| |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| |
as well as sext(C + x + ...) -> (D + sext(C-D + x + ...))<nuw><nsw>
similar to the equivalent transformation for zext's
if the top level addition in (D + (C-D + x * n)) could be proven to
not wrap, where the choice of D also maximizes the number of trailing
zeroes of (C-D + x * n), ensuring homogeneous behaviour of the
transformation and better canonicalization of such AddRec's
(indeed, there are 2^(2w) different expressions in `B1 + ext(B2 + Y)` form for
the same Y, but only 2^(2w - k) different expressions in the resulting `B3 +
ext((B4 * 2^k) + Y)` form, where w is the bit width of the integral type)
This patch generalizes sext(C1 + C2*X) --> sext(C1) + sext(C2*X) and
sext{C1,+,C2} --> sext(C1) + sext{0,+,C2} transformations added in
r209568 relaxing the requirements the following way:
1. C2 doesn't have to be a power of 2, it's enough if it's divisible by 2
a sufficient number of times;
2. C1 doesn't have to be less than C2, instead of extracting the entire
C1 we can split it into 2 terms: (00...0XXX + YY...Y000), keep the
second one that may cause wrapping within the extension operator, and
move the first one that doesn't affect wrapping out of the extension
operator, enabling further simplifications;
3. C1 and C2 don't have to be positive, splitting C1 like shown above
produces a sum that is guaranteed to not wrap, signed or unsigned;
4. in AddExpr case there could be more than 2 terms, and in case of
AddExpr the 2nd and following terms and in case of AddRecExpr the
Step component don't have to be in the C2*X form or constant
(respectively), they just need to have enough trailing zeros,
which in turn could be guaranteed by means other than arithmetics,
e.g. by a pointer alignment;
5. the extension operator doesn't have to be a sext, the same
transformation works and profitable for zext's as well.
Apparently, optimizations like SLPVectorizer currently fail to
vectorize even rather trivial cases like the following:
double bar(double *a, unsigned n) {
double x = 0.0;
double y = 0.0;
for (unsigned i = 0; i < n; i += 2) {
x += a[i];
y += a[i + 1];
}
return x * y;
}
If compiled with `clang -std=c11 -Wpedantic -Wall -O3 main.c -S -o - -emit-llvm`
(!{!"clang version 7.0.0 (trunk 337339) (llvm/trunk 337344)"})
it produces scalar code with the loop not unrolled with the unsigned `n` and
`i` (like shown above), but vectorized and unrolled loop with signed `n` and
`i`. With the changes made in this commit the unsigned version will be
vectorized (though not unrolled for unclear reasons).
How it all works:
Let say we have an AddExpr that looks like (C + x + y + ...), where C
is a constant and x, y, ... are arbitrary SCEVs. Let's compute the
minimum number of trailing zeroes guaranteed of that sum w/o the
constant term: (x + y + ...). If, for example, those terms look like
follows:
i
XXXX...X000
YYYY...YY00
...
ZZZZ...0000
then the rightmost non-guaranteed-zero bit (a potential one at i-th
position above) can change the bits of the sum to the left (and at
i-th position itself), but it can not possibly change the bits to the
right. So we can compute the number of trailing zeroes by taking a
minimum between the numbers of trailing zeroes of the terms.
Now let's say that our original sum with the constant is effectively
just C + X, where X = x + y + .... Let's also say that we've got 2
guaranteed trailing zeros for X:
j
CCCC...CCCC
XXXX...XX00 // this is X = (x + y + ...)
Any bit of C to the left of j may in the end cause the C + X sum to
wrap, but the rightmost 2 bits of C (at positions j and j - 1) do not
affect wrapping in any way. If the upper bits cause a wrap, it will be
a wrap regardless of the values of the 2 least significant bits of C.
If the upper bits do not cause a wrap, it won't be a wrap regardless
of the values of the 2 bits on the right (again).
