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On Thu, 9 Nov 2006, Ryan M. Lefever wrote: Sorry I never responded to this: > In the literature (see below for a reference), when a dominance frontier > is computed, it is computed from a CFG that contains a dummy entry node > and dummy exit node. Further, those dummy nodes are potential members > of the (post-)dominance frontier for a given basic block. In LLVM, I > could not

In the literature (see below for a reference), when a dominance frontier is computed, it is computed from a CFG that contains a dummy entry node and dummy exit node. Further, those dummy nodes are potential members of the (post-)dominance frontier for a given basic block. In LLVM, I could not figure out a way to determine if the dummy entry node is a member of the post-dominance frontier of

...w graph, there is a call to malloc inserted in block BB12. In order to avoid memory leak, free's are needed. The free cannot be inserted in BB23 because BB23 is not dominated by BB12. There are two ways to go I can think of here. One way is to insert a new basic block, say BB24, to connect both BB21 and BB22 and a free can be inserted into the new block BB24. The new block BB24 has to post-dominate BB12 and all the users of malloc have to happen before BB24. Another way to go is to insert a free in both BB21 and BB22. That is, a free is inserted in all the paths from BB12 to all exits after al...

...nserted in block BB12. In order to >> avoid memory leak, free's are needed. The free cannot be inserted in BB23 >> because BB23 is not dominated by BB12. There are two ways to go I can think >> of here. One way is to insert a new basic block, say BB24, to connect both >> BB21 and BB22 and a free can be inserted into the new block BB24. The new >> block BB24 has to post-dominate BB12 and all the users of malloc have to >> happen before BB24. Another way to go is to insert a free in both BB21 and >> BB22. That is, a free is inserted in all the paths from...

...call to malloc inserted in block BB12. In order to > avoid memory leak, free's are needed. The free cannot be inserted in BB23 > because BB23 is not dominated by BB12. There are two ways to go I can think > of here. One way is to insert a new basic block, say BB24, to connect both > BB21 and BB22 and a free can be inserted into the new block BB24. The new > block BB24 has to post-dominate BB12 and all the users of malloc have to > happen before BB24. Another way to go is to insert a free in both BB21 and > BB22. That is, a free is inserted in all the paths from BB12 to all...

...BB12. In order to >>> avoid memory leak, free's are needed. The free cannot be inserted in BB23 >>> because BB23 is not dominated by BB12. There are two ways to go I can think >>> of here. One way is to insert a new basic block, say BB24, to connect both >>> BB21 and BB22 and a free can be inserted into the new block BB24. The new >>> block BB24 has to post-dominate BB12 and all the users of malloc have to >>> happen before BB24. Another way to go is to insert a free in both BB21 and >>> BB22. That is, a free is inserted in all th...

...to >>>> avoid memory leak, free's are needed. The free cannot be inserted in BB23 >>>> because BB23 is not dominated by BB12. There are two ways to go I can think >>>> of here. One way is to insert a new basic block, say BB24, to connect both >>>> BB21 and BB22 and a free can be inserted into the new block BB24. The new >>>> block BB24 has to post-dominate BB12 and all the users of malloc have to >>>> happen before BB24. Another way to go is to insert a free in both BB21 and >>>> BB22. That is, a free is insert...

...call to malloc inserted in block > BB12. In order to avoid memory leak, free's are needed. The free > cannot be inserted in BB23 because BB23 is not dominated by BB12. > There are two ways to go I can think of here. One way is to insert a > new basic block, say BB24, to connect both BB21 and BB22 and a free > can be inserted into the new block BB24. The new block BB24 has to > post-dominate BB12 and all the users of malloc have to happen before > BB24. Another way to go is to insert a free in both BB21 and BB22. > That is, a free is inserted in all the paths from BB12 t...

...to >>>> avoid memory leak, free's are needed. The free cannot be inserted in BB23 >>>> because BB23 is not dominated by BB12. There are two ways to go I can think >>>> of here. One way is to insert a new basic block, say BB24, to connect both >>>> BB21 and BB22 and a free can be inserted into the new block BB24. The new >>>> block BB24 has to post-dominate BB12 and all the users of malloc have to >>>> happen before BB24. Another way to go is to insert a free in both BB21 and >>>> BB22. That is, a free is insert...

...gt;> avoid memory leak, free's are needed. The free cannot be inserted in BB23 >>>>> because BB23 is not dominated by BB12. There are two ways to go I can think >>>>> of here. One way is to insert a new basic block, say BB24, to connect both >>>>> BB21 and BB22 and a free can be inserted into the new block BB24. The new >>>>> block BB24 has to post-dominate BB12 and all the users of malloc have to >>>>> happen before BB24. Another way to go is to insert a free in both BB21 and >>>>> BB22. That is, a fr...

Hello, This is an RFC for adding intrinsics which perform saturating signed/unsigned left shift. There is currently a patch on Phabricator here: https://reviews.llvm.org/D83216 The intrinsics are of the form i32 @llvm.sshl.sat.i32(i32, i32) i32 @llvm.ushl.sat.i32(i32, i32) <4 x i32> @llvm.sshl.sat.v4i32(<4 x i32>, <4 x i32>) <4 x i32>

...[#uses=1] %sum.1.lcssa.us = add i32 %sum.0.pn.ph.us, %tmp ; <i32> [#uses=2] %indvar.next48 = add i32 %indvar42, 1 ; <i32> [#uses=2] %exitcond49 = icmp eq i32 %indvar.next48, 10 ; <i1> [#uses=1] br i1 %exitcond49, label %bb21.split, label %bb9.outer.us bb21.split: ; preds = %bb9.outer.us ret i32 %sum.1.lcssa.us } //////////////////////////////////////////////////////////// As shown in IR above, only the inner loop is optimized (by directly adding 6). I simply changed the HL source code like this: (...

I have had a number of occasions lately where a virtual "NAT" network (such as default) has stopped working. I have attempted to stop and then restart the network from the "network details" but the only thing that seems to work is to restart all of the virtual guests which use that network. I don't believe that I should have to do that. Is there some command which

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Dear all, suppose I have a vector with elements as: Vec <- c(2,3,4,5,6) Now I want to have all possible combination of length 3 using those elements and without any repetition. Like, I want to have 1 possibility like 2-3-4 but not 3-2-4. Can somebody guide me how to achieve that in R? Thanks for your help.

...%tmp = alloca %struct.small*, align 8 + %tmp1 = call i32 (...)* @dummy(%struct.small** %tmp) nounwind + %tmp2 = load %struct.small** %tmp, align 8 + %tmp3 = ptrtoint %struct.small* %tmp2 to i64 + %tmp4 = trunc i64 %tmp3 to i32 + %tmp5 = icmp sgt i32 %tmp4, 0 + br i1 %tmp5, label %bb6, label %bb21 + +bb6: ; preds = %bb17, %bb + %tmp7 = phi %struct.small* [ %tmp19, %bb17 ], [ %tmp2, %bb ] + %tmp8 = phi i64 [ %tmp20, %bb17 ], [ 1, %bb ] + %tmp9 = phi i32 [ %tmp14, %bb17 ], [ %tmp1, %bb ] + %tmp10 = getelementptr inbounds %struct.small* %tmp7, i6...