similar to: [LLVMdev] list-td scheduler asserts on targets with implicitly defined registers

The best fix is to teach this scheduler how to deal with these dependencies. :-) If you just want a check, I think it's easier to just check register class's copy cost. -1 means it's extremely expensive to copy registers in the particular register class. Evan On Feb 6, 2009, at 2:22 AM, Christian Sayer wrote: > Hi, > > I just switched to the 2.5 release branch and

> The best fix is to teach this scheduler how to deal with these > dependencies. :-) > > If you just want a check, I think it's easier to just check register > class's copy cost. -1 means it's extremely expensive to copy registers > in the particular register class. Evan, I am not sure what you mean by "if you just want a check" - I was trying to point out

In the X86RegisterInfo.td file, RST is defined like this: // Floating point stack registers (these are not allocatable by the // register allocator - the floating point stackifier is responsible // for transforming FPn allocations to STn registers) def RST : RegisterClass<"X86", [f64], 32, [ST0, ST1, ST2, ST3, ST4, ST5, ST6, ST7]> { let MethodProtos =

2005/7/23, Chris Lattner <sabre at nondot.org>: > > What does a 'read only' register mean? Is it a constant (e.g. returns > 1.0)? Otherwise, how can it be a useful value? Yes, it's a constant register. Because the instruction cannot contain an immediate value, a constant value may be stored in a constant register, and it's defined _before_ the program starts by

On Sat, 23 Jul 2005, Tzu-Chien Chiu wrote: > 2005/7/23, Chris Lattner <sabre at nondot.org>: >> What does a 'read only' register mean? Is it a constant (e.g. returns >> 1.0)? Otherwise, how can it be a useful value? > > Yes, it's a constant register. > > Because the instruction cannot contain an immediate value, a constant > value may be stored in

I am using some non-allocatable RegisterClasses to define lists of registers that are used for various non-allocation-related processing in the back end. For example, we have a post-allocation functional unit selection pass that is guided by the register assignment, which does things like 'myRegClass.contains(Reg)' to see if a register is in the set of registers accessible by a given unit.

Hi people, the LinearScan register allocator tries to use same register for both live intervals, if the new interval is defined by a register copy whose destination reg is compatible with the source register. This is ok. However, this "check for compatibility" is wrongly done IMHO. Say I have regclass1 with reg A, and regclass2 with regs {A, B}, but regclass2 defines only

> A node should never be put into the conservatively allocatable list if there is a chance of it spilling. I can understand why the logic of NodeMetadata::isConservativelyAllocatable is necessary for the node to be allocatable, but I have not been able to convince myself this is sufficient, especially when the node degree > available registers. Cheers, Arnaud From:

On 29 March 2014 13:38, Joerg Sonnenberger <joerg at britannica.bec.de> wrote: > I disagree. It is the *easy* part to get many known users to work. This > includes a bunch of kernels, Lisp implementations etc. The rest can be > implemented on top by hand using inline asm, so this is the crucial > part. Let me re-phrase my opinion... >From all discussions on the LLVM list,

Hi Jakob, >> Say I have regclass1 with reg A, and regclass2 with regs {A, B}, but >> regclass2 defines only "B" as allocatable by RA. > > The register allocator assumes in many places that a register is > either allocatable or reserved independently of the register class. Is there any reason for this? I mean, the methods for allowing one physical reg be

On Sep 13, 2010, at 6:59 AM, Carlos Sánchez de La Lama wrote: > Hi people, > > the LinearScan register allocator tries to use same register for both > live intervals, if the new interval is defined by a register copy > whose destination reg is compatible with the source register. This is > ok. However, this "check for compatibility" is wrongly done IMHO. >

Hi, I'm writing to ask the differences between a "reserved" register and an "unallocable" register. In X86 backend, for example, the stack pointer register and instruction pointer are reserved but allocatable. In the Doxygen document of function llvm::TargetRegisterInfo::getReservedRegs, it says that a reserved register is one that *has particular uses and should be

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Hi Arnaud, The conservatively allocatable test is supposed to check two conditions, either of which would be sufficient to make a node allocatable: (1) There exists some register that is not aliased by any register option for any neighbor. This is the "safe row" test. It is straightforward, but likely to fire only rarely. (2) The sum of the maximum number of registers aliased by any

On Apr 29, 2010, at 2:06 PM, Villmow, Micah wrote: > Ping. Anyone have any idea on how to fix this? Does your getAllocatableSet() return a BitVector that is at least getNumRegs() bits long? Otherwise this doesn't work: BitVector NonAllocatableRegs = TRI->getAllocatableSet(MF); NonAllocatableRegs.flip();

On 29 March 2014 10:37, Joerg Sonnenberger <joerg at britannica.bec.de> wrote: > I'd like to separate this into two different functionalities: We will. But in reverse order. > (1) Reserve registers, so that normal allocation won't use them. > This can be done on a global or function level. This is the most controversial part of the proposal and is the least important to

While working on a back-end for a target, I've come across something I believe to be a bug in SimpleRegisterCoalescing.cpp. I'm unsure how / whether to report it because I don't think it will necessarily crash or generate incorrect code for any of the supported targets. I believe that there may be a problem in SimpleRegisterCoalescing::runOnMachineFunction where the allocatable

Hi Quentin, MyInst is a custom instruction that has implicit-defs of fixed registers. The implicit-defs are seen at the end of Instruction Selection. I'd like to add a report, but I am working on an out-of-tree backend based on 7.0. I can try to help reduce the testcase down. Filed https://bugs.llvm.org/show_bug.cgi?id=41790 Regards, Kevin On 2019-05-07 3:45 p.m., Quentin Colombet wrote:

Hi, I have run into a test case on an out-of-tree target where PBQP fails to complete register allocation after "Attempting to spill already spilled value" (the triggered assert in InlineSpiller::spill(). First, the original LiveInterval is spilled. It is a load of a symbol into a narrow register class, i.e. a subset of the class of address registers. InlineSpiller decides to

Hi, Sorry for the delay, it has taken some extra time as more than one bug showed up ☺ I continued to look into this with your viewpoint that a node that is conservatively allocatable should never be spilled. The first thing I did was therefore to add some extra code with an assert for this. I believe I then found three bugs and fixed the two: Bug 1: Incorrect transpositions in handleAddEdge()