Re: Dominance frontier example in "Engineering a Compiler"

Rayiner Hashem <rayiner@gmail.com>
Tue, 03 Jul 2007 20:59:58 -0700

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Related articles
Dominance frontier example in "Engineering a Compiler" rayiner@gmail.com (Rayiner Hashem) (2007-06-19)
Re: Dominance frontier example in "Engineering a Compiler" klvnraju@gmail.com (Raj) (2007-07-03)
Re: Dominance frontier example in "Engineering a Compiler" rayiner@gmail.com (Rayiner Hashem) (2007-07-03)
Re: Dominance frontier example in Engineering a Compiler mstrout@cs.colostate.edu (Michelle Strout) (2007-07-05)
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From: Rayiner Hashem <rayiner@gmail.com>
Newsgroups: comp.compilers
Date: Tue, 03 Jul 2007 20:59:58 -0700
Organization: Compilers Central
References: 07-06-04007-07-010
Keywords: analysis

On Jul 3, 6:12 pm, Raj <klvnr...@gmail.com> wrote:
> B1 cannot be dominance frontier of itself.
> See the definition:
> A definition at n forces a X-function at m iff
> n not DOM(m) but n DOM(p) for some p preds(m)
>
> here, in your case
> n=m=B1
> and n DOM(m) (according to definition n should not belong to DOM(m))


That definition precludes a node from being in its own dominance
frontier, but then what about a self-loop?


entry:
a = 1
loop:
a = a +1
if a < 10 goto loop, otherwise goto exit
exit:
ret a


The only two definitions of a are in entry and loop, and entry
dominates everything, so DF(entry) = 0, so the only way the phi-node
gets inserted properly is if DF(loop) = {loop}, right?


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