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[x86] Rewrite a core part of the new vector shuffle lowering to handle 

one pesky test case correctly.

This test case caused the old code to infloop occilating between solving
the low-half and the high-half. The 'side balancing' part of
single-input v8 shuffle lowering didn't handle the one pattern which can
cause it to occilate. Fortunately the fuzz testing found this case.
Unfortuately it was *terrible* to handle. I'm really sorry for the
amount and density of the code here, I'd love suggestions on how to
simplify it. I feel like there *must* be a simpler form here, but after
a lot of days I've not found it. This is the only one I've found that
even works. I've added the one pesky test case along with some nice
comments explaining the core problem that we have to solve here.

So far this has survived approximately 32k test cases. More strenuous
fuzzing commencing.

llvm-svn: 215519
This commit is contained in:
Chandler Carruth 2014-08-13 01:25:45 +00:00
parent bb853c79c0
commit b7eda21bb0
2 changed files with 141 additions and 22 deletions
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144
llvm/lib/Target/X86/X86ISelLowering.cpp
View File

@ -7327,22 +7327,126 @@ static SDValue lowerV8I16SingleInputVectorShuffle(
// Input: [a, b, c, d, e, f, g, h] -PSHUFD[0,2,1,3]-> [a, b, e, f, c, d, g, h]
// Mask: [0, 1, 2, 7, 4, 5, 6, 3] -----------------> [0, 1, 4, 7, 2, 3, 6, 5]
//
// Before we had 3-1 in the low half and 3-1 in the high half. Afterward, 2-2
// and 2-2.
auto balanceSides = [&](ArrayRef<int> ThreeInputs, int OneInput,
int ThreeInputHalfSum, int OneInputHalfOffset) {
// However in some very rare cases we have a 1-into-3 or 3-into-1 on one half
// and an existing 2-into-2 on the other half. In this case we may have to
// pre-shuffle the 2-into-2 half to avoid turning it into a 3-into-1 or
// 1-into-3 which could cause us to cycle endlessly fixing each side in turn.
// Fortunately, we don't have to handle anything but a 2-into-2 pattern
// because any other situation (including a 3-into-1 or 1-into-3 in the other
// half than the one we target for fixing) will be fixed when we re-enter this
// path. We will also combine away any sequence of PSHUFD instructions that
// result into a single instruction. Here is an example of the tricky case:
//
// Input: [a, b, c, d, e, f, g, h] -PSHUFD[0,2,1,3]-> [a, b, e, f, c, d, g, h]
// Mask: [3, 7, 1, 0, 2, 7, 3, 5] -THIS-IS-BAD!!!!-> [5, 7, 1, 0, 4, 7, 5, 3]
//
// This now has a 1-into-3 in the high half! Instead, we do two shuffles:
//
// Input: [a, b, c, d, e, f, g, h] PSHUFHW[0,2,1,3]-> [a, b, c, d, e, g, f, h]
// Mask: [3, 7, 1, 0, 2, 7, 3, 5] -----------------> [3, 7, 1, 0, 2, 7, 3, 6]
//
// Input: [a, b, c, d, e, g, f, h] -PSHUFD[0,2,1,3]-> [a, b, e, g, c, d, f, h]
// Mask: [3, 7, 1, 0, 2, 7, 3, 6] -----------------> [5, 7, 1, 0, 4, 7, 5, 6]
//
// The result is fine to be handled by the generic logic.
auto balanceSides = [&](ArrayRef<int> AToAInputs, ArrayRef<int> BToAInputs,
ArrayRef<int> BToBInputs, ArrayRef<int> AToBInputs,
int AOffset, int BOffset) {
assert(AToAInputs.size() == 3 || AToAInputs.size() == 1 &&
"Must call this with A having 3 or 1 inputs from the A half.");
assert(BToAInputs.size() == 1 || BToAInputs.size() == 3 &&
"Must call this with B having 1 or 3 inputs from the B half.");
assert(AToAInputs.size() + BToAInputs.size() == 4 &&
"Must call this with either 3:1 or 1:3 inputs (summing to 4).");
// Compute the index of dword with only one word among the three inputs in
// a half by taking the sum of the half with three inputs and subtracting
// the sum of the actual three inputs. The difference is the remaining
// slot.
int DWordA = (ThreeInputHalfSum -
std::accumulate(ThreeInputs.begin(), ThreeInputs.end(), 0)) /
2;
int DWordB = OneInputHalfOffset / 2 + (OneInput / 2 + 1) % 2;
int ADWord, BDWord;
int &TripleDWord = AToAInputs.size() == 3 ? ADWord : BDWord;
int &OneInputDWord = AToAInputs.size() == 3 ? BDWord : ADWord;
int TripleInputOffset = AToAInputs.size() == 3 ? AOffset : BOffset;
ArrayRef<int> TripleInputs = AToAInputs.size() == 3 ? AToAInputs : BToAInputs;
int OneInput = AToAInputs.size() == 3 ? BToAInputs[0] : AToAInputs[0];
int TripleInputSum = 0 + 1 + 2 + 3 + (4 * TripleInputOffset);
int TripleNonInputIdx =
TripleInputSum - std::accumulate(TripleInputs.begin(), TripleInputs.end(), 0);
TripleDWord = TripleNonInputIdx / 2;
// We use xor with one to compute the adjacent DWord to whichever one the
// OneInput is in.
OneInputDWord = (OneInput / 2) ^ 1;
// Check for one tricky case: We're fixing a 3<-1 or a 1<-3 shuffle for AToA
