1) (match_dup 2)))
(clobber (reg:CC FLAGS_REG))])]
is wrong and could be combined into a bit-test instruction. No?
That is, why not have define_insn variants for shift instructions with
explicit truncation?
Richard.
> Thanks,
> Wei.
