[aarch64-port-dev ] [aarch64-port-dev] [10] RFR: 8184049 : Matching rule for ubfiz

[Daniel Stewart]

Felix,

I'm not sure I understand the check here and exactly what you are
proposing. It appears to me that this check is for an arithmetic right
shift, not a left shift, which is what we have my proposed solution. I
could be wrong as I am not familiar with GCC, but I can't make sense of the
check if the shift is a left shift. (Perhaps GCC is using the shift value
as if it were encoded for a ubfm instruction?) Could you explain a bit more
about what exactly you'd like to see?

Thanks,
Daniel

On Tue, Aug 1, 2017 at 11:10 AM, Felix Yang <felix.yang at linaro.org> wrote:

> LGTM, but I am not an official reviewer.
>
> Here is how the latest GCC compiler do the check, which is more readable
> to me:
>
> /* Return true if the mask and a shift amount from an RTX of the form
>    (x << SHFT_AMNT) & MASK are valid to combine into a UBFIZ instruction of
>    mode MODE.  See the *andim_ashift<mode>_bfiz pattern.  */
>
> bool
> aarch64_mask_and_shift_for_ubfiz_p (machine_mode mode, rtx mask, rtx
> shft_amnt)
> {
>   return CONST_INT_P (mask) && CONST_INT_P (shft_amnt)
> && INTVAL (shft_amnt) < GET_MODE_BITSIZE (mode)
> && exact_log2 ((INTVAL (mask) >> INTVAL (shft_amnt)) + 1) >= 0
> && (INTVAL (mask) & ((1 << INTVAL (shft_amnt)) - 1)) == 0;
> }
>
> Thanks,
> Felix
>
> On 31 July 2017 at 20:22, Daniel Stewart <daniel.stewart at linaro.org>
> wrote:
>
>> Updated webrev to address Felix's issue regarding the format.
>>
>> Please review and let me know of any changes/fixes.
>>
>> http://cr.openjdk.java.net/~njian/8184049/webrev.02/
>>
>> Daniel
>>
>> On Fri, Jul 28, 2017 at 1:37 PM, Andrew Haley <aph at redhat.com> wrote:
>>
>>> On 27/07/17 21:49, Daniel Stewart wrote:
>>> > public void run(String [] args) {
>>> >         long sum = this.sum | (1 << 27);
>>> >         int n = (int)sum;
>>> >         n = xorshift32(n);
>>> >         for (int i = 0; i < 1000; i++) {
>>> >             //System.out.println((int)Math.random());
>>> >             n += testI(n);
>>> >             n += testI2(n);
>>> >             n += testI3(n);
>>> >             n += testI4(n);
>>> >             n += testI5(n);
>>> >         }
>>> >         long n1 = (long)n;
>>> >         for (int i = 0; i < 1000; i++) {
>>> >             n1 += testL(n1);
>>> >             n1 += testL2(n1);
>>> >             n1 += testL3(n1);
>>> >             n1 += testL4(n1);
>>> >             n1 += testConv(n1);
>>> >         }
>>> >         short n2 = (short)n;
>>> >         for (int i = 0; i < 1000; i++) {
>>> >             n2 += testConv2((short)n2);
>>> >         }
>>> >         n1 += n2;
>>> >         for (int i = 0; i< 1000; i++) {
>>> >             n1 += testConv3(n);
>>>
>>> This loop doesn't test the intrinsic.  I added
>>>
>>>                n = xorshift32(n);
>>>
>>> here.
>>>
>>> >         }
>>> >         this.sum += sum ^ n1;
>>>
>>>
>>> --
>>> Andrew Haley
>>> Java Platform Lead Engineer
>>> Red Hat UK Ltd. <https://www.redhat.com>
>>> EAC8 43EB D3EF DB98 CC77 2FAD A5CD 6035 332F A671
>>>
>>
>>
>>
>> --
>> Daniel Stewart
>>
>
>


-- 
Daniel Stewart