X86SchedSandyBridge.td revision 263508
1//=- X86SchedSandyBridge.td - X86 Sandy Bridge Scheduling ----*- tablegen -*-=//
2//
3//                     The LLVM Compiler Infrastructure
4//
5// This file is distributed under the University of Illinois Open Source
6// License. See LICENSE.TXT for details.
7//
8//===----------------------------------------------------------------------===//
9//
10// This file defines the machine model for Sandy Bridge to support instruction
11// scheduling and other instruction cost heuristics.
12//
13//===----------------------------------------------------------------------===//
14
15def SandyBridgeModel : SchedMachineModel {
16  // All x86 instructions are modeled as a single micro-op, and SB can decode 4
17  // instructions per cycle.
18  // FIXME: Identify instructions that aren't a single fused micro-op.
19  let IssueWidth = 4;
20  let MicroOpBufferSize = 168; // Based on the reorder buffer.
21  let LoadLatency = 4;
22  let MispredictPenalty = 16;
23
24  // FIXME: SSE4 and AVX are unimplemented. This flag is set to allow
25  // the scheduler to assign a default model to unrecognized opcodes.
26  let CompleteModel = 0;
27}
28
29let SchedModel = SandyBridgeModel in {
30
31// Sandy Bridge can issue micro-ops to 6 different ports in one cycle.
32
33// Ports 0, 1, and 5 handle all computation.
34def SBPort0 : ProcResource<1>;
35def SBPort1 : ProcResource<1>;
36def SBPort5 : ProcResource<1>;
37
38// Ports 2 and 3 are identical. They handle loads and the address half of
39// stores.
40def SBPort23 : ProcResource<2>;
41
42// Port 4 gets the data half of stores. Store data can be available later than
43// the store address, but since we don't model the latency of stores, we can
44// ignore that.
45def SBPort4 : ProcResource<1>;
46
47// Many micro-ops are capable of issuing on multiple ports.
48def SBPort05  : ProcResGroup<[SBPort0, SBPort5]>;
49def SBPort15  : ProcResGroup<[SBPort1, SBPort5]>;
50def SBPort015 : ProcResGroup<[SBPort0, SBPort1, SBPort5]>;
51
52// 54 Entry Unified Scheduler
53def SBPortAny : ProcResGroup<[SBPort0, SBPort1, SBPort23, SBPort4, SBPort5]> {
54  let BufferSize=54;
55}
56
57// Integer division issued on port 0.
58def SBDivider : ProcResource<1>;
59
60// Loads are 4 cycles, so ReadAfterLd registers needn't be available until 4
61// cycles after the memory operand.
62def : ReadAdvance<ReadAfterLd, 4>;
63
64// Many SchedWrites are defined in pairs with and without a folded load.
65// Instructions with folded loads are usually micro-fused, so they only appear
66// as two micro-ops when queued in the reservation station.
67// This multiclass defines the resource usage for variants with and without
68// folded loads.
69multiclass SBWriteResPair<X86FoldableSchedWrite SchedRW,
70                          ProcResourceKind ExePort,
71                          int Lat> {
72  // Register variant is using a single cycle on ExePort.
73  def : WriteRes<SchedRW, [ExePort]> { let Latency = Lat; }
74
75  // Memory variant also uses a cycle on port 2/3 and adds 4 cycles to the
76  // latency.
77  def : WriteRes<SchedRW.Folded, [SBPort23, ExePort]> {
78     let Latency = !add(Lat, 4);
79  }
80}
81
82// A folded store needs a cycle on port 4 for the store data, but it does not
83// need an extra port 2/3 cycle to recompute the address.
84def : WriteRes<WriteRMW, [SBPort4]>;
85
86def : WriteRes<WriteStore, [SBPort23, SBPort4]>;
87def : WriteRes<WriteLoad,  [SBPort23]> { let Latency = 4; }
88def : WriteRes<WriteMove,  [SBPort015]>;
89def : WriteRes<WriteZero,  []>;
90
91defm : SBWriteResPair<WriteALU,   SBPort015, 1>;
92defm : SBWriteResPair<WriteIMul,  SBPort1,   3>;
93def  : WriteRes<WriteIMulH, []> { let Latency = 3; }
94defm : SBWriteResPair<WriteShift, SBPort05,  1>;
95defm : SBWriteResPair<WriteJump,  SBPort5,   1>;
96
97// This is for simple LEAs with one or two input operands.
98// The complex ones can only execute on port 1, and they require two cycles on
99// the port to read all inputs. We don't model that.
100def : WriteRes<WriteLEA, [SBPort15]>;
101
102// This is quite rough, latency depends on the dividend.
103def : WriteRes<WriteIDiv, [SBPort0, SBDivider]> {
104  let Latency = 25;
105  let ResourceCycles = [1, 10];
106}
107def : WriteRes<WriteIDivLd, [SBPort23, SBPort0, SBDivider]> {
108  let Latency = 29;
109  let ResourceCycles = [1, 1, 10];
110}
111
112// Scalar and vector floating point.
113defm : SBWriteResPair<WriteFAdd,   SBPort1, 3>;
114defm : SBWriteResPair<WriteFMul,   SBPort0, 5>;
115defm : SBWriteResPair<WriteFDiv,   SBPort0, 12>; // 10-14 cycles.
116defm : SBWriteResPair<WriteFRcp,   SBPort0, 5>;
117defm : SBWriteResPair<WriteFSqrt,  SBPort0, 15>;
118defm : SBWriteResPair<WriteCvtF2I, SBPort1, 3>;
119defm : SBWriteResPair<WriteCvtI2F, SBPort1, 4>;
120defm : SBWriteResPair<WriteCvtF2F, SBPort1, 3>;
121
122// Vector integer operations.
123defm : SBWriteResPair<WriteVecShift, SBPort05,  1>;
124defm : SBWriteResPair<WriteVecLogic, SBPort015, 1>;
125defm : SBWriteResPair<WriteVecALU,   SBPort15,  1>;
126defm : SBWriteResPair<WriteVecIMul,  SBPort0,   5>;
127defm : SBWriteResPair<WriteShuffle,  SBPort15,  1>;
128
129def : WriteRes<WriteSystem,     [SBPort015]> { let Latency = 100; }
130def : WriteRes<WriteMicrocoded, [SBPort015]> { let Latency = 100; }
131} // SchedModel
132