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llvm-mirror/lib/CodeGen/RegAllocBase.cpp

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//===-- RegAllocBase.cpp - Register Allocator Base Class ------------------===//
//
// The LLVM Compiler Infrastructure
//
// This file is distributed under the University of Illinois Open Source
// License. See LICENSE.TXT for details.
//
//===----------------------------------------------------------------------===//
//
2014-02-22 20:31:28 +01:00
// This file defines the RegAllocBase class which provides common functionality
// for LiveIntervalUnion-based register allocators.
//
//===----------------------------------------------------------------------===//
#include "RegAllocBase.h"
#include "Spiller.h"
#include "llvm/ADT/Statistic.h"
#include "llvm/CodeGen/LiveIntervalAnalysis.h"
#include "llvm/CodeGen/LiveRangeEdit.h"
#include "llvm/CodeGen/LiveRegMatrix.h"
#include "llvm/CodeGen/MachineInstr.h"
#include "llvm/CodeGen/MachineRegisterInfo.h"
#include "llvm/CodeGen/VirtRegMap.h"
#include "llvm/Target/TargetRegisterInfo.h"
#ifndef NDEBUG
#include "llvm/ADT/SparseBitVector.h"
#endif
#include "llvm/Support/CommandLine.h"
#include "llvm/Support/Debug.h"
#include "llvm/Support/ErrorHandling.h"
#include "llvm/Support/raw_ostream.h"
#include "llvm/Support/Timer.h"
using namespace llvm;
#define DEBUG_TYPE "regalloc"
STATISTIC(NumNewQueued , "Number of new live ranges queued");
// Temporary verification option until we can put verification inside
// MachineVerifier.
static cl::opt<bool, true>
VerifyRegAlloc("verify-regalloc", cl::location(RegAllocBase::VerifyEnabled),
cl::desc("Verify during register allocation"));
const char RegAllocBase::TimerGroupName[] = "Register Allocation";
bool RegAllocBase::VerifyEnabled = false;
//===----------------------------------------------------------------------===//
// RegAllocBase Implementation
//===----------------------------------------------------------------------===//
// Pin the vtable to this file.
void RegAllocBase::anchor() {}
void RegAllocBase::init(VirtRegMap &vrm,
LiveIntervals &lis,
LiveRegMatrix &mat) {
TRI = &vrm.getTargetRegInfo();
MRI = &vrm.getRegInfo();
VRM = &vrm;
LIS = &lis;
Matrix = &mat;
MRI->freezeReservedRegs(vrm.getMachineFunction());
RegClassInfo.runOnMachineFunction(vrm.getMachineFunction());
}
// Visit all the live registers. If they are already assigned to a physical
// register, unify them with the corresponding LiveIntervalUnion, otherwise push
// them on the priority queue for later assignment.
void RegAllocBase::seedLiveRegs() {
NamedRegionTimer T("Seed Live Regs", TimerGroupName, TimePassesIsEnabled);
for (unsigned i = 0, e = MRI->getNumVirtRegs(); i != e; ++i) {
unsigned Reg = TargetRegisterInfo::index2VirtReg(i);
if (MRI->reg_nodbg_empty(Reg))
continue;
enqueue(&LIS->getInterval(Reg));
}
}
// Top-level driver to manage the queue of unassigned VirtRegs and call the
// selectOrSplit implementation.
void RegAllocBase::allocatePhysRegs() {
seedLiveRegs();
// Continue assigning vregs one at a time to available physical registers.
while (LiveInterval *VirtReg = dequeue()) {
assert(!VRM->hasPhys(VirtReg->reg) && "Register already assigned");
// Unused registers can appear when the spiller coalesces snippets.
if (MRI->reg_nodbg_empty(VirtReg->reg)) {
DEBUG(dbgs() << "Dropping unused " << *VirtReg << '\n');
[RegAllocGreedy] Introduce a late pass to repair broken hints. A broken hint is a copy where both ends are assigned different colors. When a variable gets evicted in the neighborhood of such copies, it is likely we can reconcile some of them. ** Context ** Copies are inserted during the register allocation via splitting. These split points are required to relax the constraints on the allocation problem. When such a point is inserted, both ends of the copy would not share the same color with respect to the current allocation problem. When variables get evicted, the allocation problem becomes different and some split point may not be required anymore. However, the related variables may already have been colored. This usually shows up in the assembly with pattern like this: def A ... save A to B def A use A restore A from B ... use B Whereas we could simply have done: def B ... def A use A ... use B ** Proposed Solution ** A variable having a broken hint is marked for late recoloring if and only if selecting a register for it evict another variable. Indeed, if no eviction happens this is pointless to look for recoloring opportunities as it means the situation was the same as the initial allocation problem where we had to break the hint. Finally, when everything has been allocated, we look for recoloring opportunities for all the identified candidates. The recoloring is performed very late to rely on accurate copy cost (all involved variables are allocated). The recoloring is simple unlike the last change recoloring. It propagates the color of the broken hint to all its copy-related variables. If the color is available for them, the recoloring uses it, otherwise it gives up on that hint even if a more complex coloring would have worked. The recoloring happens only if it is profitable. The profitability is evaluated using the expected frequency of the copies of the currently recolored variable with a) its current color and b) with the target color. If a) is greater or equal than b), then it is profitable and the recoloring happen. ** Example ** Consider the following example: BB1: a = b = BB2: ... = b = a Let us assume b gets split: BB1: a = b = BB2: c = b ... d = c = d = a Because of how the allocation work, b, c, and d may be assigned different colors. Now, if a gets evicted to make room for c, assuming b and d were assigned to something different than a. We end up with: BB1: a = st a, SpillSlot b = BB2: c = b ... d = c = d e = ld SpillSlot = e This is likely that we can assign the same register for b, c, and d, getting rid of 2 copies. ** Performances ** Both ARM64 and x86_64 show performance improvements of up to 3% for the llvm-testsuite + externals with Os and O3. There are a few regressions too that comes from the (in)accuracy of the block frequency estimate. <rdar://problem/18312047> llvm-svn: 225422
2015-01-08 02:16:39 +01:00
aboutToRemoveInterval(*VirtReg);
LIS->removeInterval(VirtReg->reg);
continue;
}
// Invalidate all interference queries, live ranges could have changed.