// and BToA inputs. If there is also such a problem with the BToB and AToB
// inputs, we don't try to fix it necessarily -- we'll recurse and see it in
// the next pass. However, if we have a 2<-2 in the BToB and AToB inputs, it
// is essential that we don't *create* a 3<-1 as then we might oscillate.
if (BToBInputs.size() == 2 && AToBInputs.size() == 2) {
// Compute how many inputs will be flipped by swapping these DWords. We
// need
// to balance this to ensure we don't form a 3-1 shuffle in the other
// half.
int NumFlippedAToBInputs =
std::count(AToBInputs.begin(), AToBInputs.end(), 2 * ADWord) +
std::count(AToBInputs.begin(), AToBInputs.end(), 2 * ADWord + 1);
int NumFlippedBToBInputs =
std::count(BToBInputs.begin(), BToBInputs.end(), 2 * BDWord) +
std::count(BToBInputs.begin(), BToBInputs.end(), 2 * BDWord + 1);
if ((NumFlippedAToBInputs == 1 &&
(NumFlippedBToBInputs == 0 || NumFlippedBToBInputs == 2)) ||
(NumFlippedBToBInputs == 1 &&
(NumFlippedAToBInputs == 0 || NumFlippedAToBInputs == 2))) {
// We choose whether to fix the A half or B half based on whether that
// half has zero flipped inputs. At zero, we may not be able to fix it
// with that half. We also bias towards fixing the B half because that
// will more commonly be the high half, and we have to bias one way.
auto FixFlippedInputs = [&V, &DL, &Mask, &DAG](int PinnedIdx, int DWord,
ArrayRef<int> Inputs) {
int FixIdx = PinnedIdx ^ 1; // The adjacent slot to the pinned slot.
bool IsFixIdxInput = std::find(Inputs.begin(), Inputs.end(),
PinnedIdx ^ 1) != Inputs.end();
// Determine whether the free index is in the flipped dword or the
// unflipped dword based on where the pinned index is. We use this bit
// in an xor to conditionally select the adjacent dword.
int FixFreeIdx = 2 * (DWord ^ (PinnedIdx / 2 == DWord));
bool IsFixFreeIdxInput = std::find(Inputs.begin(), Inputs.end(),
FixFreeIdx) != Inputs.end();
if (IsFixIdxInput == IsFixFreeIdxInput)
FixFreeIdx += 1;
IsFixFreeIdxInput = std::find(Inputs.begin(), Inputs.end(),
FixFreeIdx) != Inputs.end();
assert(IsFixIdxInput != IsFixFreeIdxInput &&
"We need to be changing the number of flipped inputs!");
int PSHUFHalfMask[] = {0, 1, 2, 3};
std::swap(PSHUFHalfMask[FixFreeIdx % 4], PSHUFHalfMask[FixIdx % 4]);
V = DAG.getNode(FixIdx < 4 ? X86ISD::PSHUFLW : X86ISD::PSHUFHW, DL,
MVT::v8i16, V,
getV4X86ShuffleImm8ForMask(PSHUFHalfMask, DAG));
for (int &M : Mask)
if (M != -1 && M == FixIdx)
M = FixFreeIdx;
else if (M != -1 && M == FixFreeIdx)
M = FixIdx;
};
if (NumFlippedBToBInputs != 0) {
int BPinnedIdx =
BToAInputs.size() == 3 ? TripleNonInputIdx : OneInput;
FixFlippedInputs(BPinnedIdx, BDWord, BToBInputs);
} else {
assert(NumFlippedAToBInputs != 0 && "Impossible given predicates!");
int APinnedIdx =
AToAInputs.size() == 3 ? TripleNonInputIdx : OneInput;
FixFlippedInputs(APinnedIdx, ADWord, AToBInputs);
}
}
}
int PSHUFDMask[] = {0, 1, 2, 3};
PSHUFDMask[DWordA] = DWordB;
PSHUFDMask[DWordB] = DWordA;
PSHUFDMask[ADWord] = BDWord;
PSHUFDMask[BDWord] = ADWord;
V = DAG.getNode(ISD::BITCAST, DL, MVT::v8i16,
DAG.getNode(X86ISD::PSHUFD, DL, MVT::v4i32,
DAG.getNode(ISD::BITCAST, DL, MVT::v4i32, V),
@ -7350,24 +7454,20 @@ static SDValue lowerV8I16SingleInputVectorShuffle(
// Adjust the mask to match the new locations of A and B.
for (int &M : Mask)
if (M != -1 && M/2 == DWordA)
M = 2 * DWordB + M % 2;
else if (M != -1 && M/2 == DWordB)
M = 2 * DWordA + M % 2;
if (M != -1 && M/2 == ADWord)
M = 2 * BDWord + M % 2;
else if (M != -1 && M/2 == BDWord)
M = 2 * ADWord + M % 2;
// Recurse back into this routine to re-compute state now that this isn't
// a 3 and 1 problem.
return DAG.getVectorShuffle(MVT::v8i16, DL, V, DAG.getUNDEF(MVT::v8i16),
Mask);
};
if (NumLToL == 3 && NumHToL == 1)
return balanceSides(LToLInputs, HToLInputs[0], 0 + 1 + 2 + 3, 4);
else if (NumLToL == 1 && NumHToL == 3)
return balanceSides(HToLInputs, LToLInputs[0], 4 + 5 + 6 + 7, 0);
else if (NumLToH == 1 && NumHToH == 3)
return balanceSides(HToHInputs, LToHInputs[0], 4 + 5 + 6 + 7, 0);
else if (NumLToH == 3 && NumHToH == 1)
return balanceSides(LToHInputs, HToHInputs[0], 0 + 1 + 2 + 3, 4);
if ((NumLToL == 3 && NumHToL == 1) || (NumLToL == 1 && NumHToL == 3))
return balanceSides(LToLInputs, HToLInputs, HToHInputs, LToHInputs, 0, 4);
else if ((NumHToH == 3 && NumLToH == 1) || (NumHToH == 1 && NumLToH == 3))
return balanceSides(HToHInputs, LToHInputs, LToLInputs, HToLInputs, 4, 0);
// At this point there are at most two inputs to the low and high halves from
// each half. That means the inputs can always be grouped into dwords and

