The 'you' here is Richi, which Richi is probably aware but maybe not the
rest of the list :')
On 09/06/2020 15:29, Andre Vieira (lists) wrote:
Hi,
So this is my rework of the code you sent me, I have not included the
'permute' code you included as I can't figure out what it is meant to
be doing. Maybe something to look at later.
I have also included three tests that show it working for some simple
cases and even a nested one.
Unfortunately it will not handle other simple cases where reassoc
doesn't put the reduction in the form of :
sum0 = a + b;
sum1 = c + sum0;
...
For instance a testcase I have been looking at is:
unsigned int u32_single_abs_sum (unsigned int * a, unsigned int * b)
{
unsigned int sub0 = a[0] - b[0];
unsigned int sub1 = a[1] - b[1];
unsigned int sub2 = a[2] - b[2];
unsigned int sub3 = a[3] - b[3];
unsigned int sum = sub0 + sub1;
sum += sub2;
sum += sub3;
return sum;
}
Unfortunately, the code that reaches slp will look like:
_1 = *a_10(D);
_2 = *b_11(D);
_3 = MEM[(unsigned int *)a_10(D) + 4B];
_4 = MEM[(unsigned int *)b_11(D) + 4B];
_5 = MEM[(unsigned int *)a_10(D) + 8B];
_6 = MEM[(unsigned int *)b_11(D) + 8B];
_7 = MEM[(unsigned int *)a_10(D) + 12B];
_8 = MEM[(unsigned int *)b_11(D) + 12B];
_28 = _1 - _2;
_29 = _3 + _28;
_30 = _29 - _4;
_31 = _5 + _30;
_32 = _31 - _6;
_33 = _7 + _32;
sum_18 = _33 - _8;
return sum_18;
Which doesn't have the format expected as I described above... I am
wondering how to teach it to support this. Maybe starting with your
suggestion of making plus_expr and minus_expr have the same hash, so
it groups all these statements together might be a start, but you'd
still need to 'rebalance' the tree somehow.... I need to give this a
bit more thought but I wanted to share what I have so far.
The code is severely lacking in comments for now btw...
Cheers,
Andre