So let's split C to 2 constants like follows:
0000...00CC = D
CCCC...CC00 = (C - D)
and represent the whole sum as D + (C - D + X). The second term of
this new sum looks like this:
CCCC...CC00
XXXX...XX00
----------- // let's add them up
YYYY...YY00
The sum above (let's call it Y)) may or may not wrap, we don't know,
so we need to keep it under a sext/zext. Adding D to that sum though
will never wrap, signed or unsigned, if performed on the original bit
width or the extended one, because all that that final add does is
setting the 2 least significant bits of Y to the bits of D:
YYYY...YY00 = Y
0000...00CC = D
----------- <nuw><nsw>
YYYY...YYCC
Which means we can safely move that D out of the sext or zext and
claim that the top-level sum neither sign wraps nor unsigned wraps.
Let's run an example, let's say we're working in i8's and the original
expression (zext's or sext's operand) is 21 + 12x + 8y. So it goes
like this:
0001 0101 // 21
XXXX XX00 // 12x
YYYY Y000 // 8y
0001 0101 // 21
ZZZZ ZZ00 // 12x + 8y
0000 0001 // D
0001 0100 // 21 - D = 20
ZZZZ ZZ00 // 12x + 8y
0000 0001 // D
WWWW WW00 // 21 - D + 12x + 8y = 20 + 12x + 8y
therefore zext(21 + 12x + 8y) = (1 + zext(20 + 12x + 8y))<nuw><nsw>
This approach could be improved if we move away from using trailing
zeroes and use KnownBits instead. For instance, with KnownBits we could
have the following picture:
i
10 1110...0011 // this is C
XX X1XX...XX00 // this is X = (x + y + ...)
Notice that some of the bits of X are known ones, also notice that
known bits of X are interspersed with unknown bits and not grouped on
the rigth or left.
We can see at the position i that C(i) and X(i) are both known ones,
therefore the (i + 1)th carry bit is guaranteed to be 1 regardless of
the bits of C to the right of i. For instance, the C(i - 1) bit only
affects the bits of the sum at positions i - 1 and i, and does not
influence if the sum is going to wrap or not. Therefore we could split
the constant C the following way:
i
00 0010...0011 = D
10 1100...0000 = (C - D)
Let's compute the KnownBits of (C - D) + X:
XX1 1 = carry bit, blanks stand for known zeroes
10 1100...0000 = (C - D)
XX X1XX...XX00 = X
--- -----------
XX X0XX...XX00
Will this add wrap or not essentially depends on bits of X. Adding D
to this sum, however, is guaranteed to not to wrap:
0 X
00 0010...0011 = D
sX X0XX...XX00 = (C - D) + X
--- -----------
sX XXXX XX11
As could be seen above, adding D preserves the sign bit of (C - D) +
X, if any, and has a guaranteed 0 carry out, as expected.
The more bits of (C - D) we constrain, the better the transformations
introduced here canonicalize expressions as it leaves less freedom to
what values the constant part of ((C - D) + x + y + ...) can take.
Reviewed By: mzolotukhin, efriedma
Differential Revision: https://reviews.llvm.org/D48853
llvm-svn: 337943
|
| |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| |
This was done through the aid of a terrible Perl creation. I will not
paste any of the horrors here. Suffice to say, it require multiple
staged rounds of replacements, state carried between, and a few
nested-construct-parsing hacks that I'm not proud of. It happens, by
luck, to be able to deal with all the TCL-quoting patterns in evidence
in the LLVM test suite.
If anyone is maintaining large out-of-tree test trees, feel free to poke
me and I'll send you the steps I used to convert things, as well as
answer any painful questions etc. IRC works best for this type of thing
I find.
Once converted, switch the LLVM lit config to use ShTests the same as
Clang. In addition to being able to delete large amounts of Python code
from 'lit', this will also simplify the entire test suite and some of
lit's architecture.
Finally, the test suite runs 33% faster on Linux now. ;]
For my 16-hardware-thread (2x 4-core xeon e5520): 36s -> 24s
llvm-svn: 159525
|