19
llvm/test/CodeGen/X86/vector-shuffle-128-v8.ll
View File

@ -451,6 +451,25 @@ define <8 x i16> @shuffle_v8i16_45630127(<8 x i16> %a, <8 x i16> %b) {
ret <8 x i16> %shuffle
}
define <8 x i16> @shuffle_v8i16_37102735(<8 x i16> %a, <8 x i16> %b) {
; SSE2-LABEL: @shuffle_v8i16_37102735
; SSE2: # BB#0:
; SSE2-NEXT: pshufhw {{.*}} # xmm0 = xmm0[0,1,2,3,4,6,5,7]
; SSE2-NEXT: pshufd {{.*}} # xmm0 = xmm0[0,2,1,3]
; SSE2-NEXT: pshufhw {{.*}} # xmm0 = xmm0[0,1,2,3,7,5,6,4]
; SSE2-NEXT: pshufd {{.*}} # xmm0 = xmm0[0,2,2,3]
; SSE2-NEXT: pshuflw {{.*}} # xmm0 = xmm0[3,2,1,0,4,5,6,7]
; SSE2-NEXT: pshufhw {{.*}} # xmm0 = xmm0[0,1,2,3,7,4,5,6]
; SSE2-NEXT: retq
;
; SSSE3-LABEL: @shuffle_v8i16_37102735
; SSSE3: # BB#0:
; SSSE3-NEXT: pshufb {{.*}} # xmm0 = xmm0[6,7,14,15,2,3,0,1,4,5,14,15,6,7,10,11]
; SSSE3-NEXT: retq
%shuffle = shufflevector <8 x i16> %a, <8 x i16> %b, <8 x i32> <i32 3, i32 7, i32 1, i32 0, i32 2, i32 7, i32 3, i32 5>
ret <8 x i16> %shuffle
}
define <8 x i16> @shuffle_v8i16_08192a3b(<8 x i16> %a, <8 x i16> %b) {
; ALL-LABEL: @shuffle_v8i16_08192a3b
; ALL: # BB#0: