// 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),
// 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
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