5 Dec 2006 06:20:50 -0500

Related articles |
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[8 earlier articles] |

Re: finding all dominator trees DrDiettrich1@aol.com (Hans-Peter Diettrich) (2006-11-29) |

Re: finding all dominator trees cfc@shell01.TheWorld.com (Chris F Clark) (2006-11-30) |

Re: finding all dominator trees amichail@gmail.com (Amir Michail) (2006-12-03) |

Re: finding all dominator trees diablovision@yahoo.com (2006-12-04) |

Re: finding all dominator trees amichail@gmail.com (Amir Michail) (2006-12-05) |

Re: finding all dominator trees cfc@shell01.TheWorld.com (Chris F Clark) (2006-12-05) |

Re: finding all dominator trees martin@gkc.org.uk (Martin Ward) (2006-12-05) |

Re: finding all dominator trees mailbox@dmitry-kazakov.de (Dmitry A. Kazakov) (2006-12-06) |

Re: finding all dominator trees cfc@shell01.TheWorld.com (Chris F Clark) (2006-12-06) |

From: | Martin Ward <martin@gkc.org.uk> |

Newsgroups: | comp.compilers |

Date: | 5 Dec 2006 06:20:50 -0500 |

Organization: | Compilers Central |

References: | 06-11-096 06-12-025 06-12-033 |

Keywords: | analysis |

On Monday 04 Dec 2006 13:28, diablovision@yahoo.com wrote:

*> Because, in truth, the dominance relation doesn't make any sense unless*

*> you have either or both of the following conditions:*

*>*

*> 1. a unique entry node.*

*> 2. an acyclic graph.*

Several people, including myself, assumed that when Amir

referred to "dominators in a graph", he meant a control

flow graph (with a unique entry node from which the meaning

of dominance can be defined). He actually wants to compute

the set of dominator trees for the set of control flow graphs

which can be constructed from a given graph by taking each node

in turn to be the entry node (and, presumably, ignoring nodes

which are not reachable from that node). Each of these CFGs

has the same nodes and edges, but a different entry node,

and therefore a different dominance relation and dominator tree.

--

Martin

martin@gkc.org.uk http://www.cse.dmu.ac.uk/~mward/ Erdos number: 4

Don't email: d3457f@gkc.org.uk

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