search for: predecessor

...ou can see, %vec.phi has wrong incoming basic blocks. It could be as below. %vec.phi = phi <4 x i64> [ %8, %for.cond.cleanup972 ], [ zeroinitializer, %vector.body ] The problem comes from "InnerLoopVectorizer::fixNonInductionPHIs()". This function has assumption about the order of predecessors as below. // The predecessor order is preserved and we can rely on mapping between // scalar and vector block predecessors. for (unsigned i = 0; i < NumIncomingValues; ++i) { It seems it assumes loop latch as first predecessor and loop preheader as second one or something like tha...

Hi, I was reading some loop related code and I don’t quite understand an assertion in LoopBase<BlockT, LoopT>::getLoopPredecessor(). /// getLoopPredecessor - If the given loop's header has exactly one unique /// predecessor outside the loop, return it. Otherwise return null. /// This is less strict that the loop "preheader" concept, which requires /// the predecessor to have exactly one successor. /// template&...

...vestigating issues where variables that one would expect to be available in a debugger are not in code that is compiled at optimisations other than -O0 The main problem appears to be with the LiveDebugValues::join() method because it does not allow variables to be propagated into blocks unless all predecessor blocks have an Outgoing Location for that variable. As a simple example in the C code: int func2( int); void func(int a) { int b = func2(10); for(int i = 1; i < a; i++) { func2(i+b); } } One would reasonable expect when stopped within the body of the fo...

...gt;> >>> IE >>> >>> A >>> / \ >>> B C >>> \ / >>> D >>> >>> As a set, B + C dominate D. >>> >>> The set you are looking for there is (i believe): >>> >>> For each predecessor, walk the idom tree until you hit NCA of all >>> predecessors. >>> >> "For each predecessor" do you mean "For each predecessor of the basic blocks in the set"? I.e. for each predecessor of B and C in this example. Thanks Hongbin > What do you mean b...

...\ >>>>> B C >>>>> \ / >>>>> D >>>>> >>>>> As a set, B + C dominate D. >>>>> >>>>> The set you are looking for there is (i believe): >>>>> >>>>> For each predecessor, walk the idom tree until you hit NCA of all >>>>> predecessors. >>>>> >>>> "For each predecessor" do you mean "For each predecessor of the basic >> blocks in the set"? I.e. for each predecessor of B and C in this example. >>...

Hi! I want to know How to count the number of predecessors for each basic block? Thank You ____________________________________________________________________________________ Never miss a thing. Make Yahoo your home page. http://www.yahoo.com/r/hs

...merge this block with it's successor and produce a >> semantically equivalent program? I'm trying to undo some of the loop >> optimizations that LLVM has applied to my program to reduce a pair of >> nested loops to a single loop. >> >> llvm::MergeBlockIntoPredecessor does not do what I want since it >> requires that the the block have a unique predecessor. > > I didn't notice anything that will do what you want out-of-box, but it should not be hard to write. llvm::FoldSingleEntryPHINodes is an example of phi node replacement. But in this case,...

...branch to itself). Is it always possible to merge this block with it's successor and produce a semantically equivalent program? I'm trying to undo some of the loop optimizations that LLVM has applied to my program to reduce a pair of nested loops to a single loop. llvm::MergeBlockIntoPredecessor does not do what I want since it requires that the the block have a unique predecessor. Andrew

(well, from me anyway. I promise) Trying to follow the "Self-referential Many-to-Many Relationships" recipe from the Rails Recipes book. This is what I''m doing: Tasks can have a predecessor relationship with each other. My Task class has this: has_and_belongs_to_many :predecessors, :class_name => "Task", :join_table => "predecessor", :association_foreign_key => "predecessor_id", :foreign_key => "task_id"...

...>>>>>> \ / >>>>>> D >>>>>> >>>>>> As a set, B + C dominate D. >>>>>> >>>>>> The set you are looking for there is (i believe): >>>>>> >>>>>> For each predecessor, walk the idom tree until you hit NCA of all >>>>>> predecessors. >>>>>> >>>>> "For each predecessor" do you mean "For each predecessor of the basic >>> blocks in the set"? I.e. for each predecessor of B and C in this e...

...= %entry store i32 3, i32* %i %0 = load i32* %i ret i32 %0 if.else: ; preds = %entry ret i32 2 if.end: ; preds = %after_ret1, %after_ ret ret i32 1 after_ret: ; No predecessors! br label %if.end after_ret1: ; No predecessors! br label %if.end after_ret2: ; No predecessors! ret i32 0 } The following IR is slightly different in that the alloca is defined not in the first basic block: defi...

...dead PHI in my test program. --------------------- static bool mergeBlockIntoSuccessor(BasicBlock *pred, BasicBlock *succ) { if (succ == pred) return false; if (pred->getFirstNonPHI() != pred->getTerminator()) return false; // Delete the terminator in the predecessor block pred->getTerminator()->eraseFromParent(); // Update predecessor PHIs for (BasicBlock::iterator it = pred->begin(); it != pred->end(); ++it) { PHINode *phi = dyn_cast<PHINode>(it); UT_ASSERT(phi); // Adjust the PH...

Hi aditya, There are two ways to cound the number of predecessors for each basic block. You can generate the control flow graph using the CallGraphScc pass with the granularity of basic block and can simply traverse the graph bottom up till the root. The number of nodes encountered would be the number of predecessors. The second way would be to use the special...

> On Sep 21, 2016, at 2:23 PM, Daniel Berlin <dberlin at dberlin.org> wrote: > > > > // For all predecessors of this MBB, find the set of VarLocs that > // can be joined. > for (auto p : MBB.predecessors()) { > auto OL = OutLocs.find(p); > // Join is null in case of empty OutLocs from any of the pred. > if (OL == OutLocs.end()) > return false; > > > This...

...gmail.com] > Subject: Re: [LLVMdev] Question about method CodeExtractor::severSplitPHINodes > Please excuse me if I'm not supposed to reply to all. You should do reply-all, to make sure the list sees all of the thread. > Are you saying all PHI nodes are required to include all its predecessor blocks > no matter they have input or not? That's the way SSA works - each input to a block must be provided by all paths leading to the block. That must be from each immediate predecessor or from a common predecessor of two or more immediate predecessors that do not mutate the value. &g...

Thanks for reply Chuck. Please excuse me if I'm not supposed to reply to all. Are you saying all PHI nodes are required to include all its predecessor blocks no matter they have input or not? What about successor blocks? Are they optional if they don't provide inputs? BTW, where should I look at to verify this, the mem2reg.cpp & PromoteMemToReg.cpp? Thanks a lot. On Sat, Mar 2, 2013 at 6:39 PM, Caldarale, Charles R < Chuck.Caldarale...

Hi all, Is there a way to walk through ALL the predecessors of a basic block in a CFG. I tried to iterate over the preds using this method for (pred_iterator PI = pred_begin(BB), E = pred_end(BB); PI != E; ++I) { BasicBlock *PredBB = *PI; } but this only gives the immediate predecessors for a basic block. For example, in this sample control flow grap...

Is the pass "Break Critical edges" acheives the same as edge-splitting SSA, i.e., a node has either multiple predecessors or multiple successors but not both. A node with multiple predecessors and multiple successors is replaced by two consecutive nodes joined together. The first node has multiple predecessors and second node as its only successor. The second node has first node as its only predecessor and has mut...

Hi Henrique, I have tried using -mergereturn and inserting a free into the predecessors of dominance frontier of malloc block and it caused double free. It is possible for multiple free's to be inserted on the path from malloc to an exit. For example, in the following CFG: BB10 (malloc) / \ BB11 BB12 ... / \ /...

Try breaking the critical edges (-break-crit-edges). This way, a new block will be created between BB13 and BB11 (call this BB11.break) and BB15 and BB12 (call this BB12.break). The predecessors of the dominance frontier will, thus, be BB11.break, BB12.break, and BB14. When we enter through a block with a call to malloc(), we will end up in one of the blocks in the dominance frontier (kind of). These blocks must have multiple predecessors, else it would not be in the dominance frontier....