Tejas Belagod <[email protected]> writes:
> Richard Sandiford wrote:
>> Tejas Belagod <[email protected]> writes:
>>> + /* This is big-endian-safe because the elements are kept in target
>>> + memory order. So, for eg. PARALLEL element value of 2 is the same in
>>> + either endian-ness. */
>>> + if (GET_CODE (src) == VEC_SELECT
>>> + && REG_P (XEXP (src, 0)) && REG_P (dst)
>>> + && REGNO (XEXP (src, 0)) == REGNO (dst))
>>> + {
>>> + rtx par = XEXP (src, 1);
>>> + int i;
>>> +
>>> + for (i = 0; i < XVECLEN (par, 0); i++)
>>> + {
>>> + rtx tem = XVECEXP (par, 0, i);
>>> + if (!CONST_INT_P (tem) || INTVAL (tem) != i)
>>> + return 0;
>>> + }
>>> + return 1;
>>> + }
>>> +
>>
>> I think for big endian it needs to be:
>>
>> INTVAL (tem) != i + base
>>
>> where base is something like:
>>
>> int base = GET_MODE_NUNITS (GET_MODE (XEXP (src, 0))) - XVECLEN (par, 0);
>>
>> E.g. a big-endian V4HI looks like:
>>
>> msb lsb
>> 0000111122223333
>>
>> and shortening it to say V2HI only gives the low 32 bits:
>>
>> msb lsb
>> 22223333
>
> But, in this case we want
>
> msb lsb
> 00001111
It depends on whether the result occupies a full register or not.
I was thinking of the case where it didn't, but I realise now you were
thinking of the case where it did. And yeah, my suggestion doesn't
cope with that...
> I was under the impression that the const vector parallel for vec_select
> represents the element indexes of the array in memory order.
>
> Therefore, in bigendian,
>
> msb lsb
> 0000 1111 2222 3333
> element a[0] a[1] a[2] a[3]
>
> and in littleendian
>
> msb lsb
> 3333 2222 1111 0000
> element a[3] a[2] a[1] a[0]
Right. But if an N-bit value is stored in a register, it's assumed to
occupy the lsb of the register and the N-1 bits above that. The other
bits in the register are don't-care.
E.g., leaving vectors to one side, if you have:
(set (reg:HI N) (truncate:SI (reg:SI N)))
on a 32-bit !TRULY_NOOP_TRUNCATION target, it shortens like this:
msb lsb
01234567
VVVV
xxxx4567
rather than:
msb lsb
01234567
VVVV
0123xxxx
for both endiannesses. The same principle applies to vectors.
The lsb of the register is always assumed to be significant.
So maybe the original patch was correct for partial-register and
full-register results on little-endian, but only for full-register
results on big-endian.
Thanks,
Richard