Matrix->invalidateVirtRegs();
// selectOrSplit requests the allocator to return an available physical
// register if possible and populate a list of new live intervals that
// result from splitting.
DEBUG(dbgs() << "\nselectOrSplit "
<< TRI->getRegClassName(MRI->getRegClass(VirtReg->reg))
<< ':' << *VirtReg << " w=" << VirtReg->weight << '\n');
typedef SmallVector<unsigned, 4> VirtRegVec;
VirtRegVec SplitVRegs;
unsigned AvailablePhysReg = selectOrSplit(*VirtReg, SplitVRegs);
if (AvailablePhysReg == ~0u) {
// selectOrSplit failed to find a register!
// Probably caused by an inline asm.
MachineInstr *MI = nullptr;
for (MachineRegisterInfo::reg_instr_iterator
I = MRI->reg_instr_begin(VirtReg->reg), E = MRI->reg_instr_end();
I != E; ) {
MachineInstr *TmpMI = &*(I++);
if (TmpMI->isInlineAsm()) {
MI = TmpMI;
break;
}
}
if (MI)
MI->emitError("inline assembly requires more registers than available");
else
report_fatal_error("ran out of registers during register allocation");
// Keep going after reporting the error.
VRM->assignVirt2Phys(VirtReg->reg,
RegClassInfo.getOrder(MRI->getRegClass(VirtReg->reg)).front());
continue;
}
if (AvailablePhysReg)
Matrix->assign(*VirtReg, AvailablePhysReg);
for (VirtRegVec::iterator I = SplitVRegs.begin(), E = SplitVRegs.end();
I != E; ++I) {
LiveInterval *SplitVirtReg = &LIS->getInterval(*I);
assert(!VRM->hasPhys(SplitVirtReg->reg) && "Register already assigned");
if (MRI->reg_nodbg_empty(SplitVirtReg->reg)) {
DEBUG(dbgs() << "not queueing unused " << *SplitVirtReg << '\n');
[RegAllocGreedy] Introduce a late pass to repair broken hints. A broken hint is a copy where both ends are assigned different colors. When a variable gets evicted in the neighborhood of such copies, it is likely we can reconcile some of them. ** Context ** Copies are inserted during the register allocation via splitting. These split points are required to relax the constraints on the allocation problem. When such a point is inserted, both ends of the copy would not share the same color with respect to the current allocation problem. When variables get evicted, the allocation problem becomes different and some split point may not be required anymore. However, the related variables may already have been colored. This usually shows up in the assembly with pattern like this: def A ... save A to B def A use A restore A from B ... use B Whereas we could simply have done: def B ... def A use A ... use B ** Proposed Solution ** A variable having a broken hint is marked for late recoloring if and only if selecting a register for it evict another variable. Indeed, if no eviction happens this is pointless to look for recoloring opportunities as it means the situation was the same as the initial allocation problem where we had to break the hint. Finally, when everything has been allocated, we look for recoloring opportunities for all the identified candidates. The recoloring is performed very late to rely on accurate copy cost (all involved variables are allocated). The recoloring is simple unlike the last change recoloring. It propagates the color of the broken hint to all its copy-related variables. If the color is available for them, the recoloring uses it, otherwise it gives up on that hint even if a more complex coloring would have worked. The recoloring happens only if it is profitable. The profitability is evaluated using the expected frequency of the copies of the currently recolored variable with a) its current color and b) with the target color. If a) is greater or equal than b), then it is profitable and the recoloring happen. ** Example ** Consider the following example: BB1: a = b = BB2: ... = b = a Let us assume b gets split: BB1: a = b = BB2: c = b ... d = c = d = a Because of how the allocation work, b, c, and d may be assigned different colors. Now, if a gets evicted to make room for c, assuming b and d were assigned to something different than a. We end up with: BB1: a = st a, SpillSlot b = BB2: c = b ... d = c = d e = ld SpillSlot = e This is likely that we can assign the same register for b, c, and d, getting rid of 2 copies. ** Performances ** Both ARM64 and x86_64 show performance improvements of up to 3% for the llvm-testsuite + externals with Os and O3. There are a few regressions too that comes from the (in)accuracy of the block frequency estimate. <rdar://problem/18312047> llvm-svn: 225422
2015-01-08 02:16:39 +01:00
aboutToRemoveInterval(*SplitVirtReg);
LIS->removeInterval(SplitVirtReg->reg);
continue;
}
DEBUG(dbgs() << "queuing new interval: " << *SplitVirtReg << "\n");
assert(TargetRegisterInfo::isVirtualRegister(SplitVirtReg->reg) &&
"expect split value in virtual register");
enqueue(SplitVirtReg);
++NumNewQueued;
}
}
}